Reverse Contents in Array

Question

I have an array of numbers that I am trying to reverse  I believe the function in my code is correct  but I cannot get the proper output   The output reads  10 9 8 7 6  Why can t I get the other half of the numbers  When I remove the   2  from count  the output reads  10 9 8 7 6 6 7 8 9 10  void reverse int     int    int main         const int SIZE   10     int arr  SIZE     1  2  3  4  5  6  7  8  9  10       reverse arr  SIZE      return 0    void reverse int arr    int count       int temp     for  int i   0  i  lt  count 2    i             arr i    temp        temp   arr count-i-1         arr count-i-1    arr i         arr i    temp         cout  lt  lt  temp  lt  lt

User · Answer

First of all what value do you have in this pice of code? int temp;? You can't tell because in every single compilation it will have different value - you should initialize your value to not have trash value from memory. Next question is: why you assign this temp value to your array? If you want to stick with your solution I would change reverse function like this:

void reverse(int arr[], int count)
{
    int temp = 0;
    for (int i = 0; i < count/2; ++i)
    {
        temp = arr[count - i - 1];
        arr[count - i - 1] = arr[i];
        arr[i] = temp;
    }

    for (int i = 0; i < count; ++i)
    {
        std::cout << arr[i] << " ";
    }
}

Now it will works but you have other options to handle this problem.

Solution using pointers:

void reverse(int arr[], int count)
{
    int* head = arr;
    int* tail = arr + count - 1;
    for (int i = 0; i < count/2; ++i)
    {
        if (head < tail)
        {
            int tmp = *tail;
            *tail = *head;
            *head = tmp;

            head++; tail--;
        }
    }

    for (int i = 0; i < count; ++i)
    {
        std::cout << arr[i] << " ";
    }
}

And ofc like Carlos Abraham says use build in function in algorithm library

User · Answer

First of all you assign temp to array elemets and you should remove arr i    temp  statment  Next problem is that you are printing temp variable which shows only half of array elements in your for loop   If you don t want to use STL vectors I would suggest this solution       include  lt iostream gt   void reverseArray int userArray    int size    void printArray int userArray    int size    int main int arg  char  argv         int arr    1 2 3 4 5 6 7 8 9 10        int sizeOfArray   sizeof arr    sizeof arr 0         reverseArray arr  sizeOfArray       printArray arr  sizeOfArray        system  pause        return 0      void reverseArray int userArray    int size         int  ptrHead   userArray      int  ptrTail   userArray    size-1        while  ptrTail  gt  ptrHead            int temp    ptrHead           ptrHead    ptrTail           ptrTail   temp           ptrHead            ptrTail--           void printArray int userArray    int size        for  int i   0  i  lt  size  i              std  cout  lt  lt  userArray i   lt  lt

User · Answer

include  stdafx h   include  lt iostream gt  using namespace std   void main         int n  i      cout  lt  lt   n          cin  gt  gt  n      int  a   new int n       int  b   new int n       for  i   0  i  lt  n  i                  cout  lt  lt   a    lt  lt  i  lt  lt                 cin  gt  gt  a i             for  i   0  i  lt  n  i                  b i    a n - 1 - i             for  i   0  i  lt  n  i                  cout  lt  lt  b i

User · Answer

my approach is swapping the first and last element of the array int i j  for   i   0 j   size - 1   i  lt  j   i   j--          int temp   A i       A i    A j       A j    temp

User · Answer

As a direct answer to your question  Your swapping is wrong  void reverse int arr    int count      int temp     for int i   0  i  lt  count 2    i         arr i    temp      lt    Wrong  Should be deleted       temp   arr count-i-1         arr count-i-1    arr i         arr i    temp            assigning arr i    temp causes error when it first enters the loop as temp initially contains garbage data and will ruin your array  remove it and the code should work well   As an advice  use built-in functions whenever possible      In the swapping you could just use swap like  std  swap arr i   arr count-i-1      For the reverse as a whole just use reverse like std  reverse arr  arr count    I am using C  14 and reverse works with arrays without any problems

User · Answer

for i 0 i lt   s3  2  i                   z s2 i       s2 i  s2  s3-1 -i       s2  s3-1 -i  z

