[java] Make a negative number positive

I have a Java method in which I'm summing a set of numbers. However, I want any negatives numbers to be treated as positives. So (1)+(2)+(1)+(-1) should equal 5.

I'm sure there is very easy way of doing this - I just don't know how.

This question is related to java negative-number

The answer is


Just call Math.abs. For example:

int x = Math.abs(-5);

Which will set x to 5.

Note that if you pass Integer.MIN_VALUE, the same value (still negative) will be returned, as the range of int does not allow the positive equivalent to be represented.


To convert negative number to positive number (this is called absolute value), uses Math.abs(). This Math.abs() method is work like this

“number = (number < 0 ? -number : number);".

In below example, Math.abs(-1) will convert the negative number 1 to positive 1.

example

public static void main(String[] args) {

    int total = 1 + 1 + 1 + 1 + (-1);
    
    //output 3
    System.out.println("Total : " + total);
    
    int total2 = 1 + 1 + 1 + 1 + Math.abs(-1);
    
    //output 5
    System.out.println("Total 2 (absolute value) : " + total2);
    
}

Output

Total : 3 Total 2 (absolute value) : 5


This code is not safe to be called on positive numbers.

int x = -20
int y = x + (2*(-1*x));
// Therefore y = -20 + (40) = 20

if(arr[i]<0)

Math.abs(arr[i]);  //1st way (taking absolute value)

arr[i]=-(arr[i]); //2nd way (taking -ve of -ve no. yields a +ve no.)

arr[i]= ~(arr[i]-1);   //3rd way (taking negation)

I needed the absolute value of a long , and looked deeply into Math.abs and found that if my argument is less than LONG.MIN_VAL which is -9223372036854775808l, then the abs function would not return an absolute value but only the minimum value. Inthis case if your code is using this abs value further then there might be an issue.


Are you asking about absolute values?

Math.abs(...) is the function you probably want.


The concept you are describing is called "absolute value", and Java has a function called Math.abs to do it for you. Or you could avoid the function call and do it yourself:

number = (number < 0 ? -number : number);

or

if (number < 0)
    number = -number;

Try this (the negative in front of the x is valid since it is a unary operator, find more here):

int answer = -x;

With this, you can turn a positive to a negative and a negative to a positive.


However, if you want to only make a negative number positive then try this:

int answer = Math.abs(x);

A little cool math trick! Squaring the number will guarantee a positive value of x^2, and then, taking the square root will get you to the absolute value of x:

int answer = Math.sqrt(Math.pow(x, 2));

Hope it helps! Good Luck!


String s = "-1139627840";
BigInteger bg1 = new BigInteger(s);
System.out.println(bg1.abs());

Alternatively:

int i = -123;
System.out.println(Math.abs(i));

Library function Math.abs() can be used.
Math.abs() returns the absolute value of the argument

  • if the argument is negative, it returns the negation of the argument.
  • if the argument is positive, it returns the number as it is.

e.g:

  1. int x=-5;
    System.out.println(Math.abs(x));

Output: 5

  1. int y=6;
    System.out.println(Math.abs(y));

Output: 6


You're looking for absolute value, mate. Math.abs(-5) returns 5...


Can you please try this one?

public static int toPositive(int number) {
    return number & 0x7fffffff;
}

Try this (the negative in front of the x is valid since it is a unary operator, find more here):

int answer = -x;

With this, you can turn a positive to a negative and a negative to a positive.


However, if you want to only make a negative number positive then try this:

int answer = Math.abs(x);

A little cool math trick! Squaring the number will guarantee a positive value of x^2, and then, taking the square root will get you to the absolute value of x:

int answer = Math.sqrt(Math.pow(x, 2));

Hope it helps! Good Luck!


I needed the absolute value of a long , and looked deeply into Math.abs and found that if my argument is less than LONG.MIN_VAL which is -9223372036854775808l, then the abs function would not return an absolute value but only the minimum value. Inthis case if your code is using this abs value further then there might be an issue.


The concept you are describing is called "absolute value", and Java has a function called Math.abs to do it for you. Or you could avoid the function call and do it yourself:

number = (number < 0 ? -number : number);

or

if (number < 0)
    number = -number;

This code is not safe to be called on positive numbers.

int x = -20
int y = x + (2*(-1*x));
// Therefore y = -20 + (40) = 20

I would recommend the following solutions:

without lib fun:

    value = (value*value)/value

(The above does not actually work.)

with lib fun:

   value = Math.abs(value);

Use the abs function:

int sum=0;
for(Integer i : container)
  sum+=Math.abs(i);

Can you please try this one?

public static int toPositive(int number) {
    return number & 0x7fffffff;
}

Why don't you multiply that number with -1?

