My solution to return largest numbers in arrays.

const largestOfFour = arr => {
    let arr2 = [];
    arr.map(e => {
        let numStart = -Infinity;
        e.forEach(num => {
            if (num > numStart) {
                numStart = num;

            }
        })
        arr2.push(numStart);
    })
    return arr2;
}

Tail-call optimization is where you are able to avoid allocating a new stack frame for a function because the calling function will simply return the value that it gets from the called function. The most common use is tail-recursion, where a recursive function written to take advantage of tail-call optimization can use constant stack space.

Scheme is one of the few programming languages that guarantee in the spec that any implementation must provide this optimization, so here are two examples of the factorial function in Scheme:

(define (fact x)
  (if (= x 0) 1
      (* x (fact (- x 1)))))

(define (fact x)
  (define (fact-tail x accum)
    (if (= x 0) accum
        (fact-tail (- x 1) (* x accum))))
  (fact-tail x 1))

The first function is not tail recursive because when the recursive call is made, the function needs to keep track of the multiplication it needs to do with the result after the call returns. As such, the stack looks as follows:

(fact 3)
(* 3 (fact 2))
(* 3 (* 2 (fact 1)))
(* 3 (* 2 (* 1 (fact 0))))
(* 3 (* 2 (* 1 1)))
(* 3 (* 2 1))
(* 3 2)
6

In contrast, the stack trace for the tail recursive factorial looks as follows:

(fact 3)
(fact-tail 3 1)
(fact-tail 2 3)
(fact-tail 1 6)
(fact-tail 0 6)
6

As you can see, we only need to keep track of the same amount of data for every call to fact-tail because we are simply returning the value we get right through to the top. This means that even if I were to call (fact 1000000), I need only the same amount of space as (fact 3). This is not the case with the non-tail-recursive fact, and as such large values may cause a stack overflow.

If you can change from using decimal to double you can use the Interlocked class. Presumably this will be a good way of swapping variables performance wise. Also slightly more readable than XOR.

var startAngle = 159.9d;
var stopAngle = 355.87d;
stopAngle = Interlocked.Exchange(ref startAngle, stopAngle);

Msdn: Interlocked.Exchange Method (Double, Double)

This might help:

eln(x) = x

and

(lm)n = lm*n

Very Simple Geometric Solution with Explanation

_{Few days ago, a fell into the same problem & had to sit with the math book. I solved the problem by combining and simplifying some basic formulas.}

Lets consider this figure-

We want to know ?, so we need to find out a and ß first. Now, for any straight line-

y = m * x + c

Let- A = (ax, ay), B = (bx, by), and O = (ox, oy). So for the line OA-

oy = m1 * ox + c   ? c = oy - m1 * ox   ...(eqn-1)

ay = m1 * ax + c   ? ay = m1 * ax + oy - m1 * ox   [from eqn-1]
                   ? ay = m1 * ax + oy - m1 * ox
                   ? m1 = (ay - oy) / (ax - ox)
                   ? tan a = (ay - oy) / (ax - ox)   [m = slope = tan ?]   ...(eqn-2)

In the same way, for line OB-

tan ß = (by - oy) / (bx - ox)   ...(eqn-3)

Now, we need ? = ß - a. In trigonometry we have a formula-

tan (ß-a) = (tan ß + tan a) / (1 - tan ß * tan a)   ...(eqn-4)

After replacing the value of tan a (from eqn-2) and tan b (from eqn-3) in eqn-4, and applying simplification we get-

tan (ß-a) = ( (ax-ox)*(by-oy)+(ay-oy)*(bx-ox) ) / ( (ax-ox)*(bx-ox)-(ay-oy)*(by-oy) )

So,

? = ß-a = tan^(-1) ( ((ax-ox)*(by-oy)+(ay-oy)*(bx-ox)) / ((ax-ox)*(bx-ox)-(ay-oy)*(by-oy)) )

That is it!

Now, take following figure-

This C# or, Java method calculates the angle (?)-

    private double calculateAngle(double P1X, double P1Y, double P2X, double P2Y,
            double P3X, double P3Y){

        double numerator = P2Y*(P1X-P3X) + P1Y*(P3X-P2X) + P3Y*(P2X-P1X);
        double denominator = (P2X-P1X)*(P1X-P3X) + (P2Y-P1Y)*(P1Y-P3Y);
        double ratio = numerator/denominator;

        double angleRad = Math.Atan(ratio);
        double angleDeg = (angleRad*180)/Math.PI;

        if(angleDeg<0){
            angleDeg = 180+angleDeg;
        }

        return angleDeg;
    }

I bypassed the bit manuplation and just compared the index values. This is less than the loop iteration value (i in this case). This will not print the duplicate pairs and duplicate array elements also.

public static void findSumHashMap(int[] arr, int key) {
    Map<Integer, Integer> valMap = new HashMap<Integer, Integer>();
    for (int i = 0; i < arr.length; i++) {
        valMap.put(arr[i], i);
    }
    for (int i = 0; i < arr.length; i++) {
        if (valMap.containsKey(key - arr[i])
                && valMap.get(key - arr[i]) != i) {
            if (valMap.get(key - arr[i]) < i) {
                int diff = key - arr[i];
                System.out.println(arr[i] + " " + diff);
            }
        }
    }
}

For positive numbers

unsigned int x, y, q;

To round up ...

q = (x + y - 1) / y;

or (avoiding overflow in x+y)

q = 1 + ((x - 1) / y); // if x != 0

I just tried in place merge algorithm for merge sort in JAVA by using the insertion sort algorithm, using following steps.
1) Two sorted arrays are available.
2) Compare the first values of each array; and place the smallest value into the first array.
3) Place the larger value into the second array by using insertion sort (traverse from left to right).
4) Then again compare the second value of first array and first value of second array, and do the same. But when swapping happens there is some clue on skip comparing the further items, but just swapping required.

I have made some optimization here; to keep lesser comparisons in insertion sort.
The only drawback i found with this solutions is it needs larger swapping of array elements in the second array.

e.g)

First___Array : 3, 7, 8, 9

Second Array : 1, 2, 4, 5

Then 7, 8, 9 makes the second array to swap(move left by one) all its elements by one each time to place himself in the last.

So the assumption here is swapping items is negligible compare to comparing of two items.

https://github.com/skanagavelu/algorithams/blob/master/src/sorting/MergeSort.java

package sorting;

import java.util.Arrays;

public class MergeSort {
    public static void main(String[] args) {
    int[] array = { 5, 6, 10, 3, 9, 2, 12, 1, 8, 7 };
    mergeSort(array, 0, array.length -1);
    System.out.println(Arrays.toString(array));

    int[] array1 = {4, 7, 2};
    System.out.println(Arrays.toString(array1));
    mergeSort(array1, 0, array1.length -1);
    System.out.println(Arrays.toString(array1));
    System.out.println("\n\n");

    int[] array2 = {4, 7, 9};
    System.out.println(Arrays.toString(array2));
    mergeSort(array2, 0, array2.length -1);
    System.out.println(Arrays.toString(array2));
    System.out.println("\n\n");

    int[] array3 = {4, 7, 5};
    System.out.println(Arrays.toString(array3));
    mergeSort(array3, 0, array3.length -1);
    System.out.println(Arrays.toString(array3));
    System.out.println("\n\n");

    int[] array4 = {7, 4, 2};
    System.out.println(Arrays.toString(array4));
    mergeSort(array4, 0, array4.length -1);
    System.out.println(Arrays.toString(array4));
    System.out.println("\n\n");

    int[] array5 = {7, 4, 9};
    System.out.println(Arrays.toString(array5));
    mergeSort(array5, 0, array5.length -1);
    System.out.println(Arrays.toString(array5));
    System.out.println("\n\n");

    int[] array6 = {7, 4, 5};
    System.out.println(Arrays.toString(array6));
    mergeSort(array6, 0, array6.length -1);
    System.out.println(Arrays.toString(array6));
    System.out.println("\n\n");

    //Handling array of size two
    int[] array7 = {7, 4};
    System.out.println(Arrays.toString(array7));
    mergeSort(array7, 0, array7.length -1);
    System.out.println(Arrays.toString(array7));
    System.out.println("\n\n");

    int input1[] = {1};
    int input2[] = {4,2};
    int input3[] = {6,2,9};
    int input4[] = {6,-1,10,4,11,14,19,12,18};
    System.out.println(Arrays.toString(input1));
    mergeSort(input1, 0, input1.length-1);
    System.out.println(Arrays.toString(input1));
    System.out.println("\n\n");

    System.out.println(Arrays.toString(input2));
    mergeSort(input2, 0, input2.length-1);
    System.out.println(Arrays.toString(input2));
    System.out.println("\n\n");

    System.out.println(Arrays.toString(input3));
    mergeSort(input3, 0, input3.length-1);
    System.out.println(Arrays.toString(input3));
    System.out.println("\n\n");

    System.out.println(Arrays.toString(input4));
    mergeSort(input4, 0, input4.length-1);
    System.out.println(Arrays.toString(input4));
    System.out.println("\n\n");
}

private static void mergeSort(int[] array, int p, int r) {
    //Both below mid finding is fine.
    int mid = (r - p)/2 + p;
    int mid1 = (r + p)/2;
    if(mid != mid1) {
        System.out.println(" Mid is mismatching:" + mid + "/" + mid1+ "  for p:"+p+"  r:"+r);
    }

    if(p < r) {
        mergeSort(array, p, mid);
        mergeSort(array, mid+1, r);
//      merge(array, p, mid, r);
        inPlaceMerge(array, p, mid, r);
        }
    }

//Regular merge
private static void merge(int[] array, int p, int mid, int r) {
    int lengthOfLeftArray = mid - p + 1; // This is important to add +1.
    int lengthOfRightArray = r - mid;

    int[] left = new int[lengthOfLeftArray];
    int[] right = new int[lengthOfRightArray];

    for(int i = p, j = 0; i <= mid; ){
        left[j++] = array[i++];
    }

    for(int i = mid + 1, j = 0; i <= r; ){
        right[j++] = array[i++];
    }

    int i = 0, j = 0;
    for(; i < left.length && j < right.length; ) {
        if(left[i] < right[j]){
            array[p++] = left[i++];
        } else {
            array[p++] = right[j++];
        }
    }
    while(j < right.length){
        array[p++] = right[j++];
    } 
    while(i < left.length){
        array[p++] = left[i++];
    }
}

//InPlaceMerge no extra array
private static void inPlaceMerge(int[] array, int p, int mid, int r) {
    int secondArrayStart = mid+1;
    int prevPlaced = mid+1;
    int q = mid+1;
    while(p < mid+1 && q <= r){
        boolean swapped = false;
        if(array[p] > array[q]) {
            swap(array, p, q);
            swapped = true;
        }   
        if(q != secondArrayStart && array[p] > array[secondArrayStart]) {
            swap(array, p, secondArrayStart);
            swapped = true;
        }
        //Check swapped value is in right place of second sorted array
        if(swapped && secondArrayStart+1 <= r && array[secondArrayStart+1] < array[secondArrayStart]) {
            prevPlaced = placeInOrder(array, secondArrayStart, prevPlaced);
        }
        p++;
        if(q < r) {     //q+1 <= r) {
            q++;
        }
    }
}

private static int placeInOrder(int[] array, int secondArrayStart, int prevPlaced) {
    int i = secondArrayStart;
    for(; i < array.length; i++) {
        //Simply swap till the prevPlaced position
        if(secondArrayStart < prevPlaced) {
            swap(array, secondArrayStart, secondArrayStart+1);
            secondArrayStart++;
            continue;
        }
        if(array[i] < array[secondArrayStart]) {
            swap(array, i, secondArrayStart);
            secondArrayStart++;
        } else if(i != secondArrayStart && array[i] > array[secondArrayStart]){
            break;
        }
    }
    return secondArrayStart;
}

private static void swap(int[] array, int m, int n){
    int temp = array[m];
    array[m] = array[n];
    array[n] = temp;
}
}

Improving the answer of @user134548, without bits arithmetic:

public static bool IsPowerOfTwo(ulong n)
{
    if (n % 2 != 0) return false;  // is odd (can't be power of 2)

    double exp = Math.Log(n, 2);
    if (exp != Math.Floor(exp)) return false;  // if exp is not integer, n can't be power
    return Math.Pow(2, exp) == n;
}

This works fine for:

IsPowerOfTwo(9223372036854775809)

You should have look at http://en.wikipedia.org/wiki/Breadth-first_search first.

Below is a quick implementation, in which I used a list of list to represent the queue of paths.

# graph is in adjacent list representation
graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

def bfs(graph, start, end):
    # maintain a queue of paths
    queue = []
    # push the first path into the queue
    queue.append([start])
    while queue:
        # get the first path from the queue
        path = queue.pop(0)
        # get the last node from the path
        node = path[-1]
        # path found
        if node == end:
            return path
        # enumerate all adjacent nodes, construct a new path and push it into the queue
        for adjacent in graph.get(node, []):
            new_path = list(path)
            new_path.append(adjacent)
            queue.append(new_path)

print bfs(graph, '1', '11')

Another approach would be maintaining a mapping from each node to its parent, and when inspecting the adjacent node, record its parent. When the search is done, simply backtrace according the parent mapping.

graph = {
        '1': ['2', '3', '4'],
        '2': ['5', '6'],
        '5': ['9', '10'],
        '4': ['7', '8'],
        '7': ['11', '12']
        }

def backtrace(parent, start, end):
    path = [end]
    while path[-1] != start:
        path.append(parent[path[-1]])
    path.reverse()
    return path


def bfs(graph, start, end):
    parent = {}
    queue = []
    queue.append(start)
    while queue:
        node = queue.pop(0)
        if node == end:
            return backtrace(parent, start, end)
        for adjacent in graph.get(node, []):
            if node not in queue :
                parent[adjacent] = node # <<<<< record its parent 
                queue.append(adjacent)

print bfs(graph, '1', '11')

The above codes are based on the assumption that there's no cycles.

He, he I know I'm a question necromancer replying to old questions, but I've just found this question searching the net for ways to implement efficient prime numbers tests.

Until now, I believe that the fastest prime number testing algorithm is Strong Probable Prime (SPRP). I am quoting from Nvidia CUDA forums:

One of the more practical niche problems in number theory has to do with identification of prime numbers. Given N, how can you efficiently determine if it is prime or not? This is not just a thoeretical problem, it may be a real one needed in code, perhaps when you need to dynamically find a prime hash table size within certain ranges. If N is something on the order of 2^30, do you really want to do 30000 division tests to search for any factors? Obviously not.

The common practical solution to this problem is a simple test called an Euler probable prime test, and a more powerful generalization called a Strong Probable Prime (SPRP). This is a test that for an integer N can probabilistically classify it as prime or not, and repeated tests can increase the correctness probability. The slow part of the test itself mostly involves computing a value similar to A^(N-1) modulo N. Anyone implementing RSA public-key encryption variants has used this algorithm. It's useful both for huge integers (like 512 bits) as well as normal 32 or 64 bit ints.

The test can be changed from a probabilistic rejection into a definitive proof of primality by precomputing certain test input parameters which are known to always succeed for ranges of N. Unfortunately the discovery of these "best known tests" is effectively a search of a huge (in fact infinite) domain. In 1980, a first list of useful tests was created by Carl Pomerance (famous for being the one to factor RSA-129 with his Quadratic Seive algorithm.) Later Jaeschke improved the results significantly in 1993. In 2004, Zhang and Tang improved the theory and limits of the search domain. Greathouse and Livingstone have released the most modern results until now on the web, at http://math.crg4.com/primes.html, the best results of a huge search domain.

See here for more info: http://primes.utm.edu/prove/prove2_3.html and http://forums.nvidia.com/index.php?showtopic=70483

If you just need a way to generate very big prime numbers and don't care to generate all prime numbers < an integer n, you can use Lucas-Lehmer test to verify Mersenne prime numbers. A Mersenne prime number is in the form of 2^p -1. I think that Lucas-Lehmer test is the fastest algorithm discovered for Mersenne prime numbers.

And if you not only want to use the fastest algorithm but also the fastest hardware, try to implement it using Nvidia CUDA, write a kernel for CUDA and run it on GPU.

You can even earn some money if you discover large enough prime numbers, EFF is giving prizes from $50K to $250K: https://www.eff.org/awards/coop

Manipulating arrays in this way is error prone, as you've discovered. A better option may be to use a LinkedList in your situation. With a linked list, and all Java collections, array management is handled internally so you don't have to worry about moving elements around. With a LinkedList you just call remove and then addLast and the you're done.

The following C++ algorithm is not the best one, but it works for numbers under a billion and its pretty fast

#include <iostream>
using namespace std;

// ------ is_prime ------
// Determines if the integer accepted is prime or not
bool is_prime(int n){
    int i,count=0;
    if(n==1 || n==2)
      return true;
    if(n%2==0)
      return false;
    for(i=1;i<=n;i++){
    if(n%i==0)
        count++;
    }
    if(count==2)
      return true;
    else
      return false;
 }
 // ------ nextPrime -------
 // Finds and returns the next prime number
 int nextPrime(int prime){
     bool a = false;
     while (a == false){
         prime++;
         if (is_prime(prime))
            a = true;
     }
  return prime;
 }
 // ----- M A I N ------
 int main(){

      int value = 13195;
      int prime = 2;
      bool done = false;

      while (done == false){
          if (value%prime == 0){
             value = value/prime;
             if (is_prime(value)){
                 done = true;
             }
          } else {
             prime = nextPrime(prime);
          }
      }
        cout << "Largest prime factor: " << value << endl;
 }

Complete java converter code for all primitive types to/from arrays http://www.daniweb.com/code/snippet216874.html

Check out Smaz:

Smaz is a simple compression library suitable for compressing very short strings.

I would just use

if sorted(lst) == lst:
    # code here

unless it's a very big list in which case you might want to create a custom function.

if you are just going to sort it if it's not sorted, then forget the check and sort it.

lst.sort()

and don't think about it too much.

if you want a custom function, you can do something like

def is_sorted(lst, key=lambda x: x):
    for i, el in enumerate(lst[1:]):
        if key(el) < key(lst[i]): # i is the index of the previous element
            return False
    return True

This will be O(n) if the list is already sorted though (and O(n) in a for loop at that!) so, unless you expect it to be not sorted (and fairly random) most of the time, I would, again, just sort the list.

The key bits of dynamic programming are "overlapping sub-problems" and "optimal substructure". These properties of a problem mean that an optimal solution is composed of the optimal solutions to its sub-problems. For instance, shortest path problems exhibit optimal substructure. The shortest path from A to C is the shortest path from A to some node B followed by the shortest path from that node B to C.

In greater detail, to solve a shortest-path problem you will:

• find the distances from the starting node to every node touching it (say from A to B and C)
• find the distances from those nodes to the nodes touching them (from B to D and E, and from C to E and F)
• we now know the shortest path from A to E: it is the shortest sum of A-x and x-E for some node x that we have visited (either B or C)
• repeat this process until we reach the final destination node

Because we are working bottom-up, we already have solutions to the sub-problems when it comes time to use them, by memoizing them.

Remember, dynamic programming problems must have both overlapping sub-problems, and optimal substructure. Generating the Fibonacci sequence is not a dynamic programming problem; it utilizes memoization because it has overlapping sub-problems, but it does not have optimal substructure (because there is no optimization problem involved).

There's a nice article which may answer your question /only it prints the paths instead of collecting them/. Please note that you can experiment with the C++/Python samples in the online IDE.

http://www.geeksforgeeks.org/find-paths-given-source-destination/

Here is one using arrays and Java 8+

import java.util.Arrays;
import java.util.stream.IntStream;

public class HelloWorld {

    public static void main(String[] args) {
        int[] arr = {1, 2, 3, 5};
        permutation(arr, new int[]{});
    }

    static void permutation(int[] arr, int[] prefix) {
        if (arr.length == 0) {
            System.out.println(Arrays.toString(prefix));
        }
        for (int i = 0; i < arr.length; i++) {
            int i2 = i;
            int[] pre = IntStream.concat(Arrays.stream(prefix), IntStream.of(arr[i])).toArray();
            int[] post = IntStream.range(0, arr.length).filter(i1 -> i1 != i2).map(v -> arr[v]).toArray();
            permutation(post, pre);
        }
    }
}

"2) Can we use Dijksra’s algorithm for shortest paths for graphs with negative weights – one idea can be, calculate the minimum weight value, add a positive value (equal to absolute value of minimum weight value) to all weights and run the Dijksra’s algorithm for the modified graph. Will this algorithm work?"

This absolutely doesn't work unless all shortest paths have same length. For example given a shortest path of length two edges, and after adding absolute value to each edge, then the total path cost is increased by 2 * |max negative weight|. On the other hand another path of length three edges, so the path cost is increased by 3 * |max negative weight|. Hence, all distinct paths are increased by different amounts.

JS version that returns one root or an array of roots each of which will have a Children array property containing the related children. Does not depend on ordered input, decently compact, and does not use recursion. Enjoy!

// creates a tree from a flat set of hierarchically related data
var MiracleGrow = function(treeData, key, parentKey)
{
    var keys = [];
    treeData.map(function(x){
        x.Children = [];
        keys.push(x[key]);
    });
    var roots = treeData.filter(function(x){return keys.indexOf(x[parentKey])==-1});
    var nodes = [];
    roots.map(function(x){nodes.push(x)});
    while(nodes.length > 0)
    {

        var node = nodes.pop();
        var children =  treeData.filter(function(x){return x[parentKey] == node[key]});
        children.map(function(x){
            node.Children.push(x);
            nodes.push(x)
        });
    }
    if (roots.length==1) return roots[0];
    return roots;
}


// demo/test data
var treeData = [

    {id:9, name:'Led Zep', parent:null},
    {id:10, name:'Jimmy', parent:9},
    {id:11, name:'Robert', parent:9},
    {id:12, name:'John', parent:9},

    {id:8, name:'Elec Gtr Strings', parent:5},
    {id:1, name:'Rush', parent:null},
    {id:2, name:'Alex', parent:1},
    {id:3, name:'Geddy', parent:1},
    {id:4, name:'Neil', parent:1},
    {id:5, name:'Gibson Les Paul', parent:2},
    {id:6, name:'Pearl Kit', parent:4},
    {id:7, name:'Rickenbacker', parent:3},

    {id:100, name:'Santa', parent:99},
    {id:101, name:'Elf', parent:100},

];
var root = MiracleGrow(treeData, "id", "parent")
console.log(root)

I hope I can contribute something new to this problem. I noticed that all of the answers neglect the fact that there are two points where you can perform preprocessing, without slowing down your overall laundry performance.

Also, we don't need to assume a large number of socks, even for large families. Socks are taken out of the drawer and are worn, and then they are tossed in a place (maybe a bin) where they stay before being laundered. While I wouldn't call said bin a LIFO-Stack, I'd say it is safe to assume that

1. people toss both of their socks roughly in the same area of the bin,
2. the bin is not randomized at any point, and therefore
3. any subset taken from the top of this bin generally contains both socks of a pair.

Since all washing machines I know about are limited in size (regardless of how many socks you have to wash), and the actual randomizing occurs in the washing machine, no matter how many socks we have, we always have small subsets which contain almost no singletons.

Our two preprocessing stages are "putting the socks on the clothesline" and "Taking the socks from the clothesline", which we have to do, in order to get socks which are not only clean but also dry. As with washing machines, clotheslines are finite, and I assume that we have the whole part of the line where we put our socks in sight.

Here's the algorithm for put_socks_on_line():

while (socks left in basket) {
 take_sock();
 if (cluster of similar socks is present) { 
   Add sock to cluster (if possible, next to the matching pair)
 } else {
  Hang it somewhere on the line, this is now a new cluster of similar-looking socks.      
  Leave enough space around this sock to add other socks later on 
 }
}

Don't waste your time moving socks around or looking for the best match, this all should be done in O(n), which we would also need for just putting them on the line unsorted. The socks aren't paired yet, we only have several similarity clusters on the line. It's helpful that we have a limited set of socks here, as this helps us to create "good" clusters (for example, if there are only black socks in the set of socks, clustering by colours would not be the way to go)

Here's the algorithm for take_socks_from_line():

while(socks left on line) {
 take_next_sock();
 if (matching pair visible on line or in basket) {
   Take it as well, pair 'em and put 'em away
 } else {
   put the sock in the basket
 }

I should point out that in order to improve the speed of the remaining steps, it is wise not to randomly pick the next sock, but to sequentially take sock after sock from each cluster. Both preprocessing steps don't take more time than just putting the socks on the line or in the basket, which we have to do no matter what, so this should greatly enhance the laundry performance.

After this, it's easy to do the hash partitioning algorithm. Usually, about 75% of the socks are already paired, leaving me with a very small subset of socks, and this subset is already (somewhat) clustered (I don't introduce much entropy into my basket after the preprocessing steps). Another thing is that the remaining clusters tend to be small enough to be handled at once, so it is possible to take a whole cluster out of the basket.

Here's the algorithm for sort_remaining_clusters():

while(clusters present in basket) {
  Take out the cluster and spread it
  Process it immediately
  Leave remaining socks where they are
}

After that, there are only a few socks left. This is where I introduce previously unpaired socks into the system and process the remaining socks without any special algorithm - the remaining socks are very few and can be processed visually very fast.

For all remaining socks, I assume that their counterparts are still unwashed and put them away for the next iteration. If you register a growth of unpaired socks over time (a "sock leak"), you should check your bin - it might get randomized (do you have cats which sleep in there?)

I know that these algorithms take a lot of assumptions: a bin which acts as some sort of LIFO stack, a limited, normal washing machine, and a limited, normal clothesline - but this still works with very large numbers of socks.

About parallelism: As long as you toss both socks into the same bin, you can easily parallelize all of those steps.

A simple way using std::next_permutation:

#include <iostream>
#include <algorithm>
#include <vector>

int main() {
    int n, r;
    std::cin >> n;
    std::cin >> r;

    std::vector<bool> v(n);
    std::fill(v.end() - r, v.end(), true);

    do {
        for (int i = 0; i < n; ++i) {
            if (v[i]) {
                std::cout << (i + 1) << " ";
            }
        }
        std::cout << "\n";
    } while (std::next_permutation(v.begin(), v.end()));
    return 0;
}

or a slight variation that outputs the results in an easier to follow order:

#include <iostream>
#include <algorithm>
#include <vector>

int main() {
   int n, r;
   std::cin >> n;
   std::cin >> r;

   std::vector<bool> v(n);
   std::fill(v.begin(), v.begin() + r, true);

   do {
       for (int i = 0; i < n; ++i) {
           if (v[i]) {
               std::cout << (i + 1) << " ";
           }
       }
       std::cout << "\n";
   } while (std::prev_permutation(v.begin(), v.end()));
   return 0;
}

A bit of explanation:

It works by creating a "selection array" (v), where we place r selectors, then we create all permutations of these selectors, and print the corresponding set member if it is selected in in the current permutation of v.

You can implement it if you note that for each level r you select a number from 1 to n.