User · Answer

I would try to using pointers to solve this problem    include  lt iostream gt   void displayArray int table    int size    void rev int table    int size     int main int argc  char   argv         int a 10      1 2 3 4 5 6 7 8 9 10         rev a  10       displayArray a  10        return 0     void displayArray int table    int size        for  int i   0  i  lt  size  i              std  cout  lt  lt  table i   lt  lt                 std  cout  lt  lt  std  endl     void rev int table    int size         int  start   table      int  end   table    size - 1        for  int i   0  i  lt  size  i               if  start  lt  end                int temp    end               end    start               start   temp                     start            end--

User · Answer

The line  arr i    temp    is wrong    On the first iteration of your loop it sets arr i  to an undefined value  further iterations set it to an incorrect value   If you remove this line  your array should be reversed correctly   After that  you should move the code which prints the reversed array into a new loop which iterates over the whole list   Your current code only prints the first count 2 elements   int temp  i  for  i   0  i  lt  count 2    i        temp   arr count-i-1       arr count-i-1    arr i       arr i    temp    for  i   0  i  lt  count    i        cout  lt  lt  arr i   lt  lt

User · Answer

Both answers look correct to me    The first arr i    temp  should be removed You should do a second loop to print all elements  not just half the array  The loop that does the reverse doesn t need to print it

User · Answer

You are not printing the array  you are printing the value of temp - which is only half the array

User · Answer

You can use the following example to reverse the contents in an array    include  lt iostream gt   int main         int n  x          order value for var x     cin  gt  gt  x            create array and the value for array is value var x     int arr x           loop for insert values for array by reverse       for int i x  i  gt  0  i--                var i is number of elements in array          cin  gt  gt  n          arr i - 1    n                show element in array     for int l   0  l  lt  x  l              cout lt  lt arr l  lt  lt endl             return 0

User · Answer

The solution to this question is very easy  Vectors  std  vector lt int gt  vector  for int i   0  i  lt  10 i          vector push back i     std  reverse vector begin    vector end       Voila  You are done      Solution details   This is the most efficent solution  Swap can t swap 3 values but reverse definitely can   Remember to include algorithm   This is so simple that the compiled code is definitely not needed   I think this solves the OP s problem   If you think there are any errors and problems with this solution please comment below

User · Answer

Procedure     1 Take an array    2 Then by default function reverse array name  array name   size      reverse array name  array name   size  function exits in algorithm h header file    3 Now print the array     N B  Here we use new and delete for dynamic memory allocation      C   implementation       include lt bits stdc   h gt  using namespace std    int main        int n     cin gt  gt n      int  arr   new int n        for int i 0  i lt n  i     cin gt  gt arr i       reverse arr  arr n       for int i 0  i lt n  i       cout lt  lt arr i  lt  lt          delete   arr      return 0

User · Answer

void reverse int     int   void printarray int     int    int main          const int SIZE   10      int arr  SIZE     1  2  3  4  5  6  7  8  9  10        cout lt  lt  Before reverse n       printarray arr  SIZE       reverse arr  SIZE       cout lt  lt  After reverse n       printarray arr  SIZE        return 0     void printarray int arr    int count        for int i   0  i  lt  count    i          cout lt  lt arr i  lt  lt           cout lt  lt   n      void reverse int arr    int count       int temp     for  int i   0  i  lt  count 2    i             temp   arr i         arr i    arr count-i-1         arr count-i-1    temp

User · Answer

Your loop will run only for count 2 times  So it will not print the whole array   Also  temp ar i  should be used instead of ar i  temp as value of ar i  is not getting stored anywhere in the latter statement  hence it is getting destroyed

User · Answer

This would be my approach    include  lt algorithm gt   include  lt iterator gt   int main       const int SIZE   10    int arr  SIZE     1  2  3  4  5  6  7  8  9  10     std  reverse std  begin arr   std  end arr

User · Answer

try this can you better fell comparing to another codes   using namespace std  int main         int a 5   4 6 3 5 9       for int i 4 i gt  0 i--            cout lt  lt   n  lt  lt a i

User · Answer

I would use the reverse   function from the  lt algorithm gt  library   Run it online  repl it  abranhe Reverse-Array   include  lt iostream gt   include  lt algorithm gt  using namespace std   int main       int arr  10     1  2  3  4  5  6  7  8  9  10      reverse begin arr   end arr       for auto item arr          cout  lt  lt  item  lt  lt               Output   10 9 8 7 6 5 4 3 2 1   Hope you like this approach

[c++] Reverse Contents in Array

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