Like This:

//Given x as the number, if x is less than 0, return 0 - x, otherwise return x:
return (x <= 0.0F) ? 0.0F - x : x;

Are you asking about absolute values?

Math.abs(...) is the function you probably want.


Try this in the for loop:

sum += Math.abs(arr[i])

The easiest, if verbose way to do this is to wrap each number in a Math.abs() call, so you would add:

Math.abs(1) + Math.abs(2) + Math.abs(1) + Math.abs(-1)

with logic changes to reflect how your code is structured. Verbose, perhaps, but it does what you want.


I see people are saying that Math.abs(number) but this method is not full proof.

This fails when you try to wrap Math.abs(Integer.MIN_VALUE) (see ref. https://youtu.be/IWrpDP-ad7g)

If you are not sure whether you are going to receive the Integer.MIN_VALUE in the input. It is always recommended to check for that number and handle it manually.


You're looking for absolute value, mate. Math.abs(-5) returns 5...


Library function Math.abs() can be used.
Math.abs() returns the absolute value of the argument

  • if the argument is negative, it returns the negation of the argument.
  • if the argument is positive, it returns the number as it is.

e.g:

  1. int x=-5;
    System.out.println(Math.abs(x));

Output: 5

  1. int y=6;
    System.out.println(Math.abs(y));

Output: 6


String s = "-1139627840";
BigInteger bg1 = new BigInteger(s);
System.out.println(bg1.abs());

Alternatively:

int i = -123;
System.out.println(Math.abs(i));

If you're interested in the mechanics of two's complement, here's the absolutely inefficient, but illustrative low-level way this is made:

private static int makeAbsolute(int number){
     if(number >=0){
        return number;
     } else{
        return (~number)+1;
     }
}

The easiest, if verbose way to do this is to wrap each number in a Math.abs() call, so you would add:

Math.abs(1) + Math.abs(2) + Math.abs(1) + Math.abs(-1)

with logic changes to reflect how your code is structured. Verbose, perhaps, but it does what you want.


if(arr[i]<0)

Math.abs(arr[i]);  //1st way (taking absolute value)

arr[i]=-(arr[i]); //2nd way (taking -ve of -ve no. yields a +ve no.)

arr[i]= ~(arr[i]-1);   //3rd way (taking negation)

Use the abs function:

int sum=0;
for(Integer i : container)
  sum+=Math.abs(i);

You're looking for absolute value, mate. Math.abs(-5) returns 5...


The concept you are describing is called "absolute value", and Java has a function called Math.abs to do it for you. Or you could avoid the function call and do it yourself:

number = (number < 0 ? -number : number);

or

if (number < 0)
    number = -number;

dont do this

number = (number < 0 ? -number : number);

or

if (number < 0) number = -number;

this will be an bug when you run find bug on your code it will report it as RV_NEGATING_RESULT_OF


Why don't you multiply that number with -1?

Like This:

//Given x as the number, if x is less than 0, return 0 - x, otherwise return x:
return (x <= 0.0F) ? 0.0F - x : x;

You want to wrap each number into Math.abs(). e.g.

System.out.println(Math.abs(-1));

prints out "1".

If you want to avoid writing the Math.-part, you can include the Math util statically. Just write

import static java.lang.Math.abs;

along with your imports, and you can refer to the abs()-function just by writing

System.out.println(abs(-1));

In kotlin you can use unaryPlus

input = input.unaryPlus()

https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-plus.html


You want to wrap each number into Math.abs(). e.g.

System.out.println(Math.abs(-1));

prints out "1".

If you want to avoid writing the Math.-part, you can include the Math util statically. Just write

import static java.lang.Math.abs;

along with your imports, and you can refer to the abs()-function just by writing

System.out.println(abs(-1));

The easiest, if verbose way to do this is to wrap each number in a Math.abs() call, so you would add:

Math.abs(1) + Math.abs(2) + Math.abs(1) + Math.abs(-1)

with logic changes to reflect how your code is structured. Verbose, perhaps, but it does what you want.


Try this in the for loop:

sum += Math.abs(arr[i])

Use the abs function:

int sum=0;
for(Integer i : container)
  sum+=Math.abs(i);

Are you asking about absolute values?

Math.abs(...) is the function you probably want.