In C++, we need to 'manually' keep the state between calls that produces results (a combination): so, we build a class that on construction initialize the state, and has a member that on each call returns the combination while there are solutions: for instance

#include <iostream>
#include <iterator>
#include <vector>
#include <cstdlib>

using namespace std;

struct combinations
{
    typedef vector<int> combination_t;

    // initialize status
   combinations(int N, int R) :
       completed(N < 1 || R > N),
       generated(0),
       N(N), R(R)
   {
       for (int c = 1; c <= R; ++c)
           curr.push_back(c);
   }

   // true while there are more solutions
   bool completed;

   // count how many generated
   int generated;

   // get current and compute next combination
   combination_t next()
   {
       combination_t ret = curr;

       // find what to increment
       completed = true;
       for (int i = R - 1; i >= 0; --i)
           if (curr[i] < N - R + i + 1)
           {
               int j = curr[i] + 1;
               while (i <= R-1)
                   curr[i++] = j++;
               completed = false;
               ++generated;
               break;
           }

       return ret;
   }

private:

   int N, R;
   combination_t curr;
};

int main(int argc, char **argv)
{
    int N = argc >= 2 ? atoi(argv[1]) : 5;
    int R = argc >= 3 ? atoi(argv[2]) : 2;
    combinations cs(N, R);
    while (!cs.completed)
    {
        combinations::combination_t c = cs.next();
        copy(c.begin(), c.end(), ostream_iterator<int>(cout, ","));
        cout << endl;
    }
    return cs.generated;
}

test output:

1,2,
1,3,
1,4,
1,5,
2,3,
2,4,
2,5,
3,4,
3,5,
4,5,

bool pointInRectangle(Point A, Point B, Point C, Point D, Point m ) {
    Point AB = vect2d(A, B);  float C1 = -1 * (AB.y*A.x + AB.x*A.y); float  D1 = (AB.y*m.x + AB.x*m.y) + C1;
    Point AD = vect2d(A, D);  float C2 = -1 * (AD.y*A.x + AD.x*A.y); float D2 = (AD.y*m.x + AD.x*m.y) + C2;
    Point BC = vect2d(B, C);  float C3 = -1 * (BC.y*B.x + BC.x*B.y); float D3 = (BC.y*m.x + BC.x*m.y) + C3;
    Point CD = vect2d(C, D);  float C4 = -1 * (CD.y*C.x + CD.x*C.y); float D4 = (CD.y*m.x + CD.x*m.y) + C4;
    return     0 >= D1 && 0 >= D4 && 0 <= D2 && 0 >= D3;}





Point vect2d(Point p1, Point p2) {
    Point temp;
    temp.x = (p2.x - p1.x);
    temp.y = -1 * (p2.y - p1.y);
    return temp;}

I just implemented AnT's Answer using c++. I used this code to check whether the pixel's coordination(X,Y) lies inside the shape or not.

ES6 style

function gcd(...numbers) {
  return numbers.reduce((a, b) => b === 0 ? a : gcd(b, a % b));
}

function lcm(...numbers) {
  return numbers.reduce((a, b) => Math.abs(a * b) / gcd(a, b));
}

As far as I know, Perl does not optimize tail-recursive calls, but you can fake it.

sub f{
  my($l,$r) = @_;

  if( $l >= $r ){
    return $l;
  } else {

    # return f( $l+1, $r );

    @_ = ( $l+1, $r );
    goto &f;

  }
}

When first called it will allocate space on the stack. Then it will change its arguments, and restart the subroutine, without adding anything more to the stack. It will therefore pretend that it never called its self, changing it into an iterative process.

Note that there is no "my @_;" or "local @_;", if you did it would no longer work.

I don't know if homework counts...

During my studies we rolled our own program to solve the Traveling Salesman problem.

The idea was to make a comparison on several criteria (difficulty to map the problem, performance, etc) and we also used other techniques such as Simulated annealing.

It worked pretty well, but it took us a while to understand how to do the 'reproduction' phase correctly: modeling the problem at hand into something suitable for Genetic programming really struck me as the hardest part...

It was an interesting course since we also dabbled with neural networks and the like.

I'd like to know if anyone used this kind of programming in 'production' code.

One important application of Kruskal's algorithm is in single link clustering.

Consider n vertices and you have a complete graph.To obtain a k clusters of those n points.Run Kruskal's algorithm over the first n-(k-1) edges of the sorted set of edges.You obtain k-cluster of the graph with maximum spacing.

Finding the missing number from a series of numbers. IMP points to remember.

1. the array should be sorted..
2. the Function do not work on multiple missings.

3. the sequence must be an AP.

       public int execute2(int[] array) {
       int diff = Math.min(array[1]-array[0], array[2]-array[1]);
       int min = 0, max = arr.length-1;
       boolean missingNum = true;
       while(min<max) {
           int mid = (min + max) >>> 1;
           int leftDiff = array[mid] - array[min];
           if(leftDiff > diff * (mid - min)) {
               if(mid-min == 1)
                   return (array[mid] + array[min])/2;
               max = mid;
               missingNum = false;
               continue;
           }
           int rightDiff = array[max] - array[mid];
           if(rightDiff > diff * (max - mid)) {
               if(max-mid == 1)
                   return (array[max] + array[mid])/2;
               min = mid;
               missingNum = false;
               continue;
           }
           if(missingNum)
               break;
       }
       return -1;
   }
   

Simple solution using the itertools and string modules

# modules to easily set characters and iterate over them
import itertools, string 

# character limit so you don't run out of ram
maxChar = int(input('Character limit for password: '))  

# file to save output to, so you can look over the output without using so much ram
output_file = open('insert filepath here', 'a+') 

# this is the part that actually iterates over the valid characters, and stops at the 
# character limit.
x = list(map(''.join, itertools.permutations(string.ascii_lowercase, maxChar))) 

# writes the output of the above line to a file 
output_file.write(str(x)) 

# saves the output to the file and closes it to preserve ram
output_file.close() 

I piped the output to a file to save ram, and used the input function so you can set the character limit to something like "hiiworld". Below is the same script but with a more fluid character set using letters, numbers, symbols, and spaces.

import itertools, string

maxChar = int(input('Character limit for password: '))
output_file = open('insert filepath here', 'a+')

x = list(map(''.join, itertools.permutations(string.printable, maxChar)))
x.write(str(x))
x.close()

Here is c++ solution

/**
*array sorting needed to verify if first arrays n'th element is greater than sencond arrays
*some element then all elements following n will do the same
*/
#include<stdio.h>
#include<iostream>
using namespace std;
int countInversions(int array[],int size);
int merge(int arr1[],int size1,int arr2[],int size2,int[]);
int main()
{
    int array[] = {2, 4, 1, 3, 5};
    int size = sizeof(array) / sizeof(array[0]);
    int x = countInversions(array,size);
    printf("number of inversions = %d",x);
}

int countInversions(int array[],int size)
{
    if(size > 1 )
    {
    int mid = size / 2;
    int count1 = countInversions(array,mid);
    int count2 = countInversions(array+mid,size-mid);
    int temp[size];
    int count3 = merge(array,mid,array+mid,size-mid,temp);
    for(int x =0;x<size ;x++)
    {
        array[x] = temp[x];
    }
    return count1 + count2 + count3;
    }else{
        return 0;
    }
}

int merge(int arr1[],int size1,int arr2[],int size2,int temp[])
{
    int count  = 0;
    int a = 0;
    int b = 0;
    int c = 0;
    while(a < size1 && b < size2)
    {
        if(arr1[a] < arr2[b])
        {
            temp[c] = arr1[a];
            c++;
            a++;
        }else{
            temp[c] = arr2[b];
            b++;
            c++;
            count = count + size1 -a;
        }
    }

    while(a < size1)
    {
        temp[c] = arr1[a];
        c++;a++;
    }

while(b < size2)
    {
        temp[c] = arr2[b];
        c++;b++;
    }

    return count;
}

The upside-down A symbol is the universal quantifier from predicate logic. (Also see the more complete discussion of the first-order predicate calculus.) As others noted, it means that the stated assertions holds "for all instances" of the given variable (here, s). You'll soon run into its sibling, the backwards capital E, which is the existential quantifier, meaning "there exists at least one" of the given variable conforming to the related assertion.

If you're interested in logic, you might enjoy the book Logic and Databases: The Roots of Relational Theory by C.J. Date. There are several chapters covering these quantifiers and their logical implications. You don't have to be working with databases to benefit from this book's coverage of logic.

In matrix multiplication there are 3 for loop, we are using since execution of each for loop requires time complexity O(n). So for three loops it becomes O(n^3)

I was able to do this with a single loop. The time complexity seems like O(K) where K is all items of the array. Here's how I did it in JavaScript:

First off, we represent the n^2 matrix with a single array. Then, iterate through it like this:

/**
 * Rotates matrix 90 degrees clockwise
 * @param arr: the source array
 * @param n: the array side (array is square n^2)
 */
function rotate (arr, n) {
  var rotated = [], indexes = []

  for (var i = 0; i < arr.length; i++) {
    if (i < n)
      indexes[i] = i * n + (n - 1)
    else
      indexes[i] = indexes[i - n] - 1

    rotated[indexes[i]] = arr[i]
  }
  return rotated
}

Basically, we transform the source array indexes:

[0,1,2,3,4,5,6,7,8] => [2,5,8,1,4,7,0,3,6]

Then, using this transformed indexes array, we place the actual values in the final rotated array.

Here are some test cases:

//n=3
rotate([
  1, 2, 3,
  4, 5, 6,
  7, 8, 9], 3))

//result:
[7, 4, 1,
 8, 5, 2,
 9, 6, 3]


//n=4
rotate([
  1,  2,  3,  4,
  5,  6,  7,  8,
  9,  10, 11, 12,
  13, 14, 15, 16], 4))

//result:
[13,  9,  5,  1,
 14, 10,  6,  2,
 15, 11,  7,  3,
 16, 12,  8,  4]


//n=5
rotate([
  1,  2,  3,  4,  5,
  6,  7,  8,  9,  10,
  11, 12, 13, 14, 15,
  16, 17, 18, 19, 20,
  21, 22, 23, 24, 25], 5))

//result:
[21, 16, 11,  6,  1, 
 22, 17, 12,  7,  2, 
 23, 18, 13,  8,  3, 
 24, 19, 14,  9,  4, 
 25, 20, 15, 10,  5]

In general the square root of an integer (like 2, for example) can only be approximated (not because of problems with floating point arithmetic, but because they're irrational numbers which can't be calculated exactly).

Of course, some approximations are better than others. I mean, of course, that the value 1.732 is a better approximation to the square root of 3, than 1.7

The method used by the code at that link you gave works by taking a first approximation and using it to calculate a better approximation.

This is called Newton's Method, and you can repeat the calculation with each new approximation until it's accurate enough for you.

In fact there must be some way to decide when to stop the repetition or it will run forever.

Usually you would stop when the difference between approximations is less than a value you decide.

EDIT: I don't think there can be a simpler implementation than the two you already found.

Another use of BST over Heap; because of an important difference :

• finding successor and predecessor in a BST will take O(h) time. ( O(logn) in balanced BST)
• while in Heap, would take O(n) time to find successor or predecessor of some element.

Use of BST over a Heap: Now, Lets say we use a data structure to store landing time of flights. We cannot schedule a flight to land if difference in landing times is less than 'd'. And assume many flights have been scheduled to land in a data structure(BST or Heap).

Now, we want to schedule another Flight which will land at t. Hence, we need to calculate difference of t with its successor and predecessor (should be >d). Thus, we will need a BST for this, which does it fast i.e. in O(logn) if balanced.

EDITed:

Sorting BST takes O(n) time to print elements in sorted order (Inorder traversal), while Heap can do it in O(n logn) time. Heap extracts min element and re-heapifies the array, which makes it do the sorting in O(n logn) time.

First of all, the space complexity of this loop is O(1) (the input is customarily not included when calculating how much storage is required by an algorithm).

So the question that I have is if its possible that an algorithm has different time complexity from space complexity?

Yes, it is. In general, the time and the space complexity of an algorithm are not related to each other.

Sometimes one can be increased at the expense of the other. This is called space-time tradeoff.

NP-Complete is a class of problems.

The class P consists of those problems that are solvable in polynomial time. For example, they could be solved in O(n^k) for some constant k, where n is the size of the input. Simply put, you can write a program that will run in reasonable time.

The class NP consists of those problems that are verifiable in polynomial time. That is, if we are given a potential solution, then we could check if the given solution is correct in polynomial time.

Some examples are the Boolean Satisfiability (or SAT) problem, or the Hamiltonian-cycle problem. There are many problems that are known to be in the class NP.

NP-Complete means the problem is at least as hard as any problem in NP.

It is important to computer science because it has been proven that any problem in NP can be transformed into another problem in NP-complete. That means that a solution to any one NP-complete problem is a solution to all NP problems.

Many algorithms in security depends on the fact that no known solutions exist for NP hard problems. It would definitely have a significant impact on computing if a solution were found.

Someone above had the following.

bool check_prime(int num) {
for (int i = num - 1; i > 1; i--) {
    if ((num % i) == 0)
        return false;
}
return true;
}

This mostly worked. I just tested it in Visual Studio 2017. It would say that anything less than 2 was also prime (so 1, 0, -1, etc.)

Here is a slight modification to correct this.

bool check_prime(int number)
{
    if (number > 1)
    {
        for (int i = number - 1; i > 1; i--)
        {
            if ((number % i) == 0)
                return false;
        }
        return true;
    }
    return false;
}

If owner names differ, sort by them. Otherwise, use publication name for tiebreaker.

function mysortfunction(a, b) {

  var o1 = a[3].toLowerCase();
  var o2 = b[3].toLowerCase();

  var p1 = a[1].toLowerCase();
  var p2 = b[1].toLowerCase();

  if (o1 < o2) return -1;
  if (o1 > o2) return 1;
  if (p1 < p2) return -1;
  if (p1 > p2) return 1;
  return 0;
}

Given your constraints, you're stuck with brute force search (or creating an index if the search will be repeated). Can you elaborate any on how the ArrayList is generated--perhaps there is some wiggle room there.

If all you're looking for is prettier code, consider using the Apache Commons Collections classes, in particular CollectionUtils.find(), for ready-made syntactic sugar:

ArrayList haystack = // ...
final Object needleField1 = // ...
final Object needleField2 = // ...

Object found = CollectionUtils.find(haystack, new Predicate() {
   public boolean evaluate(Object input) {
      return needleField1.equals(input.field1) && 
             needleField2.equals(input.field2);
   }
});

You can go for recursive approach:

Here is the pseudo code:

Node* reverse(Node* root)
{
    if(!root) return NULL;

    if(!(root->next)) temp = root;
    else
    {
        reverse(root->next);
        root->next->next = root;
        root->next = NULL;
    }

    return temp;
}

After the call is made to the function, it returns the new root[temp] of the linked list. As it is very clear that it makes use of only two pointers.

Three ways:

1.

int main () 
{
    for (int i=2; i<100; i++) 
        for (int j=2; j*j<=i; j++)
        {
            if (i % j == 0) 
                break;
            else if (j+1 > sqrt(i)) {
                cout << i << " ";

            }

        }   

    return 0;
}

2.

int main () 
{
    for (int i=2; i<100; i++) 
    {
        bool prime=true;
        for (int j=2; j*j<=i; j++)
        {
            if (i % j == 0) 
            {
                prime=false;
                break;    
            }
        }   
        if(prime) cout << i << " ";
    }
    return 0;
}

3.

#include <vector>
int main()
{
    std::vector<int> primes;
    primes.push_back(2);
    for(int i=3; i < 100; i++)
    {
        bool prime=true;
        for(int j=0;j<primes.size() && primes[j]*primes[j] <= i;j++)
        {
            if(i % primes[j] == 0)
            {
                prime=false;
                break;
            }
        }
        if(prime) 
        {
            primes.push_back(i);
            cout << i << " ";
        }
    }

    return 0;
}

Edit: In the third example, we keep track of all of our previously calculated primes. If a number is divisible by a non-prime number, there is also some prime <= that divisor which it is also divisble by. This reduces computation by a factor of primes_in_range/total_range.

DFS(analysis):

• Setting/getting a vertex/edge label takes O(1) time
• Each vertex is labeled twice
  • once as UNEXPLORED
  • once as VISITED
• Each edge is labeled twice
  • once as UNEXPLORED
  • once as DISCOVERY or BACK
• Method incidentEdges is called once for each vertex
• DFS runs in O(n + m) time provided the graph is represented by the adjacency list structure
• Recall that Sv deg(v) = 2m

BFS(analysis):

• Setting/getting a vertex/edge label takes O(1) time
• Each vertex is labeled twice
  • once as UNEXPLORED
  • once as VISITED
• Each edge is labeled twice
  • once as UNEXPLORED
  • once as DISCOVERY or CROSS
• Each vertex is inserted once into a sequence Li
• Method incidentEdges is called once for each vertex
• BFS runs in O(n + m) time provided the graph is represented by the adjacency list structure
• Recall that Sv deg(v) = 2m

Quicksort is usually the fastest on average, but It has some pretty nasty worst-case behaviors. So if you have to guarantee no bad data gives you O(N^2), you should avoid it.

Merge-sort uses extra memory, but is particularly suitable for external sorting (i.e. huge files that don't fit into memory).

Heap-sort can sort in-place and doesn't have the worst case quadratic behavior, but on average is slower than quicksort in most cases.

Where only integers in a restricted range are involved, you can use some kind of radix sort to make it very fast.

In 99% of the cases, you'll be fine with the library sorts, which are usually based on quicksort.

public class MinStack<E>{

    private final LinkedList<E> mainStack = new LinkedList<E>();
    private final LinkedList<E> minStack = new LinkedList<E>();
    private final Comparator<E> comparator;

    public MinStack(Comparator<E> comparator) 
    {
        this.comparator = comparator;
    }

    /**
     * Pushes an element onto the stack.
     *
     *
     * @param e the element to push
     */
    public void push(E e) {
        mainStack.push(e);
        if(minStack.isEmpty())
        {
            minStack.push(e);
        }
        else if(comparator.compare(e, minStack.peek())<=0)
        {
            minStack.push(e);
        }
        else
        {
            minStack.push(minStack.peek());
        }
    }

    /**
     * Pops an element from the stack.
     *
     *
     * @throws NoSuchElementException if this stack is empty
     */
    public E pop() {
       minStack.pop();
       return  mainStack.pop();
    }

    /**
     * Returns but not remove smallest element from the stack. Return null if stack is empty.
     *
     */
    public E getMinimum()
    {
        return minStack.peek();
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append("Main stack{");
        for (E e : mainStack) {         
            sb.append(e.toString()).append(",");
        }
        sb.append("}");

        sb.append(" Min stack{");
        for (E e : minStack) {          
            sb.append(e.toString()).append(",");
        }
        sb.append("}");

        sb.append(" Minimum = ").append(getMinimum());
        return sb.toString();
    }

    public static void main(String[] args) {
        MinStack<Integer> st = new MinStack<Integer>(Comparators.INTEGERS);

        st.push(2);
        Assert.assertTrue("2 is min in stack {2}", st.getMinimum().equals(2));
        System.out.println(st);

        st.push(6);
        Assert.assertTrue("2 is min in stack {2,6}", st.getMinimum().equals(2));
        System.out.println(st);

        st.push(4);
        Assert.assertTrue("2 is min in stack {2,6,4}", st.getMinimum().equals(2));
        System.out.println(st);

        st.push(1);
        Assert.assertTrue("1 is min in stack {2,6,4,1}", st.getMinimum().equals(1));
        System.out.println(st);

        st.push(5);
        Assert.assertTrue("1 is min in stack {2,6,4,1,5}", st.getMinimum().equals(1));
        System.out.println(st);

        st.pop();
        Assert.assertTrue("1 is min in stack {2,6,4,1}", st.getMinimum().equals(1));
        System.out.println(st);

        st.pop();
        Assert.assertTrue("2 is min in stack {2,6,4}", st.getMinimum().equals(2));
        System.out.println(st);

        st.pop();
        Assert.assertTrue("2 is min in stack {2,6}", st.getMinimum().equals(2));
        System.out.println(st);

        st.pop();
        Assert.assertTrue("2 is min in stack {2}", st.getMinimum().equals(2));
        System.out.println(st);

        st.pop();
        Assert.assertTrue("null is min in stack {}", st.getMinimum()==null);
        System.out.println(st);
    }
}

Following solution also provide array of subset which provide specific sum (here sum = 9)

array = [1, 3, 4, 2, 7, 8, 9]

(0..array.size).map { |i| array.combination(i).to_a.select { |a| a.sum == 9 } }.flatten(1)

return array of subsets which return sum of 9

 => [[9], [1, 8], [2, 7], [3, 4, 2]] 

If you have php support, you can do something like this:

<img src="scriptname.php">

And the scriptname.php:

<?php
header("Content-type: image/png");
// Your email address which will be shown in the image
$email    =    "[email protected]";
$length    =    (strlen($email)*8);
$im = @ImageCreate ($length, 20)
     or die ("Kann keinen neuen GD-Bild-Stream erzeugen");
$background_color = ImageColorAllocate ($im, 255, 255, 255); // White: 255,255,255
$text_color = ImageColorAllocate ($im, 55, 103, 122);
imagestring($im, 3,5,2,$email, $text_color);
imagepng ($im);
?>

// Two stacks s1 Original and s2 as Temp one
    private Stack<Integer> s1 = new Stack<Integer>();
    private Stack<Integer> s2 = new Stack<Integer>();

    /*
     * Here we insert the data into the stack and if data all ready exist on
     * stack than we copy the entire stack s1 to s2 recursively and push the new
     * element data onto s1 and than again recursively call the s2 to pop on s1.
     * 
     * Note here we can use either way ie We can keep pushing on s1 and than
     * while popping we can remove the first element from s2 by copying
     * recursively the data and removing the first index element.
     */
    public void insert( int data )
    {
        if( s1.size() == 0 )
        {
            s1.push( data );
        }
        else
        {
            while( !s1.isEmpty() )
            {
                s2.push( s1.pop() );
            }
            s1.push( data );
            while( !s2.isEmpty() )
            {
                s1.push( s2.pop() );
            }
        }
    }

    public void remove()
    {
        if( s1.isEmpty() )
        {
            System.out.println( "Empty" );
        }
        else
        {
            s1.pop();

        }
    }

Just another answer, although it's similar to Supr's one - just a different color space.

The thing is: Humans perceive the difference in color not uniformly and the RGB color space is ignoring this. As a result if you use the RGB color space and just compute the euclidean distance between 2 colors you may get a difference which is mathematically absolutely correct, but wouldn't coincide with what humans would tell you.

This may not be a problem - the difference is not that large I think, but if you want to solve this "better" you should convert your RGB colors into a color space that was specifically designed to avoid the above problem. There are several ones, improvements from earlier models (since this is based on human perception we need to measure the "correct" values based on experimental data). There's the Lab colorspace which I think would be the best although a bit complicated to convert it to. Simpler would be the CIE XYZ one.

Here's a site that lists the formula's to convert between different color spaces so you can experiment a bit.

Here is my code:

int multiply(int a[],int n,int nextproduct,int i)
{
    int prevproduct=1;
    if(i>=n)
        return prevproduct;
    prevproduct=multiply(a,n,nextproduct*a[i],i+1);
    printf(" i=%d > %d\n",i,prevproduct*nextproduct);
    return prevproduct*a[i];
}

int main()
{
    int a[]={2,4,1,3,5};
    multiply(a,5,1,0);
    return 0;
}

I think converting to set is the easiest way to remove duplicate:

list1 = [1,2,1]
list1 = list(set(list1))
print list1

If you want to find both the min and max at the same time, the loop can be modified as follows:

int min = int.maxValue;
int max = int.minValue;

foreach num in someArray {
  if(num < min)
    min = num;
  if(num > max)
    max = num;
}

This should get achieve O(n) timing.

If you want the driving distance/route (posting it here because this is the first result for the distance between two points on google but for most people the driving distance is more useful), you can use Google Maps Distance Matrix Service:

getDrivingDistanceBetweenTwoLatLong(origin, destination) {

 return new Observable(subscriber => {
  let service = new google.maps.DistanceMatrixService();
  service.getDistanceMatrix(
    {
      origins: [new google.maps.LatLng(origin.lat, origin.long)],
      destinations: [new google.maps.LatLng(destination.lat, destination.long)],
      travelMode: 'DRIVING'
    }, (response, status) => {
      if (status !== google.maps.DistanceMatrixStatus.OK) {
        console.log('Error:', status);
        subscriber.error({error: status, status: status});
      } else {
        console.log(response);
        try {
          let valueInMeters = response.rows[0].elements[0].distance.value;
          let valueInKms = valueInMeters / 1000;
          subscriber.next(valueInKms);
          subscriber.complete();
        }
       catch(error) {
        subscriber.error({error: error, status: status});
       }
      }
    });
});
}

float roundUp(float number, float fixedBase) {
    if (fixedBase != 0 && number != 0) {
        float sign = number > 0 ? 1 : -1;
        number *= sign;
        number /= fixedBase;
        int fixedPoint = (int) ceil(number);
        number = fixedPoint * fixedBase;
        number *= sign;
    }
    return number;
}

This works for any float number or base (e.g. you can round -4 to the nearest 6.75). In essence it is converting to fixed point, rounding there, then converting back. It handles negatives by rounding AWAY from 0. It also handles a negative round to value by essentially turning the function into roundDown.

An int specific version looks like:

int roundUp(int number, int fixedBase) {
    if (fixedBase != 0 && number != 0) {
        int sign = number > 0 ? 1 : -1;
        int baseSign = fixedBase > 0 ? 1 : 0;
        number *= sign;
        int fixedPoint = (number + baseSign * (fixedBase - 1)) / fixedBase;
        number = fixedPoint * fixedBase;
        number *= sign;
    }
    return number;
}

Which is more or less plinth's answer, with the added negative input support.

You could try computing sin(pi/2) (or cos(pi/2) for that matter) using the (fairly) quickly converging power series for sin and cos. (Even better: use various doubling formulas to compute nearer x=0 for faster convergence.)

BTW, better than using series for tan(x) is, with computing say cos(x) as a black box (e.g. you could use taylor series as above) is to do root finding via Newton. There certainly are better algorithms out there, but if you don't want to verify tons of digits this should suffice (and it's not that tricky to implement, and you only need a bit of calculus to understand why it works.)

I realize that this is an old question, with an established answer. The reason I'm posting is that is the accepted answer has many elements of k-NN (k-nearest neighbors), a different algorithm.

Both k-NN and NaiveBayes are classification algorithms. Conceptually, k-NN uses the idea of "nearness" to classify new entities. In k-NN 'nearness' is modeled with ideas such as Euclidean Distance or Cosine Distance. By contrast, in NaiveBayes, the concept of 'probability' is used to classify new entities.

Since the question is about Naive Bayes, here's how I'd describe the ideas and steps to someone. I'll try to do it with as few equations and in plain English as much as possible.

First, Conditional Probability & Bayes' Rule

Before someone can understand and appreciate the nuances of Naive Bayes', they need to know a couple of related concepts first, namely, the idea of Conditional Probability, and Bayes' Rule. (If you are familiar with these concepts, skip to the section titled Getting to Naive Bayes')

Conditional Probability in plain English: What is the probability that something will happen, given that something else has already happened.

Let's say that there is some Outcome O. And some Evidence E. From the way these probabilities are defined: The Probability of having both the Outcome O and Evidence E is: (Probability of O occurring) multiplied by the (Prob of E given that O happened)

One Example to understand Conditional Probability:

Let say we have a collection of US Senators. Senators could be Democrats or Republicans. They are also either male or female.

If we select one senator completely randomly, what is the probability that this person is a female Democrat? Conditional Probability can help us answer that.

Probability of (Democrat and Female Senator)= Prob(Senator is Democrat) multiplied by Conditional Probability of Being Female given that they are a Democrat.

  P(Democrat & Female) = P(Democrat) * P(Female | Democrat) 

We could compute the exact same thing, the reverse way:

  P(Democrat & Female) = P(Female) * P(Democrat | Female) 

Understanding Bayes Rule

Conceptually, this is a way to go from P(Evidence| Known Outcome) to P(Outcome|Known Evidence). Often, we know how frequently some particular evidence is observed, given a known outcome. We have to use this known fact to compute the reverse, to compute the chance of that outcome happening, given the evidence.

P(Outcome given that we know some Evidence) = P(Evidence given that we know the Outcome) times Prob(Outcome), scaled by the P(Evidence)

The classic example to understand Bayes' Rule:

Probability of Disease D given Test-positive = 

               P(Test is positive|Disease) * P(Disease)
     _______________________________________________________________
     (scaled by) P(Testing Positive, with or without the disease)

Now, all this was just preamble, to get to Naive Bayes.

Getting to Naive Bayes'

So far, we have talked only about one piece of evidence. In reality, we have to predict an outcome given multiple evidence. In that case, the math gets very complicated. To get around that complication, one approach is to 'uncouple' multiple pieces of evidence, and to treat each of piece of evidence as independent. This approach is why this is called naive Bayes.

P(Outcome|Multiple Evidence) = 
P(Evidence1|Outcome) * P(Evidence2|outcome) * ... * P(EvidenceN|outcome) * P(Outcome)
scaled by P(Multiple Evidence)

Many people choose to remember this as:

                      P(Likelihood of Evidence) * Prior prob of outcome
P(outcome|evidence) = _________________________________________________
                                         P(Evidence)

Notice a few things about this equation:

• If the Prob(evidence|outcome) is 1, then we are just multiplying by 1.
• If the Prob(some particular evidence|outcome) is 0, then the whole prob. becomes 0. If you see contradicting evidence, we can rule out that outcome.
• Since we divide everything by P(Evidence), we can even get away without calculating it.
• The intuition behind multiplying by the prior is so that we give high probability to more common outcomes, and low probabilities to unlikely outcomes. These are also called base rates and they are a way to scale our predicted probabilities.

How to Apply NaiveBayes to Predict an Outcome?

Just run the formula above for each possible outcome. Since we are trying to classify, each outcome is called a class and it has a class label. Our job is to look at the evidence, to consider how likely it is to be this class or that class, and assign a label to each entity. Again, we take a very simple approach: The class that has the highest probability is declared the "winner" and that class label gets assigned to that combination of evidences.

Fruit Example

Let's try it out on an example to increase our understanding: The OP asked for a 'fruit' identification example.

Let's say that we have data on 1000 pieces of fruit. They happen to be Banana, Orange or some Other Fruit. We know 3 characteristics about each fruit:

1. Whether it is Long
2. Whether it is Sweet and
3. If its color is Yellow.

This is our 'training set.' We will use this to predict the type of any new fruit we encounter.

Type           Long | Not Long || Sweet | Not Sweet || Yellow |Not Yellow|Total
             ___________________________________________________________________
Banana      |  400  |    100   || 350   |    150    ||  450   |  50      |  500
Orange      |    0  |    300   || 150   |    150    ||  300   |   0      |  300
Other Fruit |  100  |    100   || 150   |     50    ||   50   | 150      |  200
            ____________________________________________________________________
Total       |  500  |    500   || 650   |    350    ||  800   | 200      | 1000
             ___________________________________________________________________

We can pre-compute a lot of things about our fruit collection.

The so-called "Prior" probabilities. (If we didn't know any of the fruit attributes, this would be our guess.) These are our base rates.

 P(Banana)      = 0.5 (500/1000)
 P(Orange)      = 0.3
 P(Other Fruit) = 0.2

Probability of "Evidence"

p(Long)   = 0.5
P(Sweet)  = 0.65
P(Yellow) = 0.8

Probability of "Likelihood"

P(Long|Banana) = 0.8
P(Long|Orange) = 0  [Oranges are never long in all the fruit we have seen.]
 ....

P(Yellow|Other Fruit)     =  50/200 = 0.25
P(Not Yellow|Other Fruit) = 0.75

Given a Fruit, how to classify it?

Let's say that we are given the properties of an unknown fruit, and asked to classify it. We are told that the fruit is Long, Sweet and Yellow. Is it a Banana? Is it an Orange? Or Is it some Other Fruit?

We can simply run the numbers for each of the 3 outcomes, one by one. Then we choose the highest probability and 'classify' our unknown fruit as belonging to the class that had the highest probability based on our prior evidence (our 1000 fruit training set):

P(Banana|Long, Sweet and Yellow) 
      P(Long|Banana) * P(Sweet|Banana) * P(Yellow|Banana) * P(banana)
    = _______________________________________________________________
                      P(Long) * P(Sweet) * P(Yellow)
                      
    = 0.8 * 0.7 * 0.9 * 0.5 / P(evidence)

    = 0.252 / P(evidence)


P(Orange|Long, Sweet and Yellow) = 0


P(Other Fruit|Long, Sweet and Yellow)
      P(Long|Other fruit) * P(Sweet|Other fruit) * P(Yellow|Other fruit) * P(Other Fruit)
    = ____________________________________________________________________________________
                                          P(evidence)

    = (100/200 * 150/200 * 50/200 * 200/1000) / P(evidence)

    = 0.01875 / P(evidence)

By an overwhelming margin (0.252 >> 0.01875), we classify this Sweet/Long/Yellow fruit as likely to be a Banana.

Why is Bayes Classifier so popular?

Look at what it eventually comes down to. Just some counting and multiplication. We can pre-compute all these terms, and so classifying becomes easy, quick and efficient.

Let z = 1 / P(evidence). Now we quickly compute the following three quantities.

P(Banana|evidence) = z * Prob(Banana) * Prob(Evidence1|Banana) * Prob(Evidence2|Banana) ...
P(Orange|Evidence) = z * Prob(Orange) * Prob(Evidence1|Orange) * Prob(Evidence2|Orange) ...
P(Other|Evidence)  = z * Prob(Other)  * Prob(Evidence1|Other)  * Prob(Evidence2|Other)  ...

Assign the class label of whichever is the highest number, and you are done.

Despite the name, Naive Bayes turns out to be excellent in certain applications. Text classification is one area where it really shines.

Hope that helps in understanding the concepts behind the Naive Bayes algorithm.

Check out Date.js

Date.today().previous().monday()

There is one thing that many people don't realize right away when dealing with loops and invariants. They get confused between the loop invariant, and the loop conditional ( the condition which controls termination of the loop ).

As people point out, the loop invariant must be true

1. before the loop starts
2. before each iteration of the loop
3. after the loop terminates

( although it can temporarily be false during the body of the loop ). On the other hand the loop conditional must be false after the loop terminates, otherwise the loop would never terminate.

Thus the loop invariant and the loop conditional must be different conditions.

A good example of a complex loop invariant is for binary search.

bsearch(type A[], type a) {
start = 1, end = length(A)

    while ( start <= end ) {
        mid = floor(start + end / 2)

        if ( A[mid] == a ) return mid
        if ( A[mid] > a ) end = mid - 1
        if ( A[mid] < a ) start = mid + 1

    }
    return -1

}

So the loop conditional seems pretty straight forward - when start > end the loop terminates. But why is the loop correct? What is the loop invariant which proves it's correctness?

The invariant is the logical statement:

if ( A[mid] == a ) then ( start <= mid <= end )

This statement is a logical tautology - it is always true in the context of the specific loop / algorithm we are trying to prove. And it provides useful information about the correctness of the loop after it terminates.

If we return because we found the element in the array then the statement is clearly true, since if A[mid] == a then a is in the array and mid must be between start and end. And if the loop terminates because start > end then there can be no number such that start <= mid and mid <= end and therefore we know that the statement A[mid] == a must be false. However, as a result the overall logical statement is still true in the null sense. ( In logic the statement if ( false ) then ( something ) is always true. )

Now what about what I said about the loop conditional necessarily being false when the loop terminates? It looks like when the element is found in the array then the loop conditional is true when the loop terminates!? It's actually not, because the implied loop conditional is really while ( A[mid] != a && start <= end ) but we shorten the actual test since the first part is implied. This conditional is clearly false after the loop regardless of how the loop terminates.

There are two major purposes of hashing functions:

• to disperse data points uniformly into n bits.
• to securely identify the input data.

It's impossible to recommend a hash without knowing what you're using it for.

If you're just making a hash table in a program, then you don't need to worry about how reversible or hackable the algorithm is... SHA-1 or AES is completely unnecessary for this, you'd be better off using a variation of FNV. FNV achieves better dispersion (and thus fewer collisions) than a simple prime mod like you mentioned, and it's more adaptable to varying input sizes.

If you're using the hashes to hide and authenticate public information (such as hashing a password, or a document), then you should use one of the major hashing algorithms vetted by public scrutiny. The Hash Function Lounge is a good place to start.

• Divide and Conquer
  • They broke into non-overlapping sub-problems
  • Example: factorial numbers i.e. fact(n) = n*fact(n-1)

fact(5) = 5* fact(4) = 5 * (4 * fact(3))= 5 * 4 * (3 *fact(2))= 5 * 4 * 3 * 2 * (fact(1))

As we can see above, no fact(x) is repeated so factorial has non overlapping problems.

• Dynamic Programming
  • They Broke into overlapping sub-problems
  • Example: Fibonacci numbers i.e. fib(n) = fib(n-1) + fib(n-2)

fib(5) = fib(4) + fib(3) = (fib(3)+fib(2)) + (fib(2)+fib(1))

As we can see above, fib(4) and fib(3) both use fib(2). similarly so many fib(x) gets repeated. that's why Fibonacci has overlapping sub-problems.

• As a result of the repetition of sub-problem in DP, we can keep such results in a table and save computation effort. this is called as memoization

this turns the recursion in to a loop:

def checkNextID(ID):
    global numOfRuns, curRes, lastResult
    while ID < lastResult:
        try:
            numOfRuns += 1
            if numOfRuns % 10 == 0:
                time.sleep(3) # sleep every 10 iterations
            if isValid(ID + 8):
                parseHTML(curRes)
                ID = ID + 8
            elif isValid(ID + 18):
                parseHTML(curRes)
                ID = ID + 18
            elif isValid(ID + 7):
                parseHTML(curRes)
                ID = ID + 7
            elif isValid(ID + 17):
                parseHTML(curRes)
                ID = ID + 17
            elif isValid(ID+6):
                parseHTML(curRes)
                ID = ID + 6
            elif isValid(ID + 16):
                parseHTML(curRes)
                ID = ID + 16
            else:
                ID = ID + 1
        except Exception, e:
            print "somethin went wrong: " + str(e)

This is what I though and it works. It will run in o(log n )

public static int FindkThSmallestElemet(Node root, int k)
    {
        int count = 0;
        Node current = root;

        while (current != null)
        {
            count++;
            current = current.left;
        }
        current = root;

        while (current != null)
        {
            if (count == k)
                return current.data;
            else
            {
                current = current.left;
                count--;
            }
        }

        return -1;


    } // end of function FindkThSmallestElemet

Here's some absolutely straightforward C++ code to solve the problem which did ask for all the combinations to be shown.

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    if (argc != 2)
    {
        printf("usage: change amount-in-cents\n");
        return 1;
    }

    int total = atoi(argv[1]);

    printf("quarter\tdime\tnickle\tpenny\tto make %d\n", total);

    int combos = 0;

    for (int q = 0; q <= total / 25; q++)
    {
        int total_less_q = total - q * 25;
        for (int d = 0; d <= total_less_q / 10; d++)
        {
            int total_less_q_d = total_less_q - d * 10;
            for (int n = 0; n <= total_less_q_d / 5; n++)
            {
                int p = total_less_q_d - n * 5;
                printf("%d\t%d\t%d\t%d\n", q, d, n, p);
                combos++;
            }
        }
    }

    printf("%d combinations\n", combos);

    return 0;
}

But I'm quite intrigued about the sub problem of just calculating the number of combinations. I suspect there's a closed-form equation for it.

Part 1 - height

As starblue says, height is just recursive. In pseudo-code:

height(node) = max(height(node.L), height(node.R)) + 1

Now height could be defined in two ways. It could be the number of nodes in the path from the root to that node, or it could be the number of links. According to the page you referenced, the most common definition is for the number of links. In which case the complete pseudo code would be:

height(node): 
   if node == null:
        return -1
   else:
        return max(height(node.L), height(node.R)) + 1

If you wanted the number of nodes the code would be:

height(node): 
   if node == null:
        return 0
   else:
        return max(height(node.L), height(node.R)) + 1

Either way, the rebalancing algorithm I think should work the same.

However, your tree will be much more efficient (O(ln(n))) if you store and update height information in the tree, rather than calculating it each time. (O(n))

Part 2 - balancing

When it says "If the balance factor of R is 1", it is talking about the balance factor of the right branch, when the balance factor at the top is 2. It is telling you how to choose whether to do a single rotation or a double rotation. In (python like) Pseudo-code:

if balance factor(top) = 2: // right is imbalanced
     if balance factor(R) = 1: // 
          do a left rotation
     else if balance factor(R) = -1:
          do a double rotation
else: // must be -2, left is imbalanced
     if balance factor(L) = 1: // 
          do a left rotation
     else if balance factor(L) = -1:
          do a double rotation

I hope this makes sense

If A and B can be expressed with 2 bytes, you can combine them on 4 bytes. Put A on the most significant half and B on the least significant half.

In C language this gives (assuming sizeof(short)=2 and sizeof(int)=4):

int combine(short A, short B)
{
    return A<<16 | B;
}

short getA(int C)
{
    return C>>16;
}

short getB(int C)
{
    return C & 0xFFFF;
}

In my opinion, the most understandable algorithm for detecting cycle in a directed graph is the graph-coloring-algorithm.

Basically, the graph coloring algorithm walks the graph in a DFS manner (Depth First Search, which means that it explores a path completely before exploring another path). When it finds a back edge, it marks the graph as containing a loop.

For an in depth explanation of the graph coloring algorithm, please read this article: http://www.geeksforgeeks.org/detect-cycle-direct-graph-using-colors/

Also, I provide an implementation of graph coloring in JavaScript https://github.com/dexcodeinc/graph_algorithm.js/blob/master/graph_algorithm.js

There is also a pure Python recursive option:

def checkEqual(lst):
    if len(lst)==2 :
        return lst[0]==lst[1]
    else:
        return lst[0]==lst[1] and checkEqual(lst[1:])

However for some reason it is in some cases two orders of magnitude slower than other options. Coming from C language mentality, I expected this to be faster, but it is not!

The other disadvantage is that there is recursion limit in Python which needs to be adjusted in this case. For example using this.

the major difference between greedy method and dynamic programming is in greedy method only one optimal decision sequence is ever generated and in dynamic programming more than one optimal decision sequence may be generated.

My answer is similar as Josh Greifer but generalised to sample covariance. Sample variance is just sample covariance but with the two inputs identical. This includes Bessel's correlation.

    template <class Iter> typename Iter::value_type cov(const Iter &x, const Iter &y)
    {
        double sum_x = std::accumulate(std::begin(x), std::end(x), 0.0);
        double sum_y = std::accumulate(std::begin(y), std::end(y), 0.0);

        double mx =  sum_x / x.size();
        double my =  sum_y / y.size();

        double accum = 0.0;

        for (auto i = 0; i < x.size(); i++)
        {
            accum += (x.at(i) - mx) * (y.at(i) - my);
        }

        return accum / (x.size() - 1);
    }

This is pure speculation on my part, but I suppose that they may use an influence map data structure overlaying the directed map in order to narrow the search domain. This would allow the search algorithm to direct the path to major routes when the desired trip is long.

Given that this is a Google app, it's also reasonable to suppose that a lot of the magic is done via extensive caching. :) I wouldn't be surprised if caching the top 5% most common Google Map route requests allowed for a large chunk (20%? 50%?) of requests to be answered by a simple look-up.

Depends on the how you implement hashing, in the worst case it can go to O(n), in best case it is 0(1) (generally you can achieve if your DS is not that big easily)

A simple example of O(1) might be return 23; -- whatever the input, this will return in a fixed, finite time.

A typical example of O(N log N) would be sorting an input array with a good algorithm (e.g. mergesort).

A typical example if O(log N) would be looking up a value in a sorted input array by bisection.

Stack<Node> stack = new Stack<>();
stack.add(root);
while (!stack.isEmpty()) {
    Node node = stack.pop();
    System.out.print(node.getData() + " ");

    Node right = node.getRight();
    if (right != null) {
        stack.push(right);
    }

    Node left = node.getLeft();
    if (left != null) {
        stack.push(left);
    }
}

Here is a better version with better output formatting and C++ 11 features:

void subset_sum_rec(std::vector<int> & nums, const int & target, std::vector<int> & partialNums) 
{
    int currentSum = std::accumulate(partialNums.begin(), partialNums.end(), 0);
    if (currentSum > target)
        return;
    if (currentSum == target) 
    {
        std::cout << "sum([";
        for (auto it = partialNums.begin(); it != std::prev(partialNums.end()); ++it)
            cout << *it << ",";
        cout << *std::prev(partialNums.end());
        std::cout << "])=" << target << std::endl;
    }
    for (auto it = nums.begin(); it != nums.end(); ++it) 
    {
        std::vector<int> remaining;
        for (auto it2 = std::next(it); it2 != nums.end(); ++it2)
            remaining.push_back(*it2);

        std::vector<int> partial = partialNums;
        partial.push_back(*it);
        subset_sum_rec(remaining, target, partial);
    }
}

Note: This answer is for the specific question asked above. If you are here from Google and just looking for a way to get a Cartesian product in Python, itertools.product or a simple list comprehension may be what you are looking for - see the other answers.

Suppose len(list1) >= len(list2). Then what you appear to want is to take all permutations of length len(list2) from list1 and match them with items from list2. In python:

import itertools
list1=['a','b','c']
list2=[1,2]

[list(zip(x,list2)) for x in itertools.permutations(list1,len(list2))]

Returns

[[('a', 1), ('b', 2)], [('a', 1), ('c', 2)], [('b', 1), ('a', 2)], [('b', 1), ('c', 2)], [('c', 1), ('a', 2)], [('c', 1), ('b', 2)]]

Kaz Kylheku here

I benchmarked two approaches for this over 63 bit numbers (the long long type on gcc x86_64), staying away from the sign bit.

(I happen to need this "find highest bit" for something, you see.)

I implemented the data-driven binary search (closely based on one of the above answers). I also implemented a completely unrolled decision tree by hand, which is just code with immediate operands. No loops, no tables.

The decision tree (highest_bit_unrolled) benchmarked to be 69% faster, except for the n = 0 case for which the binary search has an explicit test.

The binary-search's special test for 0 case is only 48% faster than the decision tree, which does not have a special test.

Compiler, machine: (GCC 4.5.2, -O3, x86-64, 2867 Mhz Intel Core i5).

int highest_bit_unrolled(long long n)
{
  if (n & 0x7FFFFFFF00000000) {
    if (n & 0x7FFF000000000000) {
      if (n & 0x7F00000000000000) {
        if (n & 0x7000000000000000) {
          if (n & 0x4000000000000000)
            return 63;
          else
            return (n & 0x2000000000000000) ? 62 : 61;
        } else {
          if (n & 0x0C00000000000000)
            return (n & 0x0800000000000000) ? 60 : 59;
          else
            return (n & 0x0200000000000000) ? 58 : 57;
        }
      } else {
        if (n & 0x00F0000000000000) {
          if (n & 0x00C0000000000000)
            return (n & 0x0080000000000000) ? 56 : 55;
          else
            return (n & 0x0020000000000000) ? 54 : 53;
        } else {
          if (n & 0x000C000000000000)
            return (n & 0x0008000000000000) ? 52 : 51;
          else
            return (n & 0x0002000000000000) ? 50 : 49;
        }
      }
    } else {
      if (n & 0x0000FF0000000000) {
        if (n & 0x0000F00000000000) {
          if (n & 0x0000C00000000000)
            return (n & 0x0000800000000000) ? 48 : 47;
          else
            return (n & 0x0000200000000000) ? 46 : 45;
        } else {
          if (n & 0x00000C0000000000)
            return (n & 0x0000080000000000) ? 44 : 43;
          else
            return (n & 0x0000020000000000) ? 42 : 41;
        }
      } else {
        if (n & 0x000000F000000000) {
          if (n & 0x000000C000000000)
            return (n & 0x0000008000000000) ? 40 : 39;
          else
            return (n & 0x0000002000000000) ? 38 : 37;
        } else {
          if (n & 0x0000000C00000000)
            return (n & 0x0000000800000000) ? 36 : 35;
          else
            return (n & 0x0000000200000000) ? 34 : 33;
        }
      }
    }
  } else {
    if (n & 0x00000000FFFF0000) {
      if (n & 0x00000000FF000000) {
        if (n & 0x00000000F0000000) {
          if (n & 0x00000000C0000000)
            return (n & 0x0000000080000000) ? 32 : 31;
          else
            return (n & 0x0000000020000000) ? 30 : 29;
        } else {
          if (n & 0x000000000C000000)
            return (n & 0x0000000008000000) ? 28 : 27;
          else
            return (n & 0x0000000002000000) ? 26 : 25;
        }
      } else {
        if (n & 0x0000000000F00000) {
          if (n & 0x0000000000C00000)
            return (n & 0x0000000000800000) ? 24 : 23;
          else
            return (n & 0x0000000000200000) ? 22 : 21;
        } else {
          if (n & 0x00000000000C0000)
            return (n & 0x0000000000080000) ? 20 : 19;
          else
            return (n & 0x0000000000020000) ? 18 : 17;
        }
      }
    } else {
      if (n & 0x000000000000FF00) {
        if (n & 0x000000000000F000) {
          if (n & 0x000000000000C000)
            return (n & 0x0000000000008000) ? 16 : 15;
          else
            return (n & 0x0000000000002000) ? 14 : 13;
        } else {
          if (n & 0x0000000000000C00)
            return (n & 0x0000000000000800) ? 12 : 11;
          else
            return (n & 0x0000000000000200) ? 10 : 9;
        }
      } else {
        if (n & 0x00000000000000F0) {
          if (n & 0x00000000000000C0)
            return (n & 0x0000000000000080) ? 8 : 7;
          else
            return (n & 0x0000000000000020) ? 6 : 5;
        } else {
          if (n & 0x000000000000000C)
            return (n & 0x0000000000000008) ? 4 : 3;
          else
            return (n & 0x0000000000000002) ? 2 : (n ? 1 : 0);
        }
      }
    }
  }
}

int highest_bit(long long n)
{
  const long long mask[] = {
    0x000000007FFFFFFF,
    0x000000000000FFFF,
    0x00000000000000FF,
    0x000000000000000F,
    0x0000000000000003,
    0x0000000000000001
  };
  int hi = 64;
  int lo = 0;
  int i = 0;

  if (n == 0)
    return 0;

  for (i = 0; i < sizeof mask / sizeof mask[0]; i++) {
    int mi = lo + (hi - lo) / 2;

    if ((n >> mi) != 0)
      lo = mi;
    else if ((n & (mask[i] << lo)) != 0)
      hi = mi;
  }

  return lo + 1;
}

Quick and dirty test program:

#include <stdio.h>
#include <time.h>
#include <stdlib.h>

int highest_bit_unrolled(long long n);
int highest_bit(long long n);

main(int argc, char **argv)
{
  long long n = strtoull(argv[1], NULL, 0);
  int b1, b2;
  long i;
  clock_t start = clock(), mid, end;

  for (i = 0; i < 1000000000; i++)
    b1 = highest_bit_unrolled(n);

  mid = clock();

  for (i = 0; i < 1000000000; i++)
    b2 = highest_bit(n);

  end = clock();

  printf("highest bit of 0x%llx/%lld = %d, %d\n", n, n, b1, b2);

  printf("time1 = %d\n", (int) (mid - start));
  printf("time2 = %d\n", (int) (end - mid));
  return 0;
}

Using only -O2, the difference becomes greater. The decision tree is almost four times faster.

I also benchmarked against the naive bit shifting code:

int highest_bit_shift(long long n)
{
  int i = 0;
  for (; n; n >>= 1, i++)
    ; /* empty */
  return i;
}

This is only fast for small numbers, as one would expect. In determining that the highest bit is 1 for n == 1, it benchmarked more than 80% faster. However, half of randomly chosen numbers in the 63 bit space have the 63rd bit set!

On the input 0x3FFFFFFFFFFFFFFF, the decision tree version is quite a bit faster than it is on 1, and shows to be 1120% faster (12.2 times) than the bit shifter.

I will also benchmark the decision tree against the GCC builtins, and also try a mixture of inputs rather than repeating against the same number. There may be some sticking branch prediction going on and perhaps some unrealistic caching scenarios which makes it artificially faster on repetitions.

This ain't no job for a human! ... but perfect for a machine

This is 2015, 6 years from when this question was first asked. Compilers have since become our masters, and our job as humans is only to help them. So what's the best way to give our intentions to the machine?

Bit-reversal is so common that you have to wonder why the x86's ever growing ISA doesn't include an instruction to do it one go.

The reason: if you give your true concise intent to the compiler, bit reversal should only take ~20 CPU cycles. Let me show you how to craft reverse() and use it:

#include <inttypes.h>
#include <stdio.h>

uint64_t reverse(const uint64_t n,
                 const uint64_t k)
{
        uint64_t r, i;
        for (r = 0, i = 0; i < k; ++i)
                r |= ((n >> i) & 1) << (k - i - 1);
        return r;
}

int main()
{
        const uint64_t size = 64;
        uint64_t sum = 0;
        uint64_t a;
        for (a = 0; a < (uint64_t)1 << 30; ++a)
                sum += reverse(a, size);
        printf("%" PRIu64 "\n", sum);
        return 0;
}

Compiling this sample program with Clang version >= 3.6, -O3, -march=native (tested with Haswell), gives artwork-quality code using the new AVX2 instructions, with a runtime of 11 seconds processing ~1 billion reverse()s. That's ~10 ns per reverse(), with .5 ns CPU cycle assuming 2 GHz puts us at the sweet 20 CPU cycles.

• You can fit 10 reverse()s in the time it takes to access RAM once for a single large array!
• You can fit 1 reverse() in the time it takes to access an L2 cache LUT twice.

Caveat: this sample code should hold as a decent benchmark for a few years, but it will eventually start to show its age once compilers are smart enough to optimize main() to just printf the final result instead of really computing anything. But for now it works in showcasing reverse().

The module timeit is useful for this and is included in the standard Python distribution.

Example:

import timeit
timeit.Timer('for i in xrange(10): oct(i)').timeit()

Here is the algorithm:

1. Explode the string with space and make an array of words.
2. Reverse the array.
3. Implode the array by space.

So to check whether a number N is Prime or not. We need to only check if N is divisible by numbers<=SQROOT(N). This is because, if we factor N into any 2 factors say X and Y, ie. N=XY. Each of X and Y cannot be less than SQROOT(N) because then, XY < N Each of X and Y cannot be greater than SQROOT(N) because then, X*Y > N

Therefore one factor must be less than or equal to SQROOT(N) ( while the other factor is greater than or equal to SQROOT(N) ). So to check if N is Prime we need only check those numbers <= SQROOT(N).

Thinking out of the box for a second, if you are making this call many many times, it is vastly more efficient to use an associative array a Map to do lookups using a hash function.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Map

Here is the method in Java

private int ipow(int base, int exp)
{
    int result = 1;
    while (exp != 0)
    {
        if ((exp & 1) == 1)
            result *= base;
        exp >>= 1;
        base *= base;
    }

    return result;
}

This was written quickly, and is neither pretty nor efficient (plus it autoboxes alot, converting between int and Integer is annoying!), but it works.

It probably breaks the rules since I'm creating my own objects but hey I'm doing this as a diversion from real work :)

This also assumes that the resultSet/table is completely read into some sort of structure before you start building Nodes, which wouldn't be the best solution if you have hundreds of thousands of rows.

public class Node {

    private Node parent = null;

    private List<Node> children;

    private String name;

    private int id = -1;

    public Node(Node parent, int id, String name) {
        this.parent = parent;
        this.children = new ArrayList<Node>();
        this.name = name;
        this.id = id;
    }

    public int getId() {
        return this.id;
    }

    public String getName() {
        return this.name;
    }

    public void addChild(Node child) {
        children.add(child);
    }

    public List<Node> getChildren() {
        return children;
    }

    public boolean isRoot() {
        return (this.parent == null);
    }

    @Override
    public String toString() {
        return "id=" + id + ", name=" + name + ", parent=" + parent;
    }
}

public class NodeBuilder {

    public static Node build(List<Map<String, String>> input) {

        // maps id of a node to it's Node object
        Map<Integer, Node> nodeMap = new HashMap<Integer, Node>();

        // maps id of a node to the id of it's parent
        Map<Integer, Integer> childParentMap = new HashMap<Integer, Integer>();

        // create special 'root' Node with id=0
        Node root = new Node(null, 0, "root");
        nodeMap.put(root.getId(), root);

        // iterate thru the input
        for (Map<String, String> map : input) {

            // expect each Map to have keys for "id", "name", "parent" ... a
            // real implementation would read from a SQL object or resultset
            int id = Integer.parseInt(map.get("id"));
            String name = map.get("name");
            int parent = Integer.parseInt(map.get("parent"));

            Node node = new Node(null, id, name);
            nodeMap.put(id, node);

            childParentMap.put(id, parent);
        }

        // now that each Node is created, setup the child-parent relationships
        for (Map.Entry<Integer, Integer> entry : childParentMap.entrySet()) {
            int nodeId = entry.getKey();
            int parentId = entry.getValue();

            Node child = nodeMap.get(nodeId);
            Node parent = nodeMap.get(parentId);
            parent.addChild(child);
        }

        return root;
    }
}

public class NodePrinter {

    static void printRootNode(Node root) {
        printNodes(root, 0);
    }

    static void printNodes(Node node, int indentLevel) {

        printNode(node, indentLevel);
        // recurse
        for (Node child : node.getChildren()) {
            printNodes(child, indentLevel + 1);
        }
    }

    static void printNode(Node node, int indentLevel) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < indentLevel; i++) {
            sb.append("\t");
        }
        sb.append(node);

        System.out.println(sb.toString());
    }

    public static void main(String[] args) {

        // setup dummy data
        List<Map<String, String>> resultSet = new ArrayList<Map<String, String>>();
        resultSet.add(newMap("1", "Node 1", "0"));
        resultSet.add(newMap("2", "Node 1.1", "1"));
        resultSet.add(newMap("3", "Node 2", "0"));
        resultSet.add(newMap("4", "Node 1.1.1", "2"));
        resultSet.add(newMap("5", "Node 2.1", "3"));
        resultSet.add(newMap("6", "Node 1.2", "1"));

        Node root = NodeBuilder.build(resultSet);
        printRootNode(root);

    }

    //convenience method for creating our dummy data
    private static Map<String, String> newMap(String id, String name, String parentId) {
        Map<String, String> row = new HashMap<String, String>();
        row.put("id", id);
        row.put("name", name);
        row.put("parent", parentId);
        return row;
    }
}

An Euler path is a path that passes through every edge exactly once. If it ends at the initial vertex then it is an Euler cycle.

A Hamiltonian path is a path that passes through every vertex exactly once (NOT every edge). If it ends at the initial vertex then it is a Hamiltonian cycle.

In an Euler path you might pass through a vertex more than once.

In a Hamiltonian path you may not pass through all edges.

If we're allowed to embed the class Node, I would solve the problem as I've implemented it below. hasLoop() runs in O(n) time, and takes only the space of counter. Does this seem like an appropriate solution? Or is there a way to do it without embedding Node? (Obviously, in a real implementation there would be more methods, like RemoveNode(Node n), etc.)

public class LinkedNodeList {
    Node first;
    Int count;

    LinkedNodeList(){
        first = null;
        count = 0;
    }

    LinkedNodeList(Node n){
        if (n.next != null){
            throw new error("must start with single node!");
        } else {
            first = n;
            count = 1;
        }
    }

    public void addNode(Node n){
        Node lookingAt = first;

        while(lookingAt.next != null){
            lookingAt = lookingAt.next;
        }

        lookingAt.next = n;
        count++;
    }

    public boolean hasLoop(){

        int counter = 0;
        Node lookingAt = first;

        while(lookingAt.next != null){
            counter++;
            if (count < counter){
                return false;
            } else {
               lookingAt = lookingAt.next;
            }
        }

        return true;

    }



    private class Node{
        Node next;
        ....
    }

}

In Python 3:

def is_prime(a):
    if a < 2:
        return False
    elif a!=2 and a % 2 == 0:
        return False
    else:
        return all (a % i for i in range(3, int(a**0.5)+1))

Explanation: A prime number is a number only divisible by itself and 1. Ex: 2,3,5,7...

1) if a<2: if "a" is less than 2 it is not a prime.

2) elif a!=2 and a % 2 == 0: if "a" is divisible by 2 then its definitely not a prime. But if a=2 we don't want to evaluate that as it is a prime number. Hence the condition a!=2

3) return all (a % i for i in range(3, int(a0.5)+1) ):** First look at what all() command does in python. Starting from 3 we divide "a" till its square root (a**0.5). If "a" is divisible the output will be False. Why square root? Let's say a=16. The square root of 16 = 4. We don't need to evaluate till 15. We only need to check till 4 to say that it's not a prime.

Extra: A loop for finding all the prime number within a range.

for i in range(1,100):
    if is_prime(i):
        print("{} is a prime number".format(i))

It represents the speed of an algorithm in the long run.

To take a literal analogy, you don't care how fast a runner can sprint a 100m dash, or even a 5k run. You care more about marathoners, and preferably ultra marathoners (beyond which the analogy to running breaks down and you have to revert to the metaphorical meaning of "the long run").

You can safely stop reading here.

I'm adding this answer because I'm surprised how mathematical and technical the rest of the answers are. The notion of the "long run" in first sentence is related to the arbitrarily time-consuming computational tasks. Unlike running, which is limited by human capacity, computational tasks can take even more than millions of years for certain algorithms to complete.

What about all those mathematical logarithms and polynomials? It turns out that algorithms are intrinsically related to these mathematical terms. If you are measuring the heights of all the kids on the block, it will take you as much time as there are kids. This is intrinsically related to the notion of n^1 or just n where n is nothing more than the number of kids on the block. In the ultra-marathon case, you are measuring the heights of all the kids in your city, but you then have to ignore travel times and assume they are all available to you in a line (otherwise we jump ahead of the current explanation).

Suppose then you are trying to arrange the list that you made of of kids heights in order of shortest height to longest height. If it is just the kids in your neighborhood you might just eyeball it and come up with the ordered list. This is the "sprint" analogy, and we truly don't care about sprints in computer science because why use a computer when you can eyeball something?

But if you were arranging the list of the heights of all kids in your city, or better yet, your country, then you will find that how you do it is intrinsically tied to the mathematical log and n^2. Going through your list to find the shortest kid, writing his name in a separate notebook, and crossing it out from the original notebook is intrinsically tied to the mathematical n^2. If you think of arranging half your notebook, then the other half, and then combining the results, you will arrive at a method that is intrinsically tied to the logarithm.

Finally, suppose you first had to go to the store to buy a measuring tape. This is an example of an effort that is of consequence in short sprints, such as measuring the kids on the block, but when you are measuring all the kids in the city you can safely ignore this cost. This is the intrinsic connection to the mathematical dropping of say lower order polynomial terms.

I hope I have explained that the big-O notation is merely about the long run, that the mathematics is inherently connected to ways of computation, and that the dropping of mathematical terms and other simplifications are connected to the long run in a rather common sense way.

Once you realize this, you'll find the big-O is really super-easy because all the hard high school math just drops out easily. The only difficult part is analyzing an algorithm to identify the mathematical terms, but with some practice you can start dropping terms during the analysis itself and safely ignore chunks of the algorithm to focus only on the part that is relevant to the big-O. I. e. you should be able to eyeball most situations.

Happy big-O-ing, it was my favorite thing about Computer Science -- finding that something was way easier than I thought, and then being able to show off at Google interviews when the uninitiated would be intimidated, lol.

A big expectation from hash functions is that their result's uniform randomness survives an operation such as hash(x) % N where N is an arbitrary number (and in many cases, a power of two), one reason being that such operations are used commonly in hash tables for determining slots. Using prime number multipliers when computing the hash decreases the probability that your multiplier and the N share divisors, which would make the result of the operation less uniformly random.

Others have pointed out the nice property that multiplication by 31 can be done by a multiplication and a subtraction. I just want to point out that there is a mathematical term for such primes: Mersenne Prime

All mersenne primes are one less than a power of two so we can write them as:

p = 2^n - 1

Multiplying x by p:

x * p = x * (2^n - 1) = x * 2^n - x = (x << n) - x

Shifts (SAL/SHL) and subtractions (SUB) are generally faster than multiplications (MUL) on many machines. See instruction tables from Agner Fog

That's why GCC seems to optimize multiplications by mersenne primes by replacing them with shifts and subs, see here.

However, in my opinion, such a small prime is a bad choice for a hash function. With a relatively good hash function, you would expect to have randomness at the higher bits of the hash. However, with the Java hash function, there is almost no randomness at the higher bits with shorter strings (and still highly questionable randomness at the lower bits). This makes it more difficult to build efficient hash tables. See this nice trick you couldn't do with the Java hash function.

Some answers mention that they believe it is good that 31 fits into a byte. This is actually useless since:

(1) We execute shifts instead of multiplications, so the size of the multiplier does not matter.

(2) As far as I know, there is no specific x86 instruction to multiply an 8 byte value with a 1 byte value so you would have needed to convert "31" to a 8 byte value anyway even if you were multiplying. See here, you multiply entire 64bit registers.

(And 127 is actually the largest mersenne prime that could fit in a byte.)

Does a smaller value increase randomness in the middle-lower bits? Maybe, but it also seems to greatly increase the possible collisions :).

One could list many different issues but they generally boil down to two core principles not being fulfilled well: Confusion and Diffusion

But is it fast? Probably, since it doesn't do much. However, if performance is really the focus here, one character per loop is quite inefficient. Why not do 4 characters at a time (8 bytes) per loop iteration for longer strings, like this? Well, that would be difficult to do with the current definition of hash where you need to multiply every character individually (please tell me if there is a bit hack to solve this :D).

PHP, assuming "set" is an array:

$foo = array("alpha", "bravo", "charlie");
$index = array_rand($foo);
$val = $foo[$index];

The Mersenne Twister functions are better but there's no MT equivalent of array_rand in PHP.

Calculating point around circumference of circle given distance travelled.
For comparison... This may be useful in Game AI when moving around a solid object in a direct path.

public static Point DestinationCoordinatesArc(Int32 startingPointX, Int32 startingPointY,
    Int32 circleOriginX, Int32 circleOriginY, float distanceToMove,
    ClockDirection clockDirection, float radius)
{
    // Note: distanceToMove and radius parameters are float type to avoid integer division
    // which will discard remainder

    var theta = (distanceToMove / radius) * (clockDirection == ClockDirection.Clockwise ? 1 : -1);
    var destinationX = circleOriginX + (startingPointX - circleOriginX) * Math.Cos(theta) - (startingPointY - circleOriginY) * Math.Sin(theta);
    var destinationY = circleOriginY + (startingPointX - circleOriginX) * Math.Sin(theta) + (startingPointY - circleOriginY) * Math.Cos(theta);

    // Round to avoid integer conversion truncation
    return new Point((Int32)Math.Round(destinationX), (Int32)Math.Round(destinationY));
}

/// <summary>
/// Possible clock directions.
/// </summary>
public enum ClockDirection
{
    [Description("Time moving forwards.")]
    Clockwise,
    [Description("Time moving moving backwards.")]
    CounterClockwise
}

private void ButtonArcDemo_Click(object sender, EventArgs e)
{
    Brush aBrush = (Brush)Brushes.Black;
    Graphics g = this.CreateGraphics();

    var startingPointX = 125;
    var startingPointY = 75;
    for (var count = 0; count < 62; count++)
    {
        var point = DestinationCoordinatesArc(
            startingPointX: startingPointX, startingPointY: startingPointY,
            circleOriginX: 75, circleOriginY: 75,
            distanceToMove: 5,
            clockDirection: ClockDirection.Clockwise, radius: 50);
        g.FillRectangle(aBrush, point.X, point.Y, 1, 1);

        startingPointX = point.X;
        startingPointY = point.Y;

        // Pause to visually observe/confirm clock direction
        System.Threading.Thread.Sleep(35);

        Debug.WriteLine($"DestinationCoordinatesArc({point.X}, {point.Y}");
    }
}

I had a lecturer who once suggested generating a random array, checking if it was sorted and then checking if the data was the same as the array to be sorted.

Best case O(N) (first time baby!) Worst case O(Never)

a complete circle in radians is 2*pi. A complete circle in degrees is 360. To go from degrees to radians, it's (d/360) * 2*pi, or d*pi/180.

A simple method to figure out whether the testString is an anagram of the baseString.

private static boolean isAnagram(String baseString, String testString){
    //Assume that there are no empty spaces in either string.

    if(baseString.length() != testString.length()){
        System.out.println("The 2 given words cannot be anagram since their lengths are different");
        return false;
    }
    else{
        if(baseString.length() == testString.length()){
            if(baseString.equalsIgnoreCase(testString)){
                System.out.println("The 2 given words are anagram since they are identical.");
                return true;
            }
            else{
                List<Character> list = new ArrayList<>();

                for(Character ch : baseString.toLowerCase().toCharArray()){
                    list.add(ch);
                }
                System.out.println("List is : "+ list);

                for(Character ch : testString.toLowerCase().toCharArray()){
                    if(list.contains(ch)){
                        list.remove(ch);
                    }
                }

                if(list.isEmpty()){
                    System.out.println("The 2 words are anagrams");
                    return true;
                }
            }
        }
    }
    return false;
}

We can do the Q1 and Q2 in O(log n) most of the time.

Suppose our memory chip consists of array of n number of test tubes. And a number x in the the test tube is represented by x milliliter of chemical-liquid.

Suppose our processor is a laser light. When we light up the laser it traverses all the tubes perpendicularly to it's length. Every-time it passes through the chemical liquid, the luminosity is reduced by 1. And passing the light at certain milliliter mark is an operation of O(1).

Now if we light our laser at the middle of the test-tube and get the output of luminosity

• equals to a pre-calculated value(calculated when no numbers were missing), then the missing numbers are greater than n/2 .
• If our output is smaller, then there is at least one missing number that is smaller than n/2 . We can also check if the luminosity is reduced by 1 or 2. if it is reduced by 1 then one missing number is smaller than n/2 and other is bigger than n/2 . If it is reduced by 2 then both numbers are smaller than n/2 .

We can repeat the above process again and again narrowing down our problem domain. In each step, we make the domain smaller by half. And finally we can get to our result.

Parallel algorithms that are worth mentioning(because they are interesting),

• sorting by some parallel algorithm, for example, parallel merge can be done in O(log^3 n) time. And then the missing number can be found by binary search in O(log n) time.
• Theoretically, if we have n processors then each process can check one of the inputs and set some flag that identifies the number(conveniently in an array). And in the next step each process can check each flag and finally output the number that is not flagged. The whole process will take O(1) time. It has additional O(n) space/memory requirement.

Note, that the two parallel algorithms provided above may need additional space as mentioned in the comment.

If you're looking to replace more than a single character, and are dealing only with std::string, then this snippet would work, replacing sNeedle in sHaystack with sReplace, and sNeedle and sReplace do not need to be the same size. This routine uses the while loop to replace all occurrences, rather than just the first one found from left to right.

while(sHaystack.find(sNeedle) != std::string::npos) {
  sHaystack.replace(sHaystack.find(sNeedle),sNeedle.size(),sReplace);
}

python implementation

def getPermutation(s, prefix=''):
        if len(s) == 0:
                print prefix
        for i in range(len(s)):
                getPermutation(s[0:i]+s[i+1:len(s)],prefix+s[i] )



getPermutation('abcd','')

Very good and remarkable question with a very interesting discussion! I also was using the Array.sort() function after pushing a single element in an array with some thousands of objects.

I had to extend your locationOf function for my purpose because of having complex objects and therefore the need for a compare function like in Array.sort():

function locationOf(element, array, comparer, start, end) {
    if (array.length === 0)
        return -1;

    start = start || 0;
    end = end || array.length;
    var pivot = (start + end) >> 1;  // should be faster than dividing by 2

    var c = comparer(element, array[pivot]);
    if (end - start <= 1) return c == -1 ? pivot - 1 : pivot;

    switch (c) {
        case -1: return locationOf(element, array, comparer, start, pivot);
        case 0: return pivot;
        case 1: return locationOf(element, array, comparer, pivot, end);
    };
};

// sample for objects like {lastName: 'Miller', ...}
var patientCompare = function (a, b) {
    if (a.lastName < b.lastName) return -1;
    if (a.lastName > b.lastName) return 1;
    return 0;
};

Math.min uses a simple comparison to do its thing. The only advantage to not using Math.min is to save the extra function calls, but that is a negligible saving.

If you have more than just three numbers, having a minimum method for any number of doubles might be valuable and would look something like:

public static double min(double ... numbers) {
    double min = numbers[0];
    for (int i=1 ; i<numbers.length ; i++) {
        min = (min <= numbers[i]) ? min : numbers[i];
    }
    return min;
}

For three numbers this is the functional equivalent of Math.min(a, Math.min(b, c)); but you save one method invocation.

XPath queries are case sensitive. Having looked at your example (which, by the way, is awesome, nobody seems to provide examples anymore!), I can get the result you want just by changing "business", to "Business"

//production[not(contains(category,'Business'))]

I have tested this by opening the XML file in Chrome, and using the Developer tools to execute that XPath queries, and it gave me just the Film category back.

Use Joda Time Library. It is very easy to handle date, time, calender and locale with it and it will be integrated to java in version 8.

DateTime#minusMonths method would help you get previous month.

DateTime month = new DateTime().minusMonths (1); 

Will it warn you if you pass a 32-bit unsigned integer to a %lu format? It should be fine since the conversion is well-defined and doesn't lose any information.

I've heard that some platforms define macros in <inttypes.h> that you can insert into the format string literal but I don't see that header on my Windows C++ compiler, which implies it may not be cross-platform.

In case someone is using Mac OSX YOSEMITE or earlier. Follow this post. It worked for me. http://forums.macrumors.com/showthread.php?t=1196027

Run these two command in Terminal:

sudo chmod 755 /Applications

sudo chmod 755 /Applications/Virtualbox.app

Took me hours to get it to work!

Does the DB know the connection has dropped, or is the session still listed in v$session? That would indicate, I think, that it's being dropped by the network. Do you know how long it can stay idle before encountering the problem, and if that bears any resemblance to the TCP idle values (net.ipv4.tcp_keepalive_time, tcp_keepalive_probes and tcp_keepalive_interval from sysctl if I recall correctly)? Can't remember whether sysctl changes persist by default, but that might be something that was modified and then reset by the reboot.

Also you might be able to reset your JDBC connections without bouncing the whole server; certainly can in WebLogic, which I realise doesn't help much, but I'm not familiar with the Tomcat equivalents.

When you say "doesn't work", how do you know it doesn't work?

You might try to figure out if the file actually exists by adding:

while [ ! -f /tmp/list.txt ]
do
  sleep 2 # or less like 0.2
done
ls -l /tmp/list.txt

You might also make sure that you're using a Bash (or related) shell by typing 'echo $SHELL'. I think that CSH and TCSH use a slightly different semantic for this loop.

For me it helped to switch the "No Common Blocks" compiler setting to NO: It pretty much seems to make sense, the setting is explained here: What is GCC_NO_COMMON_BLOCKS used for?

tl;dr

Instant.now()                           // Capture the current moment in UTC.
.atZone( ZoneId.systemDefault() )       // Adjust into the JVM's current default time zone. Same moment, different wall-clock time. Produces a `ZonedDateTime` object.
.toInstant()                            // Extract a `Instant` (always in UTC) object from the `ZonedDateTime` object.
.atZone( ZoneId.of( "Europe/Paris" ) )  // Adjust the `Instant` into a specific time zone. Renders a `ZonedDateTime` object. Same moment, different wall-clock time.
.toInstant()                            // And back to UTC again.

java.time

The modern approach uses the java.time classes that supplanted the troublesome old legacy date-time classes (Date, Calendar, etc.).

Your use of the word "local" contradicts the usage in the java.time class. In java.time, "local" means any locality or all localities, but not any one particular locality. The java.time classes with names starting with "Local…" all lack any concept of time zone or offset-from-UTC. So they do not represent a specific moment, they are not a point on the timeline, whereas your Question is all about moments, points on the timeline viewed through various wall-clock times.

Get the current system time (local time)

If you want to capture the current moment in UTC, use Instant. The Instant class represents a moment on the timeline in UTC with a resolution of nanoseconds (up to nine (9) digits of a decimal fraction).

Instant instant = Instant.now() ;  // Capture the current moment in UTC.

Adjust into a time zone by applying a ZoneId to get a ZonedDateTime. Same moment, same point on the timeline, different wall-clock time.

Specify a proper time zone name in the format of continent/region, such as America/Montreal, Africa/Casablanca, or Pacific/Auckland. Never use the 3-4 letter abbreviation such as EST or IST as they are not true time zones, not standardized, and not even unique(!).

ZoneId z = ZoneId.of( "America/Montreal" ) ;
ZonedDateTime zdt = instant.atZone( z ) ;  // Same moment, different wall-clock time.

As a shortcut, you can skip the usage of Instant to get a ZonedDateTime.

ZoneId z = ZoneId.of( "America/Montreal" ) ;
ZonedDateTime zdt = ZonedDateTime.now( z ) ;

Convert Local time to UTC // Works Fine Till here

You can adjust from the zoned date-time to UTC by extracting an Instant from a ZonedDateTime.

ZoneId z = ZoneId.of( "America/Montreal" ) ;
ZonedDateTime zdt = ZonedDateTime.now( z ) ;
Instant instant = zdt.toInstant() ;

Reverse the UTC time, back to local time.

As shown above, apply a ZoneId to adjust the same moment into another wall-clock time used by the people of a certain region (a time zone).

Instant instant = Instant.now() ;  // Capture current moment in UTC.

ZoneId zDefault = ZoneId.systemDefault() ;  // The JVM's current default time zone.
ZonedDateTime zdtDefault = instant.atZone( zDefault ) ;

ZoneId zTunis = ZoneId.of( "Africa/Tunis" ) ;  // The JVM's current default time zone.
ZonedDateTime zdtTunis = instant.atZone( zTunis ) ;

ZoneId zAuckland = ZoneId.of( "Pacific/Auckland" ) ;  // The JVM's current default time zone.
ZonedDateTime zdtAuckland = instant.atZone( zAuckland ) ;

Going back to UTC from a zoned date-time, call ZonedDateTime::toInstant. Think of it conceptually as: ZonedDateTime = Instant + ZoneId.

Instant instant = zdtAuckland.toInstant() ;

All of these objects, the Instant and the three ZonedDateTime objects all represent the very same simultaneous moment, the same point in history.

Followed 3 different approaches (listed below) but all the 3 approaches retains the time in UTC only.

Forget about trying to fix code using those awful Date, Calendar, and GregorianCalendar classes. They are a wretched mess of bad design and flaws. You need never touch them again. If you must interface with old code not yet updated to java.time, you can convert back-and-forth via new conversion methods added to the old classes.

About java.time

The java.time framework is built into Java 8 and later. These classes supplant the troublesome old legacy date-time classes such as java.util.Date, Calendar, & SimpleDateFormat.

The Joda-Time project, now in maintenance mode, advises migration to the java.time classes.

To learn more, see the Oracle Tutorial. And search Stack Overflow for many examples and explanations. Specification is JSR 310.

You may exchange java.time objects directly with your database. Use a JDBC driver compliant with JDBC 4.2 or later. No need for strings, no need for java.sql.* classes.

Where to obtain the java.time classes?

• Java SE 8, Java SE 9, Java SE 10, and later
  • Built-in.
  • Part of the standard Java API with a bundled implementation.
  • Java 9 adds some minor features and fixes.
• Java SE 6 and Java SE 7
  • Much of the java.time functionality is back-ported to Java 6 & 7 in ThreeTen-Backport.
• Android
  • Later versions of Android bundle implementations of the java.time classes.
  • For earlier Android (<26), the ThreeTenABP project adapts ThreeTen-Backport (mentioned above). See How to use ThreeTenABP….

The ThreeTen-Extra project extends java.time with additional classes. This project is a proving ground for possible future additions to java.time. You may find some useful classes here such as Interval, YearWeek, YearQuarter, and more.

Can use getElementsByTagName

var x = document.getElementsByTagName("title")[0];

alert(x.innerHTML)

// or

alert(x.textContent)

// or

document.querySelector('title')

Edits as suggested by Paul

You should read this getopts tutorial.

Example with -a switch that requires an argument :

#!/bin/bash

while getopts ":a:" opt; do
  case $opt in
    a)
      echo "-a was triggered, Parameter: $OPTARG" >&2
      ;;
    \?)
      echo "Invalid option: -$OPTARG" >&2
      exit 1
      ;;
    :)
      echo "Option -$OPTARG requires an argument." >&2
      exit 1
      ;;
  esac
done

Like greybot said(getopt != getopts) :

The external command getopt(1) is never safe to use, unless you know it is GNU getopt, you call it in a GNU-specific way, and you ensure that GETOPT_COMPATIBLE is not in the environment. Use getopts (shell builtin) instead, or simply loop over the positional parameters.

Development Time Logging

For development time logging, you can use console.log(). One important thing, if you want to disable logging in production mode, then in Root Js file of app, just assign blank function like this - console.log = {} It will disable whole log publishing throughout app altogether, which actually required in production mode as console.log consumes time.

Run Time Logging

In production mode, it is also required to see logs when real users are using your app in real time. This helps in understanding bugs, usage and unwanted cases. There are many 3rd party paid tools available in the market for this. One of them which I've used is by Logentries

The good thing is that Logentries has got React Native Module as well. So, it will take very less time for you to enable Run time logging with your mobile app.

Because i'm lazy....

import json
a = [1,2,3]
with open('test.txt', 'w') as f:
    f.write(json.dumps(a))

#Now read the file back into a Python list object
with open('test.txt', 'r') as f:
    a = json.loads(f.read())

In the Android Studio IDE, access the "Run Anything bar" by:

CTRL+CTRL +gradle CreateFullJarRelease+ENTER

After that you'll find your artefact in this folder in your project
Build > Intermediates > Full_jar > Release > CreateFullJarRelease > full.jar

OR

Gradle has already a Task for that, in the gradle side-menu, under the other folder.

Then scroll down to createFullJarRelease and click it.

After that you'll find your artefact in this folder in your project

Build > Intermediates > Full_jar > Release > CreateFullJarRelease > full.jar

I had to do something similar. Here is a simplified version of what worked for me. If a touch event is detected, remove the click binding.

$thing.on('touchstart click', function(event){
  if (event.type == "touchstart")
    $(this).off('click');

  //your code here
});

In my case the click event was bound to an <a> element so I had to remove the click binding and rebind a click event which prevented the default action for the <a> element.

$thing.on('touchstart click', function(event){
  if (event.type == "touchstart")
    $(this).off('click').on('click', function(e){ e.preventDefault(); });

  //your code here
});

I created a little HTML file that contains Angular code demonstrating the differences between them:

<!DOCTYPE html>
<html>
  <head>
    <title>Angular</title>
    <script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.3.15/angular.min.js"></script>
  </head>
  <body ng-app="myApp">
    <div ng-controller="myCtrl as VM">
      <a my-dir
        attr1="VM.sayHi('Juan')" <!-- scope: "=" -->
        attr2="VM.sayHi('Juan')" <!-- scope: "@" -->
        attr3="VM.sayHi('Juan')" <!-- scope: "&" -->
      ></a>
    </div>
    <script>
    angular.module("myApp", [])
    .controller("myCtrl", [function(){
      var vm = this;
      vm.sayHi = function(name){
        return ("Hey there, " + name);
      }
    }])
    .directive("myDir", [function(){
      return {
        scope: {
          attr1: "=",
          attr2: "@",
          attr3: "&"
        },
        link: function(scope){
          console.log(scope.attr1);   // =, logs "Hey there, Juan"
          console.log(scope.attr2);   // @, logs "VM.sayHi('Juan')"
          console.log(scope.attr3);   // &, logs "function (a){return h(c,a)}"
          console.log(scope.attr3()); // &, logs "Hey there, Juan"
        }
      }
    }]);
    </script>
  </body>
</html>

This solution worked well for me: http://www.apachefriends.org/f/viewtopic.php?f=17&t=50902&p=196185#p196185

Edit /opt/lampp/etc/extra/httpd-xampp.conf and adding Require all granted line at bottom of block <Directory "/opt/lampp/phpmyadmin"> to have the following code:

<Directory "/opt/lampp/phpmyadmin">
  AllowOverride AuthConfig Limit
  Order allow,deny
  Allow from all
  Require all granted
</Directory>

The vue documentation provides a lot of information on this on how you can deploy to different host providers.

npm run build

You can find this from the package json file. scripts section. It provides scripts for testing and development and building for production.

You can use services such as netlify which will bundle your project by linking up your github repo of the project from their site. It also provides information on how to deploy on other sites such as heroku.

You can find more details on this here

If you were like me, running maven compile deploy from eclipse's maven run configuration, the issue could be related to eclipse's own embedded maven as described in https://bugs.eclipse.org/bugs/show_bug.cgi?id=562847

The workaround is to run mvn compile deploy from CLI such as bash, or to NOT use embedded maven in the eclipse's maven run configuration, and add an external maven (mine is in /usr/share/mvn), and voila, it'll say BUILD SUCCESS.

The problem with not passing over the content-type on a GET message is that sure the content-type is irrelevant because the server side determines the content anyway. The problem that I have encountered is that there are now a lot of places that set up their webservices to be smart enough to pick up the content-type that you pass and return the response in the 'type' that you request. Eg. we are currently messaging with a place that defaults to JSON, however, they have set their webservice up so that if you pass a content-type of xml they will then return xml rather than their JSON default. Which I think going forward is a great idea

I created this prototype to prepend elements to parent element.

Node.prototype.prependChild = function (child: Node) {
    this.insertBefore(child, this.firstChild);
    return this;
};

A "const function", denoted with the keyword const after a function declaration, makes it a compiler error for this class function to change a member variable of the class. However, reading of a class variables is okay inside of the function, but writing inside of this function will generate a compiler error.

Another way of thinking about such "const function" is by viewing a class function as a normal function taking an implicit this pointer. So a method int Foo::Bar(int random_arg) (without the const at the end) results in a function like int Foo_Bar(Foo* this, int random_arg), and a call such as Foo f; f.Bar(4) will internally correspond to something like Foo f; Foo_Bar(&f, 4). Now adding the const at the end (int Foo::Bar(int random_arg) const) can then be understood as a declaration with a const this pointer: int Foo_Bar(const Foo* this, int random_arg). Since the type of this in such case is const, no modifications of member variables are possible.

It is possible to loosen the "const function" restriction of not allowing the function to write to any variable of a class. To allow some of the variables to be writable even when the function is marked as a "const function", these class variables are marked with the keyword mutable. Thus, if a class variable is marked as mutable, and a "const function" writes to this variable then the code will compile cleanly and the variable is possible to change. (C++11)

As usual when dealing with the const keyword, changing the location of the const key word in a C++ statement has entirely different meanings. The above usage of const only applies when adding const to the end of the function declaration after the parenthesis.

const is a highly overused qualifier in C++: the syntax and ordering is often not straightforward in combination with pointers. Some readings about const correctness and the const keyword:

Const correctness

The C++ 'const' Declaration: Why & How

The Syntax as specified by Microsoft for the dropping a column part of an ALTER statement is this

 DROP 
 {
     [ CONSTRAINT ] 
     { 
          constraint_name 
          [ WITH 
           ( <drop_clustered_constraint_option> [ ,...n ] ) 
          ] 
      } [ ,...n ]
      | COLUMN 
      {
          column_name 
      } [ ,...n ]
 } [ ,...n ]

Notice that the [,...n] appears after both the column name and at the end of the whole drop clause. What this means is that there are two ways to delete multiple columns. You can either do this:

ALTER TABLE TableName
    DROP COLUMN Column1, Column2, Column3

or this

ALTER TABLE TableName
    DROP 
        COLUMN Column1,
        COLUMN Column2,
        COLUMN Column3

This second syntax is useful if you want to combine the drop of a column with dropping a constraint:

ALTER TBALE TableName
    DROP
        CONSTRAINT DF_TableName_Column1,
        COLUMN Column1;

When dropping columns SQL Sever does not reclaim the space taken up by the columns dropped. For data types that are stored inline in the rows (int for example) it may even take up space on the new rows added after the alter statement. To get around this you need to create a clustered index on the table or rebuild the clustered index if it already has one. Rebuilding the index can be done with a REBUILD command after modifying the table. But be warned this can be slow on very big tables. For example:

ALTER TABLE Test
    REBUILD;

Sure. Just rename the branch locally, push the new branch, and push a deletion of the old.

The only real issue is that other users of the repository won't have local tracking branches renamed.

You'll need to download SLF4J's jars from the official site as either a zip (v1.7.4) or tar.gz (v1.7.4)

The download contains multiple jars based on how you want to use SLF4J. If you're simply trying to resolve the requirement of some other library (GWT, I assume) and don't really care about using SLF4J correctly, then I would probably pick the slf4j-api-1.7.4.jar since the Simple jar suggested by another answer does not contain, to my knowledge, the specific class you're looking for.

You can use

call script.bat

or just

script.bat

here: C:\Program Files\Java\jre7\bin it is an exe keytool.exe

Another client-side program (Windows only), is NetLimiter - http://www.netlimiter.com

My favorite way is called do the sleep to 50. here i variable need to be used inside echo statements.

for i in $(seq 1 50); do
  echo -ne "$i%\033[0K\r"
  sleep 50
done
echo "ended"

You can solve this issue be adding max-width:

#element {
   width: 100vw;
   height: 100vw;
   max-width: 100%;
}

When you using CSS to make the wrapper full width using the code width: 100vw; then you will notice a horizontal scroll in the page, and that happened because the padding and margin of html and body tags added to the wrapper size, so the solution is to add max-width: 100%

Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:

URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);

URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);

URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);

I think what you're looking for is !!val==false which can be turned to !val (even shorter):

You see:

function checkValue(value) {
    console.log(!!value);
}

checkValue(); // false
checkValue(null); // false
checkValue(undefined); // false
checkValue(false); // false
checkValue(""); // false

checkValue(true); // true
checkValue({}); // true
checkValue("any string"); // true

That works by flipping the value by using the ! operator.

If you flip null once for example like so :

console.log(!null) // that would output --> true

If you flip it twice like so :

console.log(!!null) // that would output --> false

Same with undefined or false.

Your code:

if(val==null || val===false){
  ;
}

would then become:

if(!val) {
  ;
}

That would work for all cases even when there's a string but it's length is zero. Now if you want it to also work for the number 0 (which would become false if it was double flipped) then your if would become:

if(!val && val !== 0) {
  // code runs only when val == null, undefined, false, or empty string ""
}

I suppose rgba() would work here. After all, browser support for both box-shadow and rgba() is roughly the same.

/* 50% black box shadow */
box-shadow: 10px 10px 10px rgba(0, 0, 0, 0.5);

div {_x000D_
    width: 200px;_x000D_
    height: 50px;_x000D_
    line-height: 50px;_x000D_
    text-align: center;_x000D_
    color: white;_x000D_
    background-color: red;_x000D_
    margin: 10px;_x000D_
}_x000D_
_x000D_
div.a {_x000D_
  box-shadow: 10px 10px 10px #000;_x000D_
}_x000D_
_x000D_
div.b {_x000D_
  box-shadow: 10px 10px 10px rgba(0, 0, 0, 0.5);_x000D_
}

<div class="a">100% black shadow</div>_x000D_
<div class="b">50% black shadow</div>

The Scanner class is bases on logic implemented in String next(Pattern) method. The additional API method like nextDouble() or nextFloat(). Provide the pattern inside.

Then class description says:

A simple text scanner which can parse primitive types and strings using regular expressions.

A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace. The resulting tokens may then be converted into values of different types using the various next methods.

From the description it can be sad that someone has forgot about char as it is a primitive type for sure.

But the concept of class is to find patterns, a char has no pattern is just next character. And this logic IMHO caused that nextChar has not been implemented.

If you need to read a filed char by char you can used more efficient class.

You don't want git revert. That undoes a previous commit. You want git checkout to get git's version of the file from master.

git checkout -- filename.txt

In general, when you want to perform a git operation on a single file, use -- filename.

• git-checkout Documentation
• git-revert Documentation

2020 Update

Git introduced a new command git restore in version 2.23.0. Therefore, if you have git version 2.23.0+, you can simply git restore filename.txt - which does the same thing as git checkout -- filename.txt. The docs for this command do note that it is currently experimental.

• git-restore Documentation

I have been able to use a method like this with some success:

WebElement getStaleElemById(String id) {
    try {
        return driver.findElement(By.id(id));
    } catch (StaleElementReferenceException e) {
        System.out.println("Attempting to recover from StaleElementReferenceException ...");
        return getStaleElemById(id);
    }
}

Yes, it just keeps polling the element until it's no longer considered stale (fresh?). Doesn't really get to the root of the problem, but I've found that the WebDriver can be rather picky about throwing this exception -- sometimes I get it, and sometimes I don't. Or it could be that the DOM really is changing.

So I don't quite agree with the answer above that this necessarily indicates a poorly-written test. I've got it on fresh pages which I have not interacted with in any way. I think there is some flakiness in either how the DOM is represented, or in what WebDriver considers to be stale.

If you don't mind the commas and the brackets you can simply use:

System.out.println(Arrays.deepToString(twoDm).replace("], ", "]\n"));

table{
   height:1px;
}

table > td{
   height:100%;
}

table > td > .inner{
   height:100%;
}

Confirmed working on:

• Chrome 60.0.3112.113, 63.0.3239.84
• Firefox 55.0.3, 57.0.1
• Internet Explorer 11

#!/bin/bash
echo "The number of arguments is: $#"
a=${@}
echo "The total length of all arguments is: ${#a}: "
count=0
for var in "$@"
do
    echo "The length of argument '$var' is: ${#var}"
    (( count++ ))
    (( accum += ${#var} ))
done
echo "The counted number of arguments is: $count"
echo "The accumulated length of all arguments is: $accum"

This code does not require permission at all. It is restricted only to Wi-Fi network connectivity state changes (any other network is not taken into account). The receiver is statically published in the AndroidManifest.xml file and does not need to be exported as it will be invoked by the system protected broadcast, NETWORK_STATE_CHANGED_ACTION, at every network connectivity state change.

AndroidManifest:

<receiver
    android:name=".WifiReceiver"
    android:enabled="true"
    android:exported="false">

    <intent-filter>
        <!--protected-broadcast: Special broadcast that only the system can send-->
        <!--Corresponds to: android.net.wifi.WifiManager.NETWORK_STATE_CHANGED_ACTION-->
        <action android:name="android.net.wifi.STATE_CHANGE" />
    </intent-filter>

</receiver>

BroadcastReceiver class:

public class WifiReceiver extends BroadcastReceiver {
@Override
public void onReceive(Context context, Intent intent) {
/*
 Tested (I didn't test with the WPS "Wi-Fi Protected Setup" standard):
 In API15 (ICE_CREAM_SANDWICH) this method is called when the new Wi-Fi network state is:
 DISCONNECTED, OBTAINING_IPADDR, CONNECTED or SCANNING

 In API19 (KITKAT) this method is called when the new Wi-Fi network state is:
 DISCONNECTED (twice), OBTAINING_IPADDR, VERIFYING_POOR_LINK, CAPTIVE_PORTAL_CHECK
 or CONNECTED

 (Those states can be obtained as NetworkInfo.DetailedState objects by calling
 the NetworkInfo object method: "networkInfo.getDetailedState()")
*/
    /*
     * NetworkInfo object associated with the Wi-Fi network.
     * It won't be null when "android.net.wifi.STATE_CHANGE" action intent arrives.
     */
    NetworkInfo networkInfo = intent.getParcelableExtra(WifiManager.EXTRA_NETWORK_INFO);

    if (networkInfo != null && networkInfo.isConnected()) {
        // TODO: Place the work here, like retrieving the access point's SSID

        /*
         * WifiInfo object giving information about the access point we are connected to.
         * It shouldn't be null when the new Wi-Fi network state is CONNECTED, but it got
         * null sometimes when connecting to a "virtualized Wi-Fi router" in API15.
         */
        WifiInfo wifiInfo = intent.getParcelableExtra(WifiManager.EXTRA_WIFI_INFO);
        String ssid = wifiInfo.getSSID();
    }
}
}

Permissions:

None

you can wrap the content of the <tbody> in a scrollable <div> :

html

....
<tbody>
  <tr>
    <td colspan="2">
      <div class="scrollit">
        <table>
          <tr>
            <td>January</td>
            <td>$100</td>
          </tr>
          <tr>
            <td>February</td>
            <td>$80</td>
          </tr>
          <tr>
            <td>January</td>
            <td>$100</td>
          </tr>
          <tr>
            <td>February</td>
            <td>$80</td>
          </tr>
          ...

css

.scrollit {
    overflow:scroll;
    height:100px;
}

see my jsfiddle, forked from yours: http://jsfiddle.net/VTNax/2/

After adding php directory in User Settings,

{
    "php.validate.executablePath": "C:/phpdirectory/php7.1.8/php.exe",
    "php.executablePath": "C:/phpdirectory/php7.1.8/php.exe"
}

If you still have this error, please verify you have installed :

• 64-bit or 32-bit version of php (x64 or x86), depending on your OS;

• some librairies like Visual C++ Redistributable for Visual Studio 2015 : http://www.microsoft.com/en-us/download/details.aspx?id=48145;

To test if you PHP exe is ok, open cmd.exe :

c:/prog/php-7.1.8-Win32-VC14-x64/php.exe --version

If PHP fails, a message will be prompted with the error (missing dll for example).

As mentioned here - https://kite.com/python/answers/how-to-append-a-list-as-a-row-to-a-pandas-dataframe-in-python, you'll need to first convert the list to a series then append the series to dataframe.

df = pd.DataFrame([[1, 2], [3, 4]], columns = ["a", "b"])
to_append = [5, 6]
a_series = pd.Series(to_append, index = df.columns)
df = df.append(a_series, ignore_index=True)

Here's one way:

CString str;
str.Format("%d", 5);

In your case, try _T("%d") or L"%d" rather than "%d"

Its been almost two years since I asked this question. I just came up with css calc() that resolves this issue I had and thought it would be nice to add it in case someone has the same problem. (By the way I ended up using position absolute).

http://jsfiddle.net/S8g4E/955/

Here is the css

#up { height:80px;}
#down {
    height: calc(100% - 80px);//The upper div needs to have a fixed height, 80px in this case.
}

And more information about it here: http://css-tricks.com/a-couple-of-use-cases-for-calc/

Browser support: http://caniuse.com/#feat=calc

A char in C is already a number (the character's ASCII code), no conversion required.

If you want to convert a digit to the corresponding character, you can simply add '0':

c = i +'0';

The '0' is a character in the ASCll table.

There's a ES6 shorthand import, you can reference. More readable and easy to type.

import React, { Component } from 'react';
import { arrayOf, shape, number } from 'prop-types';

class ExampleComponent extends Component {
  static propTypes = {
    annotationRanges: arrayOf(shape({
      start: number,
      end: number,
    })).isRequired,
  }

  static defaultProps = {
     annotationRanges: [],
  }
}

To resolve this issue you have to do two things :

1. You will need to set a registry entry on the target computer so that the driver can maintain a connection to the instance of Internet Explorer it creates.

2. Change few settings of Internet Explorer browser on that machine (where you desire to run automation).

1 . Setting Registry Key / Entry :

• To set registry key or entry, you need to open "Registry Editor".

• To open "Registry Editor" press windows button key + r alphabet key which will open "Run Window" and then type "regedit" and press enter.

• Or Press Windows button key and enter "regedit" at start menu and press enter. Now depending upon your OS type whether 32/64 bit follow the corresponding steps.

Windows 32 bit : go to this location - "HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Internet Explorer\Main\FeatureControl" and check for "FEATURE_BFCACHE" key.

Windows 64 bit : go to this location - HKEY_LOCAL_MACHINE\SOFTWARE\Wow6432Node\Microsoft\Internet Explorer\Main\FeatureControl and check for "FEATURE_BFCACHE" key. Please note that the FEATURE_BFCACHE subkey may or may not be present, and should be created if it is not present.

Important: Inside this key, create a DWORD value named iexplore.exe with the value of 0.

2 . Change Settings of Internet Explorer Browser :

• Click on setting button and select "Internet options".

• On "Internet options" window go to "Security" tab

• Now select "Internet" option and unchecked the "Enable Protected Mode" check box and change the "Security level" to low.

• Now select "Local Intranet" Option and change the "Security level" to low.

• Now select "Trusted Sites" Option and change the "Security level" to low.

• Now click on "Apply" button , a warning pop up may appear click on "OK" button for warning and then on "OK" button on Internet Options window.

• After this restart the browser.

This worked fine for me:

else if(requestCode == GALLERY_ACTIVITY_NEW && resultCode == Activity.RESULT_OK)
{
    Uri uri = data.getData();
    Log.i(TAG, "old uri =  " + uri);
    dumpImageMetaData(uri);

    try {
        ParcelFileDescriptor parcelFileDescriptor =
                getContentResolver().openFileDescriptor(uri, "r");
        FileDescriptor fileDescriptor = parcelFileDescriptor.getFileDescriptor();
        Log.i(TAG, "File descriptor " + fileDescriptor.toString());

        final BitmapFactory.Options options = new BitmapFactory.Options();
        options.inJustDecodeBounds = true;
        BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);

        options.inSampleSize =
           BitmapHelper.calculateInSampleSize(options,
                                              User.PICTURE_MAX_WIDTH_IN_PIXELS,
                                              User.PICTURE_MAX_HEIGHT_IN_PIXELS);
        options.inJustDecodeBounds = false;

        Bitmap bitmap = BitmapFactory.decodeFileDescriptor(fileDescriptor, null, options);
        imageViewPic.setImageBitmap(bitmap);

        ByteArrayOutputStream stream = new ByteArrayOutputStream();
        bitmap.compress(Bitmap.CompressFormat.JPEG, 100, stream);
        // get byte array here
        byte[] picData = stream.toByteArray();
        ParseFile picFile = new ParseFile(picData);
        user.setProfilePicture(picFile);
    }
    catch(FileNotFoundException exc)
    {
        Log.i(TAG, "File not found: " + exc.toString());
    }
}

Well, I have read this post and seems it is not good enough to fix this issue.

Spring based application on the tomcat 8 will not work.

Here is the solution

Step 1 ? Add the following dependency in your pom.xml

<!-- JSTL Support -->

     <dependency>
    <groupId>javax.servlet</groupId>
    <artifactId>javax.servlet-api</artifactId>
    <version>3.1.0</version>
</dependency>
<dependency>
    <groupId>javax.servlet.jsp</groupId>
    <artifactId>javax.servlet.jsp-api</artifactId>
    <version>2.3.1</version>
</dependency>
<dependency>
    <groupId>javax.servlet</groupId>
    <artifactId>jstl</artifactId>
    <version>1.2</version>
</dependency>

Step 2 ? Add All following two statement in your JSP page, ensure that you are using isELIgnored="false" attribute in <%@ page %> tag

<%@ page language="java" contentType="text/html; charset=ISO-8859-1" isELIgnored="false" pageEncoding="ISO-8859-1"%>
<%@ taglib prefix="c" uri="http://java.sun.com/jsp/jstl/core"%>

Step 3 ? Remove all the other configuration that you have done till now in web.xml or anywhere else :)

Step 4 ? Clean the Tomcat and restart Tomcat.

Side Note ? Actually, JSTL will work with only 3.x Servlet specifications on Tomcat 8.

[http://www.kriblog.com/j2ee/jsp/jstl-with-spring-4-on-tomcat-8.html]

Never use escape(). It's nothing to do with HTML-encoding. It's more like URL-encoding, but it's not even properly that. It's a bizarre non-standard encoding available only in JavaScript.

If you want an HTML encoder, you'll have to write it yourself as JavaScript doesn't give you one. For example:

function encodeHTML(s) {
    return s.replace(/&/g, '&amp;').replace(/</g, '&lt;').replace(/"/g, '&quot;');
}

However whilst this is enough to put your user_id in places like the input value, it's not enough for id because IDs can only use a limited selection of characters. (And % isn't among them, so escape() or even encodeURIComponent() is no good.)

You could invent your own encoding scheme to put any characters in an ID, for example:

function encodeID(s) {
    if (s==='') return '_';
    return s.replace(/[^a-zA-Z0-9.-]/g, function(match) {
        return '_'+match[0].charCodeAt(0).toString(16)+'_';
    });
}

But you've still got a problem if the same user_id occurs twice. And to be honest, the whole thing with throwing around HTML strings is usually a bad idea. Use DOM methods instead, and retain JavaScript references to each element, so you don't have to keep calling getElementById, or worrying about how arbitrary strings are inserted into IDs.

eg.:

function addChut(user_id) {
    var log= document.createElement('div');
    log.className= 'log';
    var textarea= document.createElement('textarea');
    var input= document.createElement('input');
    input.value= user_id;
    input.readonly= True;
    var button= document.createElement('input');
    button.type= 'button';
    button.value= 'Message';

    var chut= document.createElement('div');
    chut.className= 'chut';
    chut.appendChild(log);
    chut.appendChild(textarea);
    chut.appendChild(input);
    chut.appendChild(button);
    document.getElementById('chuts').appendChild(chut);

    button.onclick= function() {
        alert('Send '+textarea.value+' to '+user_id);
    };

    return chut;
}

You could also use a convenience function or JS framework to cut down on the lengthiness of the create-set-appends calls there.

ETA:

I'm using jQuery at the moment as a framework

OK, then consider the jQuery 1.4 creation shortcuts, eg.:

var log= $('<div>', {className: 'log'});
var input= $('<input>', {readOnly: true, val: user_id});
...

The problem I have right now is that I use JSONP to add elements and events to a page, and so I can not know whether the elements already exist or not before showing a message.

You can keep a lookup of user_id to element nodes (or wrapper objects) in JavaScript, to save putting that information in the DOM itself, where the characters that can go in an id are restricted.

var chut_lookup= {};
...

function getChut(user_id) {
    var key= '_map_'+user_id;
    if (key in chut_lookup)
        return chut_lookup[key];
    return chut_lookup[key]= addChut(user_id);
}

(The _map_ prefix is because JavaScript objects don't quite work as a mapping of arbitrary strings. The empty string and, in IE, some Object member names, confuse it.)

It sounds like you're looking for applicationHost.config, which is located in C:\Windows\System32\inetsrv\config.

Yes, it's an XML file, and yes, editing the file by hand will affect the IIS config after a restart. You can think of IIS Manager as a GUI front-end for editing applicationHost.config and web.config.

Use this to convert String to Blob. Where connection is the connection to db object.

    String strContent = s;
    byte[] byteConent = strContent.getBytes();
    Blob blob = connection.createBlob();//Where connection is the connection to db object. 
    blob.setBytes(1, byteContent);

If you save color code in the colors.xml which is under the values folder,then you should call the following:

root.setBackgroundColor(getResources().getColor(R.color.name));

name means you declare in the <color/> tag.

Maybe (from git commit man page):

git commit --no-verify

-n  
--no-verify

This option bypasses the pre-commit and commit-msg hooks. See also githooks(5).

As commented by Blaise, -n can have a different role for certain commands.
For instance, git push -n is actually a dry-run push.
Only git push --no-verify would skip the hook.

Note: Git 2.14.x/2.15 improves the --no-verify behavior:

See commit 680ee55 (14 Aug 2017) by Kevin Willford (``).
^{(Merged by Junio C Hamano -- gitster -- in commit c3e034f, 23 Aug 2017)}

commit: skip discarding the index if there is no pre-commit hook

"git commit" used to discard the index and re-read from the filesystem just in case the pre-commit hook has updated it in the middle; this has been optimized out when we know we do not run the pre-commit hook.

Davi Lima points out in the comments the git cherry-pick does not support --no-verify.
So if a cherry-pick triggers a pre-commit hook, you might, as in this blog post, have to comment/disable somehow that hook in order for your git cherry-pick to proceed.
The same process would be necessary in case of a git rebase --continue, after a merge conflict resolution.

    $routeProvider
 .when('/main' , {templateUrl: 'partials/main.html',  controller: MainController})
 .when('/login', {templateUrl: 'partials/login.html', controller: LoginController}).
 .when('/login', {templateUrl: 'partials/index.html', controller: IndexController})
 .otherwise({redirectTo: '/index'});

This guide worked:

# hostnamectl
  Operating System: CentOS Linux 7 (Core)
# git --version
git version 1.8.3.1
# sudo yum remove git*
# sudo yum -y install https://packages.endpoint.com/rhel/7/os/x86_64/endpoint-repo-1.7-1.x86_64.rpm
# sudo yum install git
# git --version
git version 2.24.1

For TSQL I like to encapsulate case statements in an outer apply. This prevents me from having to have the case statement written twice, allows reference to the case statement by alias in future joins and avoids the need for positional references.

select oa.day, 
model.name, 
attempt.type, 
oa.result
COUNT(*) MyCount 
FROM attempt attempt, prod_hw_id prod_hw_id, model model
WHERE time >= '2013-11-06 00:00:00'  
AND time < '2013-11-07 00:00:00'
AND attempt.hard_id = prod_hw_id.hard_id
AND prod_hw_id.model_id = model.model_id
OUTER APPLY (
    SELECT CURRENT_DATE-1 AS day,
     CASE WHEN attempt.result = 0 THEN 0 ELSE 1 END result
    ) oa    
group by oa.day, 
model.name, 
attempt.type, 
oa.result
order by model.name, attempt.type, oa.result;

If the replacement character can be '?' instead of a space, then I'd suggest result = text.encode('ascii', 'replace').decode():

"""Test the performance of different non-ASCII replacement methods."""


import re
from timeit import timeit


# 10_000 is typical in the project that I'm working on and most of the text
# is going to be non-ASCII.
text = 'Æ' * 10_000


print(timeit(
    """
result = ''.join([c if ord(c) < 128 else '?' for c in text])
    """,
    number=1000,
    globals=globals(),
))

print(timeit(
    """
result = text.encode('ascii', 'replace').decode()
    """,
    number=1000,
    globals=globals(),
))

Results:

0.7208260721400134
0.009975979187503592

Here is example of getting substring from 14 character to end of string. You can modify it to fit your needs

string text = "Retrieves a substring from this instance. The substring starts at a specified character position.";
//get substring where 14 is start index
string substring = text.Substring(14);

Wrap it in a scroll viewer:

<ScrollViewer>
    <TextBlock />
</ScrollViewer>

NOTE this answer applies to a TextBlock (a read-only text element) as asked for in the original question.

If you want to show scroll bars in a TextBox (an editable text element) then use the ScrollViewer attached properties:

<TextBox ScrollViewer.HorizontalScrollBarVisibility="Disabled"
         ScrollViewer.VerticalScrollBarVisibility="Auto" />

Valid values for these two properties are Disabled, Auto, Hidden and Visible.

The find one is good but I think you can use anotherway, especially if you need to now how many seconds is the file old

date -d "now - $( stat -c "%Y" $filename ) seconds" +%s

using GNU date

Here's a solution with flexbox for images with variable width and height:

.container {
  display: flex;
  flex-wrap: no-wrap;
  overflow-x: auto;
  margin: 20px;
}
img {
  flex: 0 0 auto;
  width: auto;
  height: 100px;
  max-width: 100%;
  margin-right: 10px;
}

Example: JsFiddle

Create one complex object to combine Content and Config in it as others mentioned, use dynamic and just do a .ToObject(); as:

[HttpPost]
public void StartProcessiong([FromBody] dynamic obj)
{
   var complexObj= obj.ToObject<ComplexObj>();
   var content = complexObj.Content;
   var config = complexObj.Config;
}

Inside any controller action or view, you can invoke the console by calling the console method.

For example, in a controller:

class PostsController < ApplicationController
  def new
    console
    @post = Post.new
  end
end

Or in a view:

<% console %>

<h2>New Post</h2>

This will render a console inside your view. You don't need to care about the location of the console call; it won't be rendered on the spot of its invocation but next to your HTML content.

See: http://guides.rubyonrails.org/debugging_rails_applications.html

Use MyISAM for very unimportant data or if you really need those minimal performance advantages. The read performance is not better in every case for MyISAM.

I would personally never use MyISAM at all anymore. Choose InnoDB and throw a bit more hardware if you need more performance. Another idea is to look at database systems with more features like PostgreSQL if applicable.

EDIT: For the read-performance, this link shows that innoDB often is actually not slower than MyISAM: https://www.percona.com/blog/2007/01/08/innodb-vs-myisam-vs-falcon-benchmarks-part-1/

The best solution is to just do webView.loadUrl( "javascript:window.location.reload( true )" );. This should work on all versions and doesn't introduce new history entries.

if you are on windows then you can do a right click from the folder where you want to use git bash and select "GIT BASH HERE".

Swift is statically typed so the performSelector: methods are to fall by the wayside.

Instead, use GCD to dispatch a suitable block to the relevant queue — in this case it'll presumably be the main queue since it looks like you're doing UIKit work.

EDIT: the relevant performSelector: is also notably missing from the Swift version of the NSRunLoop documentation ("1 Objective-C symbol hidden") so you can't jump straight in with that. With that and its absence from the Swiftified NSObject I'd argue it's pretty clear what Apple is thinking here.

You may want to try \r\n for carriage return / line feed

Replace this:

NSDate* now = [NSDate date];
int hour = 23 - [[now dateWithCalendarFormat:nil timeZone:nil] hourOfDay];
int min = 59 - [[now dateWithCalendarFormat:nil timeZone:nil] minuteOfHour];
int sec = 59 - [[now dateWithCalendarFormat:nil timeZone:nil] secondOfMinute];
countdownLabel.text = [NSString stringWithFormat:@"%02d:%02d:%02d", hour, min,sec];

With this:

NSDate* now = [NSDate date];
NSCalendar *gregorian = [[NSCalendar alloc] initWithCalendarIdentifier:NSGregorianCalendar];
NSDateComponents *dateComponents = [gregorian components:(NSHourCalendarUnit  | NSMinuteCalendarUnit | NSSecondCalendarUnit) fromDate:now];
NSInteger hour = [dateComponents hour];
NSInteger minute = [dateComponents minute];
NSInteger second = [dateComponents second];
[gregorian release];
countdownLabel.text = [NSString stringWithFormat:@"%02d:%02d:%02d", hour, minute, second];

To understand why it does not return the list:

sort() doesn't return any value while the sort() method just sorts the elements of a given list in a specific order - ascending or descending without returning any value.

So problem is with answer = newList.sort() where answer is none.

Instead you can just do return newList.sort().

The syntax of the sort() method is:

list.sort(key=..., reverse=...)

Alternatively, you can also use Python's in-built function sorted() for the same purpose.

sorted(list, key=..., reverse=...)

Note: The simplest difference between sort() and sorted() is: sort() doesn't return any value while, sorted() returns an iterable list.

So in your case answer = sorted(newList).

If you're connecting to a named instance and UDP is not available when connecting to it, then you may need to specify the protocol as well.

Example: tcp:192.168.1.21\SQL2K5,1443

Usually, I work with DATE columns, not the larger but more precise TIMESTAMP used by some answers.

The following will return the current UTC date as just that -- a DATE.

CAST(sys_extract_utc(SYSTIMESTAMP) AS DATE)

I often store dates like this, usually with the field name ending in _UTC to make it clear for the developer. This allows me to avoid the complexity of time zones until last-minute conversion by the user's client. Oracle can store time zone detail with some data types, but those types require more table space than DATE, and knowledge of the original time zone is not always required.

In command mode (press Esc if you are not sure) you can use:

• gg,
• :1,
• 1G,
• or 1gg.

Another alternative is if you already installed minGW and added the bin folder the to Path environment variable, you can use "mingw32-make" instead of "make".

You can also create a symlink from "make" to "mingw32-make", or copying and changing the name of the file. I would not recommend the options before, they will work until you do changes on the minGW.

var formatter = new Intl.NumberFormat("ru", {
  style: "currency",
  currency: "GBP"
});

alert( formatter.format(1234.5) ); // 1 234,5 £

https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/NumberFormat

Hadley Wickham

dplyr

packages is always a saver in case of data wrangling. To add the desired division as a third variable I would use mutate()

d <- mutate(d, new = min / count2.freq)

UPDATE: READONLY doesn't work on checkboxes

You could use disabled="disabled" but at this point checkbox's value will not appear into POST values. One of the strategy is to add an hidden field holding checkbox's value within the same form and read value back from that field

Simply change disabled to readonly

First of all you need to download Mysql connector which is compatible with your pentaho version.after that paste it to data-integration/lib folder and restart your pentaho. check this https://help.pentaho.com/Documentation/8.1/Setup/JDBC_Drivers_Reference#MY_SQL

Another option would be to set the webroot path to the angular cli dist folder. In your Program.cs when configuring the WebHostBuilder just say

.UseWebRoot(Directory.GetCurrentDirectory() + "\\Frontend\\dist")

or whatever the path to your dist dir is.

Almost all the answers here rely on either (1) require() or (2) installed.packages() to check if a given package is already installed or not.

I'm adding an answer because these are unsatisfactory for a lightweight approach to answering this question.

• require has the side effect of loading the package's namespace, which may not always be desirable
• installed.packages is a bazooka to light a candle -- it will check the universe of installed packages first, then we check if our one (or few) package(s) are "in stock" at this library. No need to build a haystack just to find a needle.

This answer was also inspired by @ArtemKlevtsov's great answer in a similar spirit on a duplicated version of this question. He noted that system.file(package=x) can have the desired affect of returning '' if the package isn't installed, and something with nchar > 1 otherwise.

If we look under the hood of how system.file accomplishes this, we can see it uses a different base function, find.package, which we could use directly:

# a package that exists
find.package('data.table', quiet=TRUE)
# [1] "/Library/Frameworks/R.framework/Versions/4.0/Resources/library/data.table"

# a package that does not
find.package('InstantaneousWorldPeace', quiet=TRUE)
# character(0)

We can also look under the hood at find.package to see how it works, but this is mainly an instructive exercise -- the only ways to slim down the function that I see would be to skip some robustness checks. But the basic idea is: look in .libPaths() -- any installed package pkg will have a DESCRIPTION file at file.path(.libPaths(), pkg), so a quick-and-dirty check is file.exists(file.path(.libPaths(), pkg, 'DESCRIPTION').

Use the std::vector class from the standard library.

std::unique only removes duplicate elements if they're neighbours: you have to sort the vector first before it will work as you intend.

std::unique is defined to be stable, so the vector will still be sorted after running unique on it.

Expanding a bit on the answers from @Guru and @Ronnis, you can hide the sequence and make it look more like an auto-increment using a trigger, and have a procedure that does the insert for you and returns the generated ID as an out parameter.

create table batch(batchid number,
    batchname varchar2(30),
    batchtype char(1),
    source char(1),
    intarea number)
/

create sequence batch_seq start with 1
/

create trigger batch_bi
before insert on batch
for each row
begin
    select batch_seq.nextval into :new.batchid from dual;
end;
/

create procedure insert_batch(v_batchname batch.batchname%TYPE,
    v_batchtype batch.batchtype%TYPE,
    v_source batch.source%TYPE,
    v_intarea batch.intarea%TYPE,
    v_batchid out batch.batchid%TYPE)
as
begin
    insert into batch(batchname, batchtype, source, intarea)
    values(v_batchname, v_batchtype, v_source, v_intarea)
    returning batchid into v_batchid;
end;
/

You can then call the procedure instead of doing a plain insert, e.g. from an anoymous block:

declare
    l_batchid batch.batchid%TYPE;
begin
    insert_batch(v_batchname => 'Batch 1',
        v_batchtype => 'A',
        v_source => 'Z',
        v_intarea => 1,
        v_batchid => l_batchid);
    dbms_output.put_line('Generated id: ' || l_batchid);

    insert_batch(v_batchname => 'Batch 99',
        v_batchtype => 'B',
        v_source => 'Y',
        v_intarea => 9,
        v_batchid => l_batchid);
    dbms_output.put_line('Generated id: ' || l_batchid);
end;
/

Generated id: 1
Generated id: 2

You can make the call without an explicit anonymous block, e.g. from SQL*Plus:

variable l_batchid number;
exec insert_batch('Batch 21', 'C', 'X', 7, :l_batchid);

... and use the bind variable :l_batchid to refer to the generated value afterwards:

print l_batchid;
insert into some_table values(:l_batch_id, ...);

In Javascript, deep-copy techniques depend on the elements in an array. Let's start there.

Three types of elements

Elements can be: literal values, literal structures, or prototypes.

// Literal values (type1)
const booleanLiteral = true;
const numberLiteral = 1;
const stringLiteral = 'true';

// Literal structures (type2)
const arrayLiteral = [];
const objectLiteral = {};

// Prototypes (type3)
const booleanPrototype = new Bool(true);
const numberPrototype = new Number(1);
const stringPrototype = new String('true');
const arrayPrototype = new Array();
const objectPrototype = new Object(); // or `new function () {}`

From these elements we can create three types of arrays.

// 1) Array of literal-values (boolean, number, string) 
const type1 = [true, 1, "true"];

// 2) Array of literal-structures (array, object)
const type2 = [[], {}];

// 3) Array of prototype-objects (function)
const type3 = [function () {}, function () {}];

Deep copy techniques depend on the three array types

Based on the types of elements in the array, we can use various techniques to deep copy.

• Array of literal-values (type1)
  The [...myArray], myArray.splice(0), myArray.slice(), and myArray.concat() techniques can be used to deep copy arrays with literal values (boolean, number, and string) only; where the Spread operator [...myArray] has the best performance (https://measurethat.net/Benchmarks/Show/4281/0/spread-array-performance-vs-slice-splice-concat).

• Array of literal-values (type1) and literal-structures (type2)
  The JSON.parse(JSON.stringify(myArray)) technique can be used to deep copy literal values (boolean, number, string) and literal structures (array, object), but not prototype objects.

• All arrays (type1, type2, type3)
  The jQuery $.extend(myArray) technique can be used to deep-copy all array-types. Libraries like Underscore and Lo-dash offer similar deep-copy functions to jQuery $.extend(), yet have lower performance. More surprisingly, $.extend() has higher performance than the JSON.parse(JSON.stringify(myArray)) technique http://jsperf.com/js-deep-copy/15.
  And for those developers that shy away from third-party libraries (like jQuery), you can use the following custom function; which has higher performance than $.extend, and deep-copies all arrays.

function copy(aObject) {
  if (!aObject) {
    return aObject;
  }

  let v;
  let bObject = Array.isArray(aObject) ? [] : {};
  for (const k in aObject) {
    v = aObject[k];
    bObject[k] = (typeof v === "object") ? copy(v) : v;
  }

  return bObject;
}

So to answer the question...

Question

var arr1 = ['a','b','c'];
var arr2 = arr1;

I realized that arr2 refers to the same array as arr1, rather than a new, independent array. How can I copy the array to get two independent arrays?

Answer

Because arr1 is an array of literal values (boolean, number, or string), you can use any deep copy technique discussed above, where the spread operator ... has the highest performance.

// Highest performance for deep copying literal values
arr2 = [...arr1];

// Any of these techniques will deep copy literal values as well,
//   but with lower performance.
arr2 = arr1.slice();
arr2 = arr1.splice(0);
arr2 = arr1.concat();
arr2 = JSON.parse(JSON.stringify(arr1));
arr2 = $.extend(true, [], arr1); // jQuery.js needed
arr2 = _.extend(arr1); // Underscore.js needed
arr2 = _.cloneDeep(arr1); // Lo-dash.js needed
arr2 = copy(arr1); // Custom-function needed - as provided above

No, you cannot do that. That's just the way Python has its syntax. Once you exit a try-block because of an exception, there is no way back in.

What about a for-loop though?

funcs = do_smth1, do_smth2

for func in funcs:
    try:
        func()
    except Exception:
        pass  # or you could use 'continue'

Note however that it is considered a bad practice to have a bare except. You should catch for a specific exception instead. I captured for Exception because that's as good as I can do without knowing what exceptions the methods might throw.

For me, it's very simple:

     private boolean mLoading = false;

     mList.setOnScrollListener(new RecyclerView.OnScrollListener() {

        @Override
        public void onScrolled(RecyclerView recyclerView, int dx, int dy) {
            super.onScrolled(recyclerView, dx, dy);

            int totalItem = mLinearLayoutManager.getItemCount();
            int lastVisibleItem = mLinearLayoutManager.findLastVisibleItemPosition();

            if (!mLoading && lastVisibleItem == totalItem - 1) {
                mLoading = true;
                // Scrolled to bottom. Do something here.
                mLoading = false;
            }
        }
    });

Be careful with asynchronous jobs: mLoading must be changed at the end of the asynchronous jobs. Hope it will be helpful!

I remote into the computer using Enter-PSsession pcName then I type Get-PSDrive

That will list all drives and space used and remaining. If you need to see all the info formated, pipe it to FL like this: Get-PSdrive | FL *

sounds like you want something like:

select PropertyID, SUM(Amount)
from MyTable
Where EndDate is null
Group by PropertyID

This is how I make forms full screen.

private void button1_Click(object sender, EventArgs e)
{
    int minx, miny, maxx, maxy;
    inx = miny = int.MaxValue;
    maxx = maxy = int.MinValue;

    foreach (Screen screen in Screen.AllScreens)
    {
        var bounds = screen.Bounds;
        minx = Math.Min(minx, bounds.X);
        miny = Math.Min(miny, bounds.Y);
        maxx = Math.Max(maxx, bounds.Right);
        maxy = Math.Max(maxy, bounds.Bottom);
    }

    Form3 fs = new Form3();
    fs.Activate();
    Rectangle tempRect = new Rectangle(1, 0, maxx, maxy);
    this.DesktopBounds = tempRect;
}

I'm guessing that your class is in a different namespace. The canonical way to solve this would be to use the resources class and a strongly typed resource:

ProjectNamespace.Properties.Resources.file

Use the IDE's resource manager to add resources.

You can call tail +[line number] [file] and pipe it to grep -n which shows the line number:

tail +[line number] [file] | grep -n /regex/

The only problem with this method is the line numbers reported by grep -n will be [line number] - 1 less than the actual line number in [file].

C style, simple ring buffer for integers. First use init than use put and get. If buffer does not contain any data it returns "0" zero.

//=====================================
// ring buffer address based
//=====================================
#define cRingBufCount   512
int     sRingBuf[cRingBufCount];    // Ring Buffer
int     sRingBufPut;                // Input index address
int     sRingBufGet;                // Output index address
Bool    sRingOverWrite;

void    GetRingBufCount(void)
{
int     r;
`       r= sRingBufPut - sRingBufGet;
        if ( r < cRingBufCount ) r+= cRingBufCount;
        return r; 
}

void    InitRingBuffer(void)
{
        sRingBufPut= 0;
        sRingBufGet= 0;
}       

void    PutRingBuffer(int d)
{
        sRingBuffer[sRingBufPut]= d;
        if (sRingBufPut==sRingBufGet)// both address are like ziro
        {
            sRingBufPut= IncRingBufferPointer(sRingBufPut);
            sRingBufGet= IncRingBufferPointer(sRingBufGet);
        }
        else //Put over write a data
        {
            sRingBufPut= IncRingBufferPointer(sRingBufPut);
            if (sRingBufPut==sRingBufGet)
            {
                sRingOverWrite= Ture;
                sRingBufGet= IncRingBufferPointer(sRingBufGet);
            }
        }
}

int     GetRingBuffer(void)
{
int     r;
        if (sRingBufGet==sRingBufPut) return 0;
        r= sRingBuf[sRingBufGet];
        sRingBufGet= IncRingBufferPointer(sRingBufGet);
        sRingOverWrite=False;
        return r;
}

int     IncRingBufferPointer(int a)
{
        a+= 1;
        if (a>= cRingBufCount) a= 0;
        return a;
}

You should have written what version of Oracle you use. You most likely use something else than Oracle 11g, that's why you can't shrink a temp tablespace.

Alternatives:

1) alter database tempfile '[your_file]' resize 128M; which will probably fail
2) Drop and recreate the tablespace. If the temporary tablespace you want to shrink is your default temporary tablespace, you may have to first create a new temporary tablespace, set it as the default temporary tablespace then drop your old default temporary tablespace and recreate it. Afterwards drop the second temporary table created. 3) For Oracle 9i and higher you could just drop the tempfile(s) and add a new one(s)

Everything is described here in great detail.

See this link: http://databaseguide.blogspot.com/2008/06/resizing-temporary-tablespace.html
It was already linked, but maybe you missed it, so here it is again.

The input date value format needs the date specified as per http://tools.ietf.org/html/rfc3339#section-5.6 full-date.

So I've ended up doing:

<input type="date" id="last-start-date" value="@string.Format("{0:yyyy-MM-dd}", Model.LastStartDate)" />

I did try doing it "properly" using:

[DataType(DataType.Date)]
[DisplayFormat(ApplyFormatInEditMode = true, DataFormatString = "{0:yyyy-MM-dd}")]
public DateTime LastStartDate
{
    get { return lastStartDate; }
    set { lastStartDate = value; }
}

with

@Html.TextBoxFor(model => model.LastStartDate, 
    new { type = "date" })

Unfortunately that always seemed to set the value attribute of the input to a standard date time so I've ended up applying the formatting directly as above.

Edit:

According to Jorn if you use

@Html.EditorFor(model => model.LastStartDate)

instead of TextBoxFor it all works fine.

Description and examples can be found in IEEE Std 1800-2017 § 11.5.1 "Vector bit-select and part-select addressing". First IEEE appearance is IEEE 1364-2001 (Verilog) § 4.2.1 "Vector bit-select and part-select addressing". Here is an direct example from the LRM:

logic [31: 0] a_vect;
logic [0 :31] b_vect;
logic [63: 0] dword;
integer sel;
a_vect[ 0 +: 8] // == a_vect[ 7 : 0]
a_vect[15 -: 8] // == a_vect[15 : 8]
b_vect[ 0 +: 8] // == b_vect[0 : 7]
b_vect[15 -: 8] // == b_vect[8 :15]
dword[8*sel +: 8] // variable part-select with fixed width

If sel is 0 then dword[8*(0) +: 8] == dword[7:0]
If sel is 7 then dword[8*(7) +: 8] == dword[63:56]

The value to the left always the starting index. The number to the right is the width and must be a positive constant. the + and - indicates to select the bits of a higher or lower index value then the starting index.

Assuming address is in little endian ([msb:lsb]) format, then if(address[2*pointer+:2]) is the equivalent of if({address[2*pointer+1],address[2*pointer]})

df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
df['period'] = df[['Year', 'quarter']].apply(lambda x: ''.join(x), axis=1)

Yields this dataframe

   Year quarter  period
0  2014      q1  2014q1
1  2015      q2  2015q2

This method generalizes to an arbitrary number of string columns by replacing df[['Year', 'quarter']] with any column slice of your dataframe, e.g. df.iloc[:,0:2].apply(lambda x: ''.join(x), axis=1).

You can check more information about apply() method here

It appears that tkFileDialog.askdirectory should work. documentation

Because C will promote floats to doubles for functions that take variable arguments. Pointers aren't promoted to anything, so you should be using %lf, %lg or %le (or %la in C99) to read in doubles.

There are two "quick" methods:

x = length_of_your_list
a = [None]*x
# or
a = [None for _ in xrange(x)]

It appears that [None]*x is faster:

>>> from timeit import timeit
>>> timeit("[None]*100",number=10000)
0.023542165756225586
>>> timeit("[None for _ in xrange(100)]",number=10000)
0.07616496086120605

But if you are ok with a range (e.g. [0,1,2,3,...,x-1]), then range(x) might be fastest:

>>> timeit("range(100)",number=10000)
0.012513160705566406

java.io.File class contains four static separator variables. For better understanding, Let's understand with the help of some code

1. separator: Platform dependent default name-separator character as String. For windows, it’s ‘\’ and for unix it’s ‘/’
2. separatorChar: Same as separator but it’s char
3. pathSeparator: Platform dependent variable for path-separator. For example PATH or CLASSPATH variable list of paths separated by ‘:’ in Unix systems and ‘;’ in Windows system
4. pathSeparatorChar: Same as pathSeparator but it’s char

Note that all of these are final variables and system dependent.

Here is the java program to print these separator variables. FileSeparator.java

import java.io.File;

public class FileSeparator {

    public static void main(String[] args) {
        System.out.println("File.separator = "+File.separator);
        System.out.println("File.separatorChar = "+File.separatorChar);
        System.out.println("File.pathSeparator = "+File.pathSeparator);
        System.out.println("File.pathSeparatorChar = "+File.pathSeparatorChar);
    }

}

Output of above program on Unix system:

File.separator = /
File.separatorChar = /
File.pathSeparator = :
File.pathSeparatorChar = :

Output of the program on Windows system:

File.separator = \
File.separatorChar = \
File.pathSeparator = ;
File.pathSeparatorChar = ;

To make our program platform independent, we should always use these separators to create file path or read any system variables like PATH, CLASSPATH.

Here is the code snippet showing how to use separators correctly.

//no platform independence, good for Unix systems
File fileUnsafe = new File("tmp/abc.txt");
//platform independent and safe to use across Unix and Windows
File fileSafe = new File("tmp"+File.separator+"abc.txt");

String interpolation is going to be included with Python 3.6 as specified in PEP 498. You will be able to do this:

name = 'Spongebob Squarepants'
print(f'Who lives in a Pineapple under the sea? \n{name}')

Note that I hate Spongebob, so writing this was slightly painful. :)

First export the schema metadata:

expdp dumpfile=filename logfile=logname directory=dir_name schemas=schema_name

and then import by using the sqlfile option (it will not import data it will just write the schema DDL to that file)

impdp dumpfile=filename logfile=logname directory=dir_name sqlfile=ddl.sql

I did this to set my root password in initial set up of MySQL in OSx. Open a terminal.

sudo sh -c 'echo /usr/local/mysql/bin > /etc/paths.d/mysql'

Close the terminal and open a new terminal. And followings are worked in Linux, to set root password.

sudo /usr/local/mysql/support-files/mysql.server stop
sudo mysqld_safe --skip-grant-tables

(sudo mysqld_safe --skip-grant-tables : This did not work for me in first time. But second try, out was success.)

Then login to MySQL

mysql -u root

FLUSH PRIVILEGES;

Now change the password:

ALTER USER 'root'@'localhost' IDENTIFIED BY 'newpassword';

Restart MySQL:

sudo /usr/local/mysql/support-files/mysql.server stop
sudo /usr/local/mysql/support-files/mysql.server start

in Kotlin, putting Context/App Context in companion object still produce warning Do not place Android context classes in static fields; this is a memory leak (and also breaks Instant Run)

or if you use something like this:

    companion object {
        lateinit var instance: MyApp
    }

It's simply fooling the lint to not discover the memory leak, the App instance still can produce memory leak, since Application class and its descendant is a Context.

Alternatively, you can use functional interface or Functional properties to help you get your app context.

Simply create an object class:

object CoreHelper {
    lateinit var contextGetter: () -> Context
}

or you could use it more safely using nullable type:

object CoreHelper {
    var contextGetter: (() -> Context)? = null
}

and in your App class add this line:

class MyApp: Application() {

    override fun onCreate() {
        super.onCreate()
        CoreHelper.contextGetter = {
            this
        }
    }
}

and in your manifest declare the app name to . MyApp

    <application
            android:name=".MyApp"

When you wanna get the context simply call:

CoreHelper.contextGetter()

// or if you use the nullable version
CoreHelper.contextGetter?.invoke()

Hope it will help.

First of all, public static non-final fields are evil. Spring does not allow injecting to such fields for a reason.

Your workaround is valid, you don't even need getter/setter, private field is enough. On the other hand try this:

@Value("${my.name}")
public void setPrivateName(String privateName) {
    Sample.name = privateName;
}  

(works with @Autowired/@Resource). But to give you some constructive advice: Create a second class with private field and getter instead of public static field.

Maybe it helps someone who comes across the question: All the Conditional Renderings in React It's an article about all the different options for conditional rendering in React.

Key takeaways of when to use which conditional rendering:

** if-else

• is the most basic conditional rendering
• beginner friendly
• use if to opt-out early from a render method by returning null

** ternary operator

• use it over an if-else statement
• it is more concise than if-else

** logical && operator

• use it when one side of the ternary operation would return null

** switch case

• verbose
• can only be inlined with self invoking function
• avoid it, use enums instead

** enums

• perfect to map different states
• perfect to map more than one condition

** multi-level/nested conditional renderings

• avoid them for the sake of readability
• split up components into more lightweight components with their own simple conditional rendering
• use HOCs

** HOCs

• use them to shield away conditional rendering
• components can focus on their main purpose

** external templating components

• avoid them and be comfortable with JSX and JavaScript

Assigning Comparison to property value

JavaScript

You could assign the comparison of the property to "1"

obj["isChecked"] = (obj["isChecked"]==="1");

This only evaluates for a String value of "1" though. Other variables evaulate to false like an actual typeof number would be false. (i.e. obj["isChecked"]=1)

If you wanted to be indiscrimate about "1" or 1, you could use:

obj["isChecked"] = (obj["isChecked"]=="1");

Example Outputs

console.log(obj["isChecked"]==="1"); // true
console.log(obj["isChecked"]===1); // false
console.log(obj["isChecked"]==1); // true
console.log(obj["isChecked"]==="0"); // false
console.log(obj["isChecked"]==="Elephant"); // false

PHP

Same concept in PHP

$obj["isChecked"] = ($obj["isChecked"] == "1");

The same operator limitations as stated above for JavaScript apply.

Double Not

The 'double not' also works. It's confusing when people first read it but it works in both languages for integer/number type values. It however does not work in JavaScript for string type values as they always evaluate to true:

JavaScript

!!"1"; //true
!!"0"; //true
!!1; //true
!!0; //false
!!parseInt("0",10); // false

PHP

echo !!"1"; //true
echo !!"0"; //false
echo !!1; //true
echo !!0; //false

Udo G. said:

• The eval() can't be used inside a function and must be called inside the global scope otherwise no functions or variables will be accessible (i.e. you can't create a include() utility function or something like that).

He's right, but there's a way to affect the global scope from a function. Improving his example:

function include(file_) {
    with (global) {
        eval(fs.readFileSync(file_) + '');
    };
};

include('somefile_with_some_declarations.js');

// the declarations are now accessible here.

Hope, that helps.

Recursive algorithm's time complexity can be better estimated by drawing recursion tree, In this case the recurrence relation for drawing recursion tree would be T(n)=T(n-1)+T(n-2)+O(1) note that each step takes O(1) meaning constant time,since it does only one comparison to check value of n in if block.Recursion tree would look like

          n
   (n-1)      (n-2)
(n-2)(n-3) (n-3)(n-4) ...so on

Here lets say each level of above tree is denoted by i hence,

i
0                        n
1            (n-1)                 (n-2)
2        (n-2)    (n-3)      (n-3)     (n-4)
3   (n-3)(n-4) (n-4)(n-5) (n-4)(n-5) (n-5)(n-6)

lets say at particular value of i, the tree ends, that case would be when n-i=1, hence i=n-1, meaning that the height of the tree is n-1. Now lets see how much work is done for each of n layers in tree.Note that each step takes O(1) time as stated in recurrence relation.

2^0=1                        n
2^1=2            (n-1)                 (n-2)
2^2=4        (n-2)    (n-3)      (n-3)     (n-4)
2^3=8   (n-3)(n-4) (n-4)(n-5) (n-4)(n-5) (n-5)(n-6)    ..so on
2^i for ith level

since i=n-1 is height of the tree work done at each level will be

i work
1 2^1
2 2^2
3 2^3..so on

Hence total work done will sum of work done at each level, hence it will be 2^0+2^1+2^2+2^3...+2^(n-1) since i=n-1. By geometric series this sum is 2^n, Hence total time complexity here is O(2^n)

@coldmind answer is correct but lacks details.

The 'NOT NULL constraint failed' occurs when something tries to set None to the 'zipcode' property, while it has not been explicitely allowed.

It usually happens when:

1) your field has Null=False by default, so that the value in the database cannot be None (i.e. undefined) when the object is created and saved in the database (this happens after a objects_set.create() call or setting the .zipcode property and doing a .save() call).

For instance, if somewhere in your code an assignement results in:

model.zipcode = None

this error is raised

2) When creating or updating the database, Django is constrained to find a default value to fill the field, because Null=False by default. It does not find any because you haven't defined any. So this error can not only happen during code execution but also when creating the database?

3) Note that the same error would be returned of you define default=None, or if your default value with an incorrect type, for instance default='00000' instead of 00000 for your field (maybe can there be automatic conversion between char and integers, but I would advise against relying on it. Besides, explicit is better than implicit). Most likely an error would also be raised if the default value violates the max_length property, e.g. 123456

So you'll have to define the field by one of the following:

models.IntegerField(_('zipcode'), max_length=5, Null=True,
   blank=True)

models.IntegerField(_('zipcode'), max_length=5, Null=False,
   blank=True, default=00000)

models.IntegerField(_('zipcode'), max_length=5, blank=True,
   default=00000)

and then make a migration (python3 manage.py makemigration ) and then migrate (python3 manage.py migrate).

For safety you can also delete the last failed migration files in <app_name>/migrations/, there are usually named after this pattern:

<NUMBER>_auto_<DATE>_<HOUR>.py

Finally, if you don't set Null=True, make sure that mode.zipcode = None is never done anywhere.

Not an exact answer to your question, but a bit of information: if your device does use NTP for time (eg. if it is a tablet with no 3G or GPS capabilities), the server can be configured in /system/etc/gps.conf - obviously this file can only be edited with root access, but is viewable on non-rooted devices.

As far as I know there is no option to create global configuration for java applications. You always create a duplicate of the configuration.

Also, if you are using PDE (for plugin development), you can create target platform using windows -> Preferences -> Plug-in development -> Target Platform. Edit has options for program/vm arguments.

Hope this helps

    String name,lname;
 name= JOptionPane.showInputDialog(null,"Enter your name");
   lname= JOptionPane.showInputDialog(null,"Enter your father name");
    if(name.equals("Ahmad")){
       JOptionPane.showMessageDialog(null,"welcome "+name);
    }
    if(lname.equals("Khan"))
   JOptionPane.showMessageDialog(null,"Name : "+name +"\nLast name :"+lname ); 

    else {
       JOptionPane.showMessageDialog(null,"try again " );
    } 
  }}

Another way to look at this. Check out the details of the exception:

In [49]: try: 
    ...:     open('file.DNE.txt') 
    ...: except Exception as  e: 
    ...:     print(dir(e)) 
    ...:                                                                                                                                    
['__cause__', '__class__', '__context__', '__delattr__', '__dict__', '__dir__', '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__', '__init_subclass__', '__le__', '__lt__', '__ne__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setstate__', '__sizeof__', '__str__', '__subclasshook__', '__suppress_context__', '__traceback__', 'args', 'characters_written', 'errno', 'filename', 'filename2', 'strerror', 'with_traceback']

There are lots of "things" to access using the 'as e' syntax.

This code was solely meant to show the details of this instance.

Interpolate:

body
    height: calc(100% - #{$body_padding})

For this case, border-box would also suffice:

body
    box-sizing: border-box
    height: 100%
    padding-top: $body_padding

If you do not really care about rounding, just added a toFixed(x) and then removing trailing 0es and the dot if necessary. It is not a fast solution.

function format(value, decimals) {
    if (value) {
        value = value.toFixed(decimals);            
    } else {
        value = "0";
    }
    if (value.indexOf(".") < 0) { value += "."; }
    var dotIdx = value.indexOf(".");
    while (value.length - dotIdx <= decimals) { value += "0"; } // add 0's

    return value;
}

#Sample data
set.seed(1)
dat <- data.frame(x = sample(0:2, 5, TRUE), y = sample(0:2, 5, TRUE))
#-----
  x y
1 0 2
2 1 2
3 1 1
4 2 1
5 0 0

#replace zeros with NA
dat[dat==0] <- NA
#-----
   x  y
1 NA  2
2  1  2
3  1  1
4  2  1
5 NA NA

Domain Driven Design is a methodology and process prescription for the development of complex systems whose focus is mapping activities, tasks, events, and data within a problem domain into the technology artifacts of a solution domain.

The emphasis of Domain Driven Design is to understand the problem domain in order to create an abstract model of the problem domain which can then be implemented in a particular set of technologies. Domain Driven Design as a methodology provides guidelines for how this model development and technology development can result in a system that meets the needs of the people using it while also being robust in the face of change in the problem domain.

The process side of Domain Driven Design involves the collaboration between domain experts, people who know the problem domain, and the design/architecture experts, people who know the solution domain. The idea is to have a shared model with shared language so that as people from these two different domains with their two different perspectives discuss the solution they are actually discussing a shared knowledge base with shared concepts.

The lack of a shared problem domain understanding between the people who need a particular system and the people who are designing and implementing the system seems to be a core impediment to successful projects. Domain Driven Design is a methodology to address this impediment.

It is more than having an object model. The focus is really about the shared communication and improving collaboration so that the actual needs within the problem domain can be discovered and an appropriate solution created to meet those needs.

Domain-Driven Design: The Good and The Challenging provides a brief overview with this comment:

DDD helps discover the top-level architecture and inform about the mechanics and dynamics of the domain that the software needs to replicate. Concretely, it means that a well done DDD analysis minimizes misunderstandings between domain experts and software architects, and it reduces the subsequent number of expensive requests for change. By splitting the domain complexity in smaller contexts, DDD avoids forcing project architects to design a bloated object model, which is where a lot of time is lost in working out implementation details — in part because the number of entities to deal with often grows beyond the size of conference-room white boards.

Also see this article Domain Driven Design for Services Architecture which provides a short example. The article provides the following thumbnail description of Domain Driven Design.

Domain Driven Design advocates modeling based on the reality of business as relevant to our use cases. As it is now getting older and hype level decreasing, many of us forget that the DDD approach really helps in understanding the problem at hand and design software towards the common understanding of the solution. When building applications, DDD talks about problems as domains and subdomains. It describes independent steps/areas of problems as bounded contexts, emphasizes a common language to talk about these problems, and adds many technical concepts, like entities, value objects and aggregate root rules to support the implementation.

Martin Fowler has written a number of articles in which Domain Driven Design as a methodology is mentioned. For instance this article, BoundedContext, provides an overview of the bounded context concept from Domain Driven Development.

In those younger days we were advised to build a unified model of the entire business, but DDD recognizes that we've learned that "total unification of the domain model for a large system will not be feasible or cost-effective" 1. So instead DDD divides up a large system into Bounded Contexts, each of which can have a unified model - essentially a way of structuring MultipleCanonicalModels.

There is one another method of temp table

create table #TempTable (
ID int,
name varchar(max)
)

insert into #TempTable (ID,name)
Select ID,Name 
from Table

SELECT * 
FROM #TempTable
WHERE ID  = 1 

Make Sure You are selecting the right database.

In visual studio, go to help | about and look at the version of Visual Studio that you're using to compile your app.

i think you should try this technique it will help to call as many as pages you like all pages will run at once independently without waiting for each page response as asynchronous.

cornjobpage.php //mainpage

    <?php

post_async("http://localhost/projectname/testpage.php", "Keywordname=testValue");
//post_async("http://localhost/projectname/testpage.php", "Keywordname=testValue2");
//post_async("http://localhost/projectname/otherpage.php", "Keywordname=anyValue");
//call as many as pages you like all pages will run at once independently without waiting for each page response as asynchronous.
            ?>
            <?php

            /*
             * Executes a PHP page asynchronously so the current page does not have to wait for it to     finish running.
             *  
             */
            function post_async($url,$params)
            {

                $post_string = $params;

                $parts=parse_url($url);

                $fp = fsockopen($parts['host'],
                    isset($parts['port'])?$parts['port']:80,
                    $errno, $errstr, 30);

                $out = "GET ".$parts['path']."?$post_string"." HTTP/1.1\r\n";//you can use POST instead of GET if you like
                $out.= "Host: ".$parts['host']."\r\n";
                $out.= "Content-Type: application/x-www-form-urlencoded\r\n";
                $out.= "Content-Length: ".strlen($post_string)."\r\n";
                $out.= "Connection: Close\r\n\r\n";
                fwrite($fp, $out);
                fclose($fp);
            }
            ?>

testpage.php

    <?
    echo $_REQUEST["Keywordname"];//case1 Output > testValue
    ?>

PS:if you want to send url parameters as loop then follow this answer :https://stackoverflow.com/a/41225209/6295712

If you found this thread trying to compare data fames in tests, then take a look at assert_frame_equal method: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.testing.assert_frame_equal.html

Same issue here, comparing the htdocs/xampp folder in 5.6.11 with 5.6.8 I saw all the files there are missing in 5.6.11. Copied the entire htdocs/xampp folder from 5.6.8 to 5.6.11 and worked fine.

There is a good, simple example of exactly this, here: http://www.queness.com/post/77/simple-jquery-modal-window-tutorial

If you need to add emails as CC or BCC, add the following part in the variable you use as for your header :

$headers .= "CC: [email protected]".PHP_EOL;
$headers .= "BCC: [email protected]".PHP_EOL;

Regards

You can use ClosedXML for this.

Store your table in a DataTable and you can export the table to excel by this simple snippet:

XLWorkbook workbook = new XLWorkbook();
DataTable table = GetYourTable();
workbook.Worksheets.Add(table );

You can read the documentation of ClosedXML to learn more. Hope this helps!

.row {
    margin-right: 15px;
}

throw this in your CSS

I realize this may be a bit late, but I stumbled upon this and was wondering how to handle situations with multiple identical values, but different keys (as per bigbearzhu's comment).

So I modified Stephan Muller's answer slightly:

A datalist with non-unique values:

<input list="answers" name="answer" id="answerInput">
<datalist id="answers">
  <option value="42">The answer</option>
  <option value="43">The answer</option>
  <option value="44">Another Answer</option>
</datalist>
<input type="hidden" name="answer" id="answerInput-hidden">

When the user selects an option, the browser replaces input.value with the value of the datalist option instead of the innerText.

The following code then checks for an option with that value, pushes that into the hidden field and replaces the input.value with the innerText.

document.querySelector('#answerInput').addEventListener('input', function(e) {
    var input = e.target,   
        list = input.getAttribute('list'),
        options = document.querySelectorAll('#' + list + ' option[value="'+input.value+'"]'),
        hiddenInput = document.getElementById(input.getAttribute('id') + '-hidden');

    if (options.length > 0) {
      hiddenInput.value = input.value;
      input.value = options[0].innerText;
      }

});

As a consequence the user sees whatever the option's innerText says, but the unique id from option.value is available upon form submit. Demo jsFiddle

There are two options available primarily; in case of imputation or filling of missing values NaN / np.nan with only numerical replacements (across column(s):

df['Amount'].fillna(value=None, method= ,axis=1,) is sufficient:

From the Documentation:

value : scalar, dict, Series, or DataFrame Value to use to fill holes (e.g. 0), alternately a dict/Series/DataFrame of values specifying which value to use for each index (for a Series) or column (for a DataFrame). (values not in the dict/Series/DataFrame will not be filled). This value cannot be a list.

Which means 'strings' or 'constants' are no longer permissable to be imputed.

For more specialized imputations use SimpleImputer():

from sklearn.impute import SimpleImputer
si = SimpleImputer(strategy='constant', missing_values=np.nan, fill_value='Replacement_Value')
df[['Col-1', 'Col-2']] = si.fit_transform(X=df[['C-1', 'C-2']])

Found an alternative at Child elements with margins within DIVs You can also add:

.parent { overflow: auto; }

or:

.parent { overflow: hidden; }

This prevents the margins to collapse. Border and padding do the same. Hence, you can also use the following to prevent a top-margin collapse:

.parent {
    padding-top: 1px;
    margin-top: -1px;
}

Update by popular request: The whole point of collapsing margins is handling textual content. For example:

h1, h2, p, ul {_x000D_
  margin-top: 1em;_x000D_
  margin-bottom: 1em;_x000D_
}

<h1>Title!</h1>_x000D_
<div class="text">_x000D_
  <h2>Title!</h2>_x000D_
  <p>Paragraph</p>_x000D_
</div>_x000D_
<div class="text">_x000D_
  <h2>Title!</h2>_x000D_
  <p>Paragraph</p>_x000D_
  <ul>_x000D_
    <li>list item</li>_x000D_
  </ul>_x000D_
</div>

Because the browser collapses margins, the text would appear as you'd expect, and the <div> wrapper tags don't influence the margins. Each element ensures it has spacing around it, but spacing won't be doubled. The margins of the <h2> and <p> won't add up, but slide into each other (they collapse). The same happens for the <p> and <ul> element.

Sadly, with modern designs this idea can bite you when you explicitly want a container. This is called a new block formatting context in CSS speak. The overflow or margin trick will give you that.

I just used this and it works fine:

System.Diagnostics.Process.Start("C:/Users/Admin/files");

I had the very same problem recently with Android Studio 1.3. The only working solution was to remove the .gradle and .idea folders and re-import the project into Android Studio.

I think that .Net Framework does this automatically but just in case. First, make sure to select what you want to erase, and then call the garbage collector:

randomClass object1 = new randomClass
...
...
// Give a null value to the code you want to delete
object1 = null;
// Then call the garbage collector to erase what you gave the null value
GC.Collect();

I think that's it.. Hope I help someone.

this procedure works pretty well

pdf2ps large.pdf very_large.ps

ps2pdf very_large.ps small.pdf

give it a try.

-Wall and -Wextra sets the stage in GCC and the subsequent -Wno-unused-variable may not take effect. For example, if you have:

CFLAGS += -std=c99 -pedantic -pedantic-errors -Werror -g0 -Os \ -fno-strict-overflow -fno-strict-aliasing \ -Wall -Wextra \ -pthread \ -Wno-unused-label \ -Wno-unused-function \ -Wno-unused-parameter \ -Wno-unused-variable \ $(INC)

then GCC sees the instruction -Wall -Wextra and seems to ignore -Wno-unused-variable

This can instead look like this below and you get the desired effect of not being stopped in your compile on the unused variable:

CFLAGS += -std=c99 -pedantic -pedantic-errors -Werror -g0 -Os \ -fno-strict-overflow -fno-strict-aliasing \ -pthread \ -Wno-unused-label \ -Wno-unused-function \ $(INC)

There is a good reason it is called a "warning" vs an "error". Failing the compile just because you code is not complete (say you are stubbing the algorithm out) can be defeating.

Splice - MDN reference - ECMA-262 spec

Syntax
array.splice(start[, deleteCount[, item1[, item2[, ...]]]])

Parameters

• start: required. Initial index.
  If start is negative it is treated as "Math.max((array.length + start), 0)" as per spec (example provided below) effectively from the end of array.
• deleteCount: optional. Number of elements to be removed (all from start if not provided).
• item1, item2, ...: optional. Elements to be added to the array from start index.

Returns: An array with deleted elements (empty array if none removed)

Mutate original array: Yes

Examples:

const array = [1,2,3,4,5];

// Remove first element
console.log('Elements deleted:', array.splice(0, 1), 'mutated array:', array);
// Elements deleted: [ 1 ] mutated array: [ 2, 3, 4, 5 ]

// array = [ 2, 3, 4, 5]
// Remove last element (start -> array.length+start = 3)
console.log('Elements deleted:', array.splice(-1, 1), 'mutated array:', array);
// Elements deleted: [ 5 ] mutated array: [ 2, 3, 4 ]

More examples in MDN Splice examples

Slice - MDN reference - ECMA-262 spec

Syntax
array.slice([begin[, end]])
Parameters

• begin: optional. Initial index (default 0).
  If begin is negative it is treated as "Math.max((array.length + begin), 0)" as per spec (example provided below) effectively from the end of array.
• end: optional. Last index for extraction but not including (default array.length). If end is negative it is treated as "Math.max((array.length + begin),0)" as per spec (example provided below) effectively from the end of array.

Returns: An array containing the extracted elements.

Mutate original: No

Examples:

const array = [1,2,3,4,5];

// Extract first element
console.log('Elements extracted:', array.slice(0, 1), 'array:', array);
// Elements extracted: [ 1 ] array: [ 1, 2, 3, 4, 5 ]

// Extract last element (start -> array.length+start = 4)
console.log('Elements extracted:', array.slice(-1), 'array:', array);
// Elements extracted: [ 5 ] array: [ 1, 2, 3, 4, 5 ]

More examples in MDN Slice examples

Performance comparison

Don't take this as absolute truth as depending on each scenario one might be performant than the other.
Performance test

You'll get this error as well if you are verifying that an extension method of an interface is called.

For example if you are mocking:

var mockValidator = new Mock<IValidator<Foo>>();
mockValidator
  .Verify(validator => validator.ValidateAndThrow(foo, null));

You will get the same exception because .ValidateAndThrow() is an extension on the IValidator<T> interface.

public static void ValidateAndThrow<T>(this IValidator<T> validator, T instance, string ruleSet = null)...

Just to give some ideas to oracle Java source developer :-)

The solution already exists in .Net and is more very more readable !

In Visual Basic .Net

Drawable drawable 
    = If(Common.getDrawableFromUrl(this, product.getMapPath())
        ,getRandomDrawable()
        )

In C#

Drawable drawable 
    = Common.getDrawableFromUrl(this, product.getMapPath() 
        ?? getRandomDrawable();

These solutions are powerful as Optional Java solution (default string is only evaluated if original value is null) without using lambda expression, just in adding a new operator.

Just to see quickly the difference with Java solution, I have added the 2 Java solutions

Using Optional in Java

Drawable drawable = 
    Optional.ofNullable(Common.getDrawableFromUrl(this, product.getMapPath()))
        .orElseGet(() -> getRandomDrawable());

Using { } in Java

Drawable drawable = Common.getDrawableFromUrl(this, product.getMapPath());
if (drawable != null)
    {
    drawable = getRandomDrawable();
    }

Personally, I like VB.Net but I prefer ?? C# or if {} solution in Java ... and you ?

I too faced the same error. One solution that worked for me was to add the following lines in your service.ts file instead of import 'rxjs/add/operator/map':

import { Observable } from 'rxjs';
import { map } from 'rxjs/operators';

For an example, my app.service.ts after debugging was like,

import { Injectable } from '@angular/core';
import { HttpClient } from '@angular/common/http';
import { Observable } from 'rxjs';
import { map } from 'rxjs/operators';

@Injectable()
export class AppService {
    constructor(private http: HttpClient) {}
    getData(): Observable<any> {
        return this.http.get('https://samples.openweathermap.org/data/2.5/history/city?q=Warren,OH&appid=b6907d289e10d714a6e88b30761fae22')
        .pipe(map(result => result));
    }
}

I had the same issue and put double quotes around the username and password and it worked: create public database link "opps" identified by "opps" using 'TEST';

I just put one simple declaration of the app

var app = angular.module("app", ["xeditable"]);

Then I built one service and two controllers

For each controller I had a line in the JS

app.controller('EditableRowCtrl', function ($scope, CRUD_OperService) {

And in the HTML I declared the app scope in a surrounding div

<div ng-app="app">

and each controller scope separately in their own surrounding div (within the app div)

<div ng-controller="EditableRowCtrl">

This worked fine

When using tag helpers in ASP.NET Core, the format needs specified in ISO format. If not specified as such, bound input data won't display properly and will show as mm/dd/yyyy with no value.

Model:

[Display(Name = "Hire")]
[DataType(DataType.Date)]
[DisplayFormat(ApplyFormatInEditMode = true, DataFormatString = "{0:yyyy-MM-dd}")]
public DateTime? HireDate { get; set; }

View:

<input asp-for="Entity.HireDate" class="form-control" />

The format can also be specified in the view using the asp-format attribute.

The resulting HTML will look as follows:

<input class="form-control" type="date" id="Entity_HireDate" 
    name="Entity.HireDate" value="2012-01-01">

I use the following format and works well.

robocopy \\SourceServer\Path \\TargetServer\Path filename.txt

to copy everything you can replace filename.txt with *.* and there are plenty of other switches to copy subfolders etc... see here: http://ss64.com/nt/robocopy.html

I typically abbreviate so that I can use several places in my code. Near the top, do something like this:

 string nl = System.Environment.NewLine;

Then I can just use "nl" instead of the full qualification everywhere when constructing strings.

You can use "ToTitleCase method"

string s = new CultureInfo("en-US").TextInfo.ToTitleCase("red house");
//result : Red House

this extention method solve every titlecase problem.

easy to usage

string str = "red house";
str.ToTitleCase();
//result : Red house

string str = "red house";
str.ToTitleCase(TitleCase.All);
//result : Red House

the Extention method

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Globalization;

namespace Test
{
    public static class StringHelper
    {
        private static CultureInfo ci = new CultureInfo("en-US");
        //Convert all first latter
        public static string ToTitleCase(this string str)
        {
            str = str.ToLower();
            var strArray = str.Split(' ');
            if (strArray.Length > 1)
            {
                strArray[0] = ci.TextInfo.ToTitleCase(strArray[0]);
                return string.Join(" ", strArray);
            }
            return ci.TextInfo.ToTitleCase(str);
        }
        public static string ToTitleCase(this string str, TitleCase tcase)
        {
            str = str.ToLower();
            switch (tcase)
            {
                case TitleCase.First:
                    var strArray = str.Split(' ');
                    if (strArray.Length > 1)
                    {
                        strArray[0] = ci.TextInfo.ToTitleCase(strArray[0]);
                        return string.Join(" ", strArray);
                    }
                    break;
                case TitleCase.All:
                    return ci.TextInfo.ToTitleCase(str);
                default:
                    break;
            }
            return ci.TextInfo.ToTitleCase(str);
        }
    }

    public enum TitleCase
    {
        First,
        All
    }
}

If you are facing problem with pip3 install pyqt5 then try pip3 install pyqt5==5.12.0
This solved the problem for me

How to generate or reverse a Dockerfile from an image?

You can.

alias dfimage="docker run -v /var/run/docker.sock:/var/run/docker.sock --rm alpine/dfimage"
dfimage -sV=1.36 nginx:latest

It will pull the target docker image automaticlaly and export Dockerfile. Parameter -sV=1.36 is not always required.

Reference: https://hub.docker.com/repository/docker/alpine/dfimage

below is the old answer, it doesn't work any more.

$ docker pull centurylink/dockerfile-from-image
$ alias dfimage="docker run -v /var/run/docker.sock:/var/run/docker.sock --rm centurylink/dockerfile-from-image"
$ dfimage --help
Usage: dockerfile-from-image.rb [options] <image_id>
    -f, --full-tree                  Generate Dockerfile for all parent layers
    -h, --help                       Show this message

        if ( params.build_deploy == '1' ) {
            println "build_deploy ? ${params.build_deploy}"
              jobB = build job: 'k8s-core-user_deploy', propagate: false, wait: true, parameters: [
                         string(name:'environment', value: "${params.environment}"),
                         string(name:'branch_name', value: "${params.branch_name}"),
                         string(name:'service_name', value: "${params.service_name}"),                      
                     ]
            println jobB.getResult()
        }

If using a class-based view:

#
# in settings.py
#
YOUR_CUSTOM_SETTING = 'some value'

#
# in views.py
#
from django.conf import settings #for getting settings vars

class YourView(DetailView): #assuming DetailView; whatever though

    # ...

    def get_context_data(self, **kwargs):

        context = super(YourView, self).get_context_data(**kwargs)
        context['YOUR_CUSTOM_SETTING'] = settings.YOUR_CUSTOM_SETTING

        return context

#
# in your_template.html, reference the setting like any other context variable
#
{{ YOUR_CUSTOM_SETTING }}

In other words, you want to pipe stdout into one filter (tee bbb.out) and stderr into another filter (tee ccc.out). There is no standard way to pipe anything other than stdout into another command, but you can work around that by juggling file descriptors.

{ { ./aaa.sh | tee bbb.out; } 2>&1 1>&3 | tee ccc.out; } 3>&1 1>&2

See also How to grep standard error stream (stderr)? and When would you use an additional file descriptor?

In bash (and ksh and zsh), but not in other POSIX shells such as dash, you can use process substitution:

./aaa.sh > >(tee bbb.out) 2> >(tee ccc.out)

Beware that in bash, this command returns as soon as ./aaa.sh finishes, even if the tee commands are still executed (ksh and zsh do wait for the subprocesses). This may be a problem if you do something like ./aaa.sh > >(tee bbb.out) 2> >(tee ccc.out); process_logs bbb.out ccc.out. In that case, use file descriptor juggling or ksh/zsh instead.

select to_char(date,'Mon') as mon,
       extract(year from date) as yyyy,
       sum("Sales") as "Sales"
from yourtable
group by 1,2

At the request of Radu, I will explain that query:

to_char(date,'Mon') as mon, : converts the "date" attribute into the defined format of the short form of month.

extract(year from date) as yyyy : Postgresql's "extract" function is used to extract the YYYY year from the "date" attribute.

sum("Sales") as "Sales" : The SUM() function adds up all the "Sales" values, and supplies a case-sensitive alias, with the case sensitivity maintained by using double-quotes.

group by 1,2 : The GROUP BY function must contain all columns from the SELECT list that are not part of the aggregate (aka, all columns not inside SUM/AVG/MIN/MAX etc functions). This tells the query that the SUM() should be applied for each unique combination of columns, which in this case are the month and year columns. The "1,2" part is a shorthand instead of using the column aliases, though it is probably best to use the full "to_char(...)" and "extract(...)" expressions for readability.

The colorbar has to have its own axes. However, you can create an axes that overlaps with the previous one. Then use the cax kwarg to tell fig.colorbar to use the new axes.

For example:

import numpy as np
import matplotlib.pyplot as plt

data = np.arange(100, 0, -1).reshape(10, 10)

fig, ax = plt.subplots()
cax = fig.add_axes([0.27, 0.8, 0.5, 0.05])

im = ax.imshow(data, cmap='gist_earth')
fig.colorbar(im, cax=cax, orientation='horizontal')
plt.show()

The Left join in this query is pointless:

UPDATE md SET md.status = '3' 
FROM pd_mounting_details AS md 
LEFT OUTER JOIN pd_order_ecolid AS oe ON md.order_data = oe.id

It would update all rows of pd_mounting_details, whether or not a matching row exists in pd_order_ecolid. If you wanted to only update matching rows, it should be an inner join.

If you want to apply some condition based on the join occurring or not, you need to add a WHERE clause and/or a CASE expression in your SET clause.

If anyone finds their way here with the same problem I was having, my problem was that I was missing the <dependencyManagement> tags around dependencies I had copied from the child pom.

In Winforms app, both methods:

System.Diagnostics.Debug.WriteLine("my string")

and

System.Console.WriteLine("my string")

write to the output window.

In AspNetCore app, only System.Diagnostics.Debug.WriteLine("my string") writes to the output window.

Maybe you can use ListBuffers in scala to create empty list and add strings later because ListBuffers are mutable. Also all the List functions are available for the ListBuffers in scala.

import scala.collection.mutable.ListBuffer 

val dm = ListBuffer[String]()
dm: scala.collection.mutable.ListBuffer[String] = ListBuffer()
dm += "text1"
dm += "text2"
dm = ListBuffer(text1, text2)

if you want you can convert this to a list by using .toList

Mockito matchers are static methods and calls to those methods, which stand in for arguments during calls to when and verify.

Hamcrest matchers (archived version) (or Hamcrest-style matchers) are stateless, general-purpose object instances that implement Matcher<T> and expose a method matches(T) that returns true if the object matches the Matcher's criteria. They are intended to be free of side effects, and are generally used in assertions such as the one below.

/* Mockito */  verify(foo).setPowerLevel(gt(9000));
/* Hamcrest */ assertThat(foo.getPowerLevel(), is(greaterThan(9000)));

Mockito matchers exist, separate from Hamcrest-style matchers, so that descriptions of matching expressions fit directly into method invocations: Mockito matchers return T where Hamcrest matcher methods return Matcher objects (of type Matcher<T>).

Mockito matchers are invoked through static methods such as eq, any, gt, and startsWith on org.mockito.Matchers and org.mockito.AdditionalMatchers. There are also adapters, which have changed across Mockito versions:

• For Mockito 1.x, Matchers featured some calls (such as intThat or argThat) are Mockito matchers that directly accept Hamcrest matchers as parameters. ArgumentMatcher<T> extended org.hamcrest.Matcher<T>, which was used in the internal Hamcrest representation and was a Hamcrest matcher base class instead of any sort of Mockito matcher.
• For Mockito 2.0+, Mockito no longer has a direct dependency on Hamcrest. Matchers calls phrased as intThat or argThat wrap ArgumentMatcher<T> objects that no longer implement org.hamcrest.Matcher<T> but are used in similar ways. Hamcrest adapters such as argThat and intThat are still available, but have moved to MockitoHamcrest instead.

Regardless of whether the matchers are Hamcrest or simply Hamcrest-style, they can be adapted like so:

/* Mockito matcher intThat adapting Hamcrest-style matcher is(greaterThan(...)) */
verify(foo).setPowerLevel(intThat(is(greaterThan(9000))));

In the above statement: foo.setPowerLevel is a method that accepts an int. is(greaterThan(9000)) returns a Matcher<Integer>, which wouldn't work as a setPowerLevel argument. The Mockito matcher intThat wraps that Hamcrest-style Matcher and returns an int so it can appear as an argument; Mockito matchers like gt(9000) would wrap that entire expression into a single call, as in the first line of example code.

What matchers do/return

when(foo.quux(3, 5)).thenReturn(true);

When not using argument matchers, Mockito records your argument values and compares them with their equals methods.

when(foo.quux(eq(3), eq(5))).thenReturn(true);    // same as above
when(foo.quux(anyInt(), gt(5))).thenReturn(true); // this one's different

When you call a matcher like any or gt (greater than), Mockito stores a matcher object that causes Mockito to skip that equality check and apply your match of choice. In the case of argumentCaptor.capture() it stores a matcher that saves its argument instead for later inspection.

Matchers return dummy values such as zero, empty collections, or null. Mockito tries to return a safe, appropriate dummy value, like 0 for anyInt() or any(Integer.class) or an empty List<String> for anyListOf(String.class). Because of type erasure, though, Mockito lacks type information to return any value but null for any() or argThat(...), which can cause a NullPointerException if trying to "auto-unbox" a null primitive value.

Matchers like eq and gt take parameter values; ideally, these values should be computed before the stubbing/verification starts. Calling a mock in the middle of mocking another call can interfere with stubbing.

Matcher methods can't be used as return values; there is no way to phrase thenReturn(anyInt()) or thenReturn(any(Foo.class)) in Mockito, for instance. Mockito needs to know exactly which instance to return in stubbing calls, and will not choose an arbitrary return value for you.

Implementation details

Matchers are stored (as Hamcrest-style object matchers) in a stack contained in a class called ArgumentMatcherStorage. MockitoCore and Matchers each own a ThreadSafeMockingProgress instance, which statically contains a ThreadLocal holding MockingProgress instances. It's this MockingProgressImpl that holds a concrete ArgumentMatcherStorageImpl. Consequently, mock and matcher state is static but thread-scoped consistently between the Mockito and Matchers classes.

Most matcher calls only add to this stack, with an exception for matchers like and, or, and not. This perfectly corresponds to (and relies on) the evaluation order of Java, which evaluates arguments left-to-right before invoking a method:

when(foo.quux(anyInt(), and(gt(10), lt(20)))).thenReturn(true);
[6]      [5]  [1]       [4] [2]     [3]

This will:

1. Add anyInt() to the stack.
2. Add gt(10) to the stack.
3. Add lt(20) to the stack.
4. Remove gt(10) and lt(20) and add and(gt(10), lt(20)).
5. Call foo.quux(0, 0), which (unless otherwise stubbed) returns the default value false. Internally Mockito marks quux(int, int) as the most recent call.
6. Call when(false), which discards its argument and prepares to stub method quux(int, int) identified in 5. The only two valid states are with stack length 0 (equality) or 2 (matchers), and there are two matchers on the stack (steps 1 and 4), so Mockito stubs the method with an any() matcher for its first argument and and(gt(10), lt(20)) for its second argument and clears the stack.

This demonstrates a few rules:

• Mockito can't tell the difference between quux(anyInt(), 0) and quux(0, anyInt()). They both look like a call to quux(0, 0) with one int matcher on the stack. Consequently, if you use one matcher, you have to match all arguments.

• Call order isn't just important, it's what makes this all work. Extracting matchers to variables generally doesn't work, because it usually changes the call order. Extracting matchers to methods, however, works great.

  int between10And20 = and(gt(10), lt(20));
  /* BAD */ when(foo.quux(anyInt(), between10And20)).thenReturn(true);
  // Mockito sees the stack as the opposite: and(gt(10), lt(20)), anyInt().
  
  public static int anyIntBetween10And20() { return and(gt(10), lt(20)); }
  /* OK */  when(foo.quux(anyInt(), anyIntBetween10And20())).thenReturn(true);
  // The helper method calls the matcher methods in the right order.
  

• The stack changes often enough that Mockito can't police it very carefully. It can only check the stack when you interact with Mockito or a mock, and has to accept matchers without knowing whether they're used immediately or abandoned accidentally. In theory, the stack should always be empty outside of a call to when or verify, but Mockito can't check that automatically. You can check manually with Mockito.validateMockitoUsage().

• In a call to when, Mockito actually calls the method in question, which will throw an exception if you've stubbed the method to throw an exception (or require non-zero or non-null values). doReturn and doAnswer (etc) do not invoke the actual method and are often a useful alternative.

• If you had called a mock method in the middle of stubbing (e.g. to calculate an answer for an eq matcher), Mockito would check the stack length against that call instead, and likely fail.

• If you try to do something bad, like stubbing/verifying a final method, Mockito will call the real method and also leave extra matchers on the stack. The final method call may not throw an exception, but you may get an InvalidUseOfMatchersException from the stray matchers when you next interact with a mock.

Common problems

• InvalidUseOfMatchersException:

  • Check that every single argument has exactly one matcher call, if you use matchers at all, and that you haven't used a matcher outside of a when or verify call. Matchers should never be used as stubbed return values or fields/variables.

  • Check that you're not calling a mock as a part of providing a matcher argument.

  • Check that you're not trying to stub/verify a final method with a matcher. It's a great way to leave a matcher on the stack, and unless your final method throws an exception, this might be the only time you realize the method you're mocking is final.

• NullPointerException with primitive arguments: (Integer) any() returns null while any(Integer.class) returns 0; this can cause a NullPointerException if you're expecting an int instead of an Integer. In any case, prefer anyInt(), which will return zero and also skip the auto-boxing step.

• NullPointerException or other exceptions: Calls to when(foo.bar(any())).thenReturn(baz) will actually call foo.bar(null), which you might have stubbed to throw an exception when receiving a null argument. Switching to doReturn(baz).when(foo).bar(any()) skips the stubbed behavior.

General troubleshooting

• Use MockitoJUnitRunner, or explicitly call validateMockitoUsage in your tearDown or @After method (which the runner would do for you automatically). This will help determine whether you've misused matchers.

• For debugging purposes, add calls to validateMockitoUsage in your code directly. This will throw if you have anything on the stack, which is a good warning of a bad symptom.

background-image: url("/your-dir/your_image.jpg");
min-height: 100%;
background-repeat: no-repeat;
background-attachment: fixed;
background-position: center;
background-size: cover;}

You can use two ways to resolve this problem, first way that i think it as best way is replace importing segment of your code with bellow one:

import Home from './layouts/Home'

or export your component without default which is called named export like this

import React, { Component } from 'react';

class Home extends Component{
    render(){
        return(
        <p className="App-intro">
          Hello Man
        </p>
        )
    }
} 

export {Home};

For spaces use "[^a-z A-Z 0-9]" this pattern

:help local-additions

Lists local plugins added.

There is a native method for UIImage since iOS 13

let image = yourImage.withTintColor(.systemRed)

No need for jQuery

JavaScript does have a native .trim() method.

var name = "    John Smith  ";
name = name.trim();

console.log(name); // "John Smith"

Example Here

String.prototype.trim()

The trim() method removes whitespace from both ends of a string. Whitespace in this context is all the whitespace characters (space, tab, no-break space, etc.) and all the line terminator characters (LF, CR, etc.).

I have also used following link as others have suggested you for bluetooth communication.

http://developer.android.com/guide/topics/connectivity/bluetooth.html

The thing is all you need is a class BluetoothChatService.java

this class has following threads:

1. Accept
2. Connecting
3. Connected

Now when you call start function of the BluetoothChatService like:

mChatService.start();

It starts accept thread which means it will start looking for connection.

Now when you call

mChatService.connect(<deviceObject>,false/true);

Here first argument is device object that you can get from paired devices list or when you scan for devices you will get all the devices in range you can pass that object to this function and 2nd argument is a boolean to make secure or insecure connection.

connect function will start connecting thread which will look for any device which is running accept thread.

When such a device is found both accept thread and connecting thread will call connected function in BluetoothChatService:

connected(mmSocket, mmDevice, mSocketType);

this method starts connected thread in both the devices: Using this socket object connected thread obtains the input and output stream to the other device. And calls read function on inputstream in a while loop so that it's always trying read from other device so that whenever other device send a message this read function returns that message.

BluetoothChatService also has a write method which takes byte[] as input and calls write method on connected thread.

mChatService.write("your message".getByte());

write method in connected thread just write this byte data to outputsream of the other device.

public void write(byte[] buffer) {
   try {
       mmOutStream.write(buffer);
    // Share the sent message back to the UI Activity
    // mHandler.obtainMessage(
    // BluetoothGameSetupActivity.MESSAGE_WRITE, -1, -1,
    // buffer).sendToTarget();
    } catch (IOException e) {
    Log.e(TAG, "Exception during write", e);
     }
}

Now to communicate between two devices just call write function on mChatService and handle the message that you will receive on the other device.

I had to implement CoordinatorLayout with toolbar scrolling and it just took me all the day messing around this. I've got it working by removing NestedScrollView at all. So I'm just using RelativeLayout at the root.

<?xml version="1.0" encoding="utf-8"?>
<RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    android:layout_width="match_parent"
    android:layout_height="match_parent">

        <android.support.v7.widget.RecyclerView
            android:id="@+id/rv_nearby"
            android:layout_width="match_parent"
            android:layout_height="match_parent"
            app:layout_behavior="@string/appbar_scrolling_view_behavior" />

</RelativeLayout>

For current date just use UNIX_TIMESTAMP() in your MySQL query.

As of 2014, you can trigger the click on a <a/> tag. However, for security reasons, you have to do it in a click event handler, or the browser will tag it as a popup (some other events may allow you to safely trigger the opening).

jsFiddle

My case: configuring the service on new server. ASP.NET 4.0 was not installed/registered properly; svc extension was not recognized.

To answer the question. stringstream basically allows you to treat a string object like a stream, and use all stream functions and operators on it.

I saw it used mainly for the formatted output/input goodness.

One good example would be c++ implementation of converting number to stream object.

Possible example:

template <class T>
string num2str(const T& num, unsigned int prec = 12) {
    string ret;
    stringstream ss;
    ios_base::fmtflags ff = ss.flags();
    ff |= ios_base::floatfield;
    ff |= ios_base::fixed;
    ss.flags(ff);
    ss.precision(prec);
    ss << num;
    ret = ss.str();
    return ret;
};

Maybe it's a bit complicated but it is quite complex. You create stringstream object ss, modify its flags, put a number into it with operator<<, and extract it via str(). I guess that operator>> could be used.

Also in this example the string buffer is hidden and not used explicitly. But it would be too long of a post to write about every possible aspect and use-case.

Note: I probably stole it from someone on SO and refined, but I don't have original author noted.

This is just additional information for this answer.

If you are using nginx, you would add proxy_set_header X-Real-IP $remote_addr; to the location block for the site. /etc/nginx/sites-available/www.example.com for example. Here is a example server block.

server {
    listen 80;
    listen [::]:80;

    server_name example.com www.example.com;

    location / {
        proxy_set_header  X-Real-IP  $remote_addr;
        proxy_pass http://127.0.1.1:3080;
        proxy_http_version 1.1;
        proxy_set_header Upgrade $http_upgrade;
        proxy_set_header Connection 'upgrade';
        proxy_set_header Host $host;
        proxy_cache_bypass $http_upgrade;
    }
}

After restarting nginx, you will be able to access the ip in your node/express application routes with req.headers['x-real-ip'] || req.connection.remoteAddress;

Here is my simple version. Every time you hit a key, delete all from console and draw as many '*' as the length of password string is.

int chr = 0;
string pass = "";
const int ENTER = 13;
const int BS = 8;

do
{
   chr = Console.ReadKey().KeyChar;
   Console.Clear(); //imediately clear the char you printed

   //if the char is not 'return' or 'backspace' add it to pass string
   if (chr != ENTER && chr != BS) pass += (char)chr;

   //if you hit backspace remove last char from pass string
   if (chr == BS) pass = pass.Remove(pass.Length-1, 1);

   for (int i = 0; i < pass.Length; i++)
   {
      Console.Write('*');
   }
} 
 while (chr != ENTER);

Console.Write("\n");
Console.Write(pass);

Console.Read(); //just to see the pass

If you pushed a file or folder before it was in .gitignore (or had no .gitignore):

• Comment it out from .gitignore
• Add it back on the filesystem
• Remove it from the folder
• git add your file && commit it
• git push

So with the code you have provided.

var bytes = Convert.FromBase64String(resizeImage.Content);
using (var imageFile = new FileStream(filePath, FileMode.Create))
{
    imageFile.Write(bytes ,0, bytes.Length);
    imageFile.Flush();
}

add this to your coursel div :

data-interval="false"

Because % is only defined for integer types. That's the modulus operator.

5.6.2 of the standard:

The operands of * and / shall have arithmetic or enumeration type; the operands of % shall have integral or enumeration type. [...]

As Oli pointed out, you can use fmod(). Don't forget to include math.h.

prefer this links about properties in objective-c in iOS...

https://techguy1996.blogspot.com/2020/02/properties-in-objective-c-ios.html

Or you can use a android.support.v7.widget.CardView like so:

<android.support.v7.widget.CardView
    xmlns:card_view="http://schemas.android.com/apk/res-auto"
    android:layout_width="wrap_content"
    android:layout_height="wrap_content"
    card_view:cardBackgroundColor="@color/white"
    card_view:cardCornerRadius="4dp">

    <!--YOUR CONTENT-->
</android.support.v7.widget.CardView>

These functions work well

In Java

private void setActionbarTextColor(ActionBar actBar, int color) {

    String title = actBar.getTitle().toString();
    Spannable spannablerTitle = new SpannableString(title);
    spannablerTitle.setSpan(new ForegroundColorSpan(color), 0, spannablerTitle.length(), Spannable.SPAN_EXCLUSIVE_EXCLUSIVE);
    actBar.setTitle(spannablerTitle);

}

then to use it just feed it your action bar and the new color i.e.

ActionBar actionBar = getActionBar();    // Or getSupportActionBar() if using appCompat
int red = Color.RED
setActionbarTextColor(actionBar, red);

In Kotlin

You can use an extension function like this:

private fun ActionBar.setTitleColor(color: Int) {
    val text = SpannableString(title ?: "")
    text.setSpan(ForegroundColorSpan(color),0,text.length, Spannable.SPAN_INCLUSIVE_INCLUSIVE)
    title = text
}

And then apply to your ActionBar with

actionBar?.setTitleColor(Color.RED)

It looks like you were on the right track - get_FOO_display() is most certainly what you want:

In templates, you don't include () in the name of a method. Do the following:

{{ person.get_gender_display }}

I ended up overriding Fragment.onResume() and grabbing the attributes from the underlying dialog, then setting width/height params there. I set the outermost layout height/width to match_parent. Note that this code seems to respect the margins I defined in the xml layout as well.

@Override
public void onResume() {
    super.onResume();
    ViewGroup.LayoutParams params = getDialog().getWindow().getAttributes();
    params.width = LayoutParams.MATCH_PARENT;
    params.height = LayoutParams.MATCH_PARENT;
    getDialog().getWindow().setAttributes((android.view.WindowManager.LayoutParams) params);
}

Just call moment as a function without any arguments:

moment()

For timezone information with moment, look at the moment-timezone package: http://momentjs.com/timezone/

Here is my implementation

Implementation 1

Object.prototype.access = function() {
    var ele = this[arguments[0]];
    if(arguments.length === 1) return ele;
    return ele.access.apply(ele, [].slice.call(arguments, 1));
}

Implementation 2 (using array reduce instead of slice)

Object.prototype.access = function() {
    var self = this;
    return [].reduce.call(arguments,function(prev,cur) {
        return prev[cur];
    }, self);
}

Examples:

var myobj = {'a':{'b':{'c':{'d':'abcd','e':[11,22,33]}}}};

myobj.access('a','b','c'); // returns: {'d':'abcd', e:[0,1,2,3]}
myobj.a.b.access('c','d'); // returns: 'abcd'
myobj.access('a','b','c','e',0); // returns: 11

it can also handle objects inside arrays as for

var myobj2 = {'a': {'b':[{'c':'ab0c'},{'d':'ab1d'}]}}
myobj2.access('a','b','1','d'); // returns: 'ab1d'

Writing your own CSV code is probably a waste of your time, just use a package such as league/csv - it deals with all the difficult stuff for you, the documentation is good and it's very stable / reliable:

http://csv.thephpleague.com/

You'll need to be using composer. If you don't know what composer is I highly recommend you have a look: https://getcomposer.org/

If you're interested in creating a new dict without using intermediary storage: (this is faster, and in my opinion, cleaner than using dict.items())

dic2 = dict(dic0, **dic1)

Or if you're happy to use one of the existing dicts:

dic0.update(dic1)

There is also solution with system insets, but it works only with API >= 21 (Android L). Say you have BottomNavigationView, which is child of LinearLayout and you need to hide it when keyboard is shown:

> LinearLayout
  > ContentView
  > BottomNavigationView

All you need to do is to extend LinearLayout in such way:

public class KeyboardAwareLinearLayout extends LinearLayout {
    public KeyboardAwareLinearLayout(Context context) {
        super(context);
    }

    public KeyboardAwareLinearLayout(Context context, @Nullable AttributeSet attrs) {
        super(context, attrs);
    }

    public KeyboardAwareLinearLayout(Context context,
                                     @Nullable AttributeSet attrs, int defStyleAttr) {
        super(context, attrs, defStyleAttr);
    }

    public KeyboardAwareLinearLayout(Context context, AttributeSet attrs,
                                     int defStyleAttr, int defStyleRes) {
        super(context, attrs, defStyleAttr, defStyleRes);
    }

    @Override
    public WindowInsets onApplyWindowInsets(WindowInsets insets) {
        int childCount = getChildCount();
        for (int index = 0; index < childCount; index++) {
            View view = getChildAt(index);
            if (view instanceof BottomNavigationView) {
                int bottom = insets.getSystemWindowInsetBottom();
                if (bottom >= ViewUtils.dpToPx(200)) {
                    // keyboard is shown
                    view.setVisibility(GONE);
                } else {
                    // keyboard is hidden
                    view.setVisibility(VISIBLE);
                }
            }
        }
        return insets;
    }
}

The idea is that when keyboard is shown, system insets are changed with pretty big .bottom value.

ResourceManager will work if your image is in a resource file. If it is just a file in your project (let's say the root) you can get it using something like this:

System.Reflection.Assembly assembly = System.Reflection.Assembly.GetExecutingAssembly();
System.IO.Stream file = assembly .GetManifestResourceStream("AssemblyName." + channel);
this.pictureBox1.Image = Image.FromStream(file);

Or if you're in WPF:

    private ImageSource GetImage(string channel)
    {
        StreamResourceInfo sri = Application.GetResourceStream(new Uri("/TestApp;component/" + channel, UriKind.Relative));
        BitmapImage bmp = new BitmapImage();
        bmp.BeginInit();
        bmp.StreamSource = sri.Stream;
        bmp.EndInit();

        return bmp;
    }

That is the release version of RHEL, or at least the release of RHEL from which the package supplying /etc/redhat-release was installed. A file like that is probably the closest you can come; you could also look at /etc/lsb-release.

It is theoretically possible to have packages installed from a mix of versions (e.g. upgrading part of the system to 5.5 while leaving other parts at 5.4), so if you depend on the versions of specific components you will need to check for those individually.

import cv2 as cv

im_color = cv.imread("lena.png", cv.IMREAD_COLOR)
im_gray = cv.cvtColor(im_color, cv.COLOR_BGR2GRAY)

At this point you have a color and a gray image. We are dealing with 8-bit, uint8 images here. That means the images can have pixel values in the range of [0, 255] and the values have to be integers.

Let's do a binary thresholding operation. It creates a black and white masked image. The black regions have value 0 and the white regions 255

_, mask = cv.threshold(im_gray, thresh=180, maxval=255, type=cv.THRESH_BINARY)
im_thresh_gray = cv.bitwise_and(im_gray, mask)

The mask can be seen below on the left. The image on it's right is the result of applying bitwise_and operation between the gray image and the mask. What happened is, the spatial locations where the mask had a pixel value zero (black), became pixel value zero in the result image. The locations where the mask had pixel value 255 (white), the resulting image retained it's original gray value.

To apply this mask to our original color image, we need to convert the mask into a 3 channel image as the original color image is a 3 channel image.

mask3 = cv.cvtColor(mask, cv.COLOR_GRAY2BGR)  # 3 channel mask

Then, we can apply this 3 channel mask to our color image using the same bitwise_and function.

im_thresh_color = cv.bitwise_and(im_color, mask3)

mask3 from the code is the image below on the left, and im_thresh_color is on its right.

You can plot the results and see for yourself.

cv.imshow("original image", im_color)
cv.imshow("binary mask", mask)
cv.imshow("3 channel mask", mask3)
cv.imshow("im_thresh_gray", im_thresh_gray)
cv.imshow("im_thresh_color", im_thresh_color)
cv.waitKey(0)

The original image is lenacolor.png that I found here.

The problem is that Elements only takes the direct child elements of whatever you call it on. If you want all descendants, use the Descendants method:

var query = from c in xmlFile.Descendants("Band")

Just for fun I took Siddharth's suggestion of separate textboxes and did comboboxes. If anybody's interested, add a userform with three comboboxes named cboDay, cboMonth and cboYear and arrange them left to right. Then paste the code below into the UserForm's code module. The required combobox properties are set in UserFormInitialization, so no additional prep should be required.

The tricky part is changing the day when it becomes invalid because of a change in year or month. This code just resets it to 01 when that happens and highlights cboDay.

I haven't coded anything like this in a while. Hopefully it will be of interest to somebody, someday. If not it was fun!

Dim Initializing As Boolean

Private Sub UserForm_Initialize()
Dim i As Long
Dim ctl As MSForms.Control
Dim cbo As MSForms.ComboBox

Initializing = True
With Me
    With .cboMonth
        '        .AddItem "month"
        For i = 1 To 12
            .AddItem Format(i, "00")
        Next i
        .Tag = "DateControl"
    End With
    With .cboDay
        '        .AddItem "day"
        For i = 1 To 31
            .AddItem Format(i, "00")
        Next i
        .Tag = "DateControl"
    End With
    With .cboYear
        '        .AddItem "year"
        For i = Year(Now()) To Year(Now()) + 12
            .AddItem i
        Next i
        .Tag = "DateControl"
    End With
    DoEvents
    For Each ctl In Me.Controls
        If ctl.Tag = "DateControl" Then
            Set cbo = ctl
            With cbo
                .ListIndex = 0
                .MatchRequired = True
                .MatchEntry = fmMatchEntryComplete
                .Style = fmStyleDropDownList
            End With
        End If
    Next ctl
End With
Initializing = False
End Sub

Private Sub cboDay_Change()
If Not Initializing Then
    If Not IsValidDate Then
        ResetMonth
    End If
End If
End Sub

Private Sub cboMonth_Change()
If Not Initializing Then
    ResetDayList
    If Not IsValidDate Then
        ResetMonth
    End If
End If
End Sub

Private Sub cboYear_Change()
If Not Initializing Then
    ResetDayList
    If Not IsValidDate Then
        ResetMonth
    End If
End If
End Sub

Function IsValidDate() As Boolean
With Me
    IsValidDate = IsDate(.cboMonth & "/" & .cboDay & "/" & .cboYear)
End With
End Function
Sub ResetDayList()
Dim i As Long
Dim StartDay As String

With Me.cboDay
    StartDay = .Text
    For i = 31 To 29 Step -1
        On Error Resume Next
        .RemoveItem i - 1
        On Error GoTo 0
    Next i
    For i = 29 To 31
        If IsDate(Me.cboMonth & "/" & i & "/" & Me.cboYear) Then
            .AddItem Format(i, "0")
        End If
    Next i
    On Error Resume Next
    .Text = StartDay
    If Err.Number <> 0 Then
        .SetFocus
        .ListIndex = 0
    End If
End With
End Sub

Sub ResetMonth()
Me.cboDay.ListIndex = 0
End Sub

Yes, and its proposal is Stage 4 now. This means that the proposal is ready for inclusion in the formal ECMAScript standard. You can already use it in recent desktop versions of Chrome, Edge and Firefox, but we will have to wait for a bit longer until this feature reaches cross-browser stability.

Have a look at the following example to demonstrate its behavior:

// note: this will work only if you're running latest versions of aforementioned browsers_x000D_
const var1 = undefined;_x000D_
const var2 = "fallback value";_x000D_
_x000D_
const result = var1 ?? var2;_x000D_
console.log(`Nullish coalescing results in: ${result}`);

Previous example is equivalent to:

const var1 = undefined;_x000D_
const var2 = "fallback value";_x000D_
_x000D_
const result = (var1 !== null && var1 !== undefined) ?_x000D_
    var1 :_x000D_
    var2;_x000D_
console.log(`Nullish coalescing results in: ${result}`);

Note that nullish coalescing will not threat falsy values the way the || operator did (it only checks for undefined or null values), hence the following snippet will act as follows:

// note: this will work only if you're running latest versions of aforementioned browsers_x000D_
const var1 = ""; // empty string_x000D_
const var2 = "fallback value";_x000D_
_x000D_
const result = var1 ?? var2;_x000D_
console.log(`Nullish coalescing results in: ${result}`);

For TypesScript users, starting off TypeScript 3.7, this feature is also available now.

Your question was already answered by the other posters, I'd just like to point out that

 delete from table where id <> 2

(or variants thereof, not id = 2 etc) will not delete rows where id is NULL.

If you also want to delete rows with id = NULL:

delete from table where id <> 2 or id is NULL

You could delete all the rows from all tables using an approach like Rubens suggested, or you could just drop and recreate all the tables. Always a good idea to have the full db creation scripts anyway so that may be the easiest/quickest method.

this worked for me in Sept 2015 - Hope this helps someone out there.

// 1
    var nav = self.navigationController?.navigationBar
    // 2 set the style 
    nav?.barStyle = UIBarStyle.Black
    nav?.tintColor = UIColor.yellowColor()
    // 3
    let imageView = UIImageView(frame: CGRect(x: 0, y: 0, width: 40, height: 40))
    imageView.contentMode = .ScaleAspectFit
    // 4
    let image = UIImage(named: "logo.png")
    imageView.image = image
    // 5
    navigationItem.titleView = imageView

Below are two methods that are superior to looping. Both handle a "no-find" case.

1. The VBA equivalent of a normal function VLOOKUP with error-handling if the variable doesn't exist (INDEX/MATCH may be a better route than VLOOKUP, ie if your two columns A and B were in reverse order, or were far apart)

2. VBAs FIND method (matching a whole string in column A given I use the xlWhole argument)

   Sub Method1()
   Dim strSearch As String
   Dim strOut As String
   Dim bFailed As Boolean
   
   strSearch = "trees"
   
   On Error Resume Next
   strOut = Application.WorksheetFunction.VLookup(strSearch, Range("A:B"), 2, False)
   If Err.Number <> 0 Then bFailed = True
   On Error GoTo 0
   
   If Not bFailed Then
   MsgBox "corresponding value is " & vbNewLine & strOut
   Else
   MsgBox strSearch & " not found"
   End If
   End Sub
   
   Sub Method2()
       Dim rng1 As Range
       Dim strSearch As String
       strSearch = "trees"
       Set rng1 = Range("A:A").Find(strSearch, , xlValues, xlWhole)
       If Not rng1 Is Nothing Then
           MsgBox "Find has matched " & strSearch & vbNewLine & "corresponding cell is " & rng1.Offset(0, 1)
       Else
           MsgBox strSearch & " not found"
       End If
   End Sub
   

You can groupby on all the columns and call size the index indicates the duplicate values:

In [28]:
df.groupby(df.columns.tolist(),as_index=False).size()

Out[28]:
one    three  two  
False  False  True     1
True   False  False    2
       True   True     1
dtype: int64

I usually have it on a Makefile:

# All .js compiled into a single one.
compiled=./path/of/js/main.js

compile:
    @find ./path/of/js -type f -name "*.js" | xargs cat > $(compiled)

Then you run:

make compile

I hope it helps.

Since the glyphicons image is a sprite, you really can't do that: fundamentally what you want is to limit the size of the background, but there's no way to specify how big the background is. Either you cut out the icon you want, size it down and use it, or use something like the input field prepend/append option (http://twitter.github.io/bootstrap/base-css.html#forms and then search for prepended inputs).

Boost Getting Started

If you want to use it from Boost TR1 instead

shared_ptr Example