I see people are saying that Math.abs(number) but this method is not full proof.

This fails when you try to wrap Math.abs(Integer.MIN_VALUE) (see ref. https://youtu.be/IWrpDP-ad7g)

If you are not sure whether you are going to receive the Integer.MIN_VALUE in the input. It is always recommended to check for that number and handle it manually.


Are you asking about absolute values?

Math.abs(...) is the function you probably want.


The easiest, if verbose way to do this is to wrap each number in a Math.abs() call, so you would add:

Math.abs(1) + Math.abs(2) + Math.abs(1) + Math.abs(-1)

with logic changes to reflect how your code is structured. Verbose, perhaps, but it does what you want.


When you need to represent a value without the concept of a loss or absence (negative value), that is called "absolute value".


The logic to obtain the absolute value is very simple: "If it's positive, maintain it. If it's negative, negate it".


What this means is that your logic and code should work like the following:

//If value is negative...
if ( value < 0 ) {
  //...negate it (make it a negative negative-value, thus a positive value).
  value = negate(value);
}

There are 2 ways you can negate a value:

  1. By, well, negating it's value: value = (-value);
  2. By multiplying it by "100% negative", or "-1": value = value * (-1);

Both are actually two sides of the same coin. It's just that you usually don't remember that value = (-value); is actually value = 1 * (-value);.


Well, as for how you actually do it in Java, it's very simple, because Java already provides a function for that, in the Math class: value = Math.abs(value);

Yes, doing it without Math.abs() is just a line of code with very simple math, but why make your code look ugly? Just use Java's provided Math.abs() function! They provide it for a reason!

If you absolutely need to skip the function, you can use value = (value < 0) ? (-value) : value;, which is simply a more compact version of the code I mentioned in the logic (3rd) section, using the Ternary operator (? :).


Additionally, there might be situations where you want to always represent loss or absence within a function that might receive both positive and negative values.

Instead of doing some complicated check, you can simply get the absolute value, and negate it: negativeValue = (-Math.abs(value));


With that in mind, and considering a case with a sum of multiple numbers such as yours, it would be a nice idea to implement a function:

int getSumOfAllAbsolutes(int[] values){
  int total = 0;
  for(int i=0; i<values.lenght; i++){
    total += Math.abs(values[i]);
  }
  return total;
}

Depending on the probability you might need related code again, it might also be a good idea to add them to your own "utils" library, splitting such functions into their core components first, and maintaining the final function simply as a nest of calls to the core components' now-split functions:

int[] makeAllAbsolute(int[] values){
  //@TIP: You can also make a reference-based version of this function, so that allocating 'absolutes[]' is not needed, thus optimizing.
  int[] absolutes = values.clone();
  for(int i=0; i<values.lenght; i++){
    absolutes[i] = Math.abs(values[i]);
  }
  return absolutes;
}

int getSumOfAllValues(int[] values){
  int total = 0;
  for(int i=0; i<values.lenght; i++){
    total += values[i];
  }
return total;
}

int getSumOfAllAbsolutes(int[] values){
  return getSumOfAllValues(makeAllAbsolute(values));
}

If you're interested in the mechanics of two's complement, here's the absolutely inefficient, but illustrative low-level way this is made:

private static int makeAbsolute(int number){
     if(number >=0){
        return number;
     } else{
        return (~number)+1;
     }
}

In kotlin you can use unaryPlus

input = input.unaryPlus()

https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-plus.html


To convert negative number to positive number (this is called absolute value), uses Math.abs(). This Math.abs() method is work like this

“number = (number < 0 ? -number : number);".

In below example, Math.abs(-1) will convert the negative number 1 to positive 1.

example

public static void main(String[] args) {

    int total = 1 + 1 + 1 + 1 + (-1);
    
    //output 3
    System.out.println("Total : " + total);
    
    int total2 = 1 + 1 + 1 + 1 + Math.abs(-1);
    
    //output 5
    System.out.println("Total 2 (absolute value) : " + total2);
    
}

Output

Total : 3 Total 2 (absolute value) : 5


Use the abs function:

int sum=0;
for(Integer i : container)
  sum+=Math.abs(i);

I would recommend the following solutions:

without lib fun:

    value = (value*value)/value

(The above does not actually work.)

with lib fun:

   value = Math.abs(value);

You're looking for absolute value, mate. Math.abs(-5) returns 5...


When you need to represent a value without the concept of a loss or absence (negative value), that is called "absolute value".


The logic to obtain the absolute value is very simple: "If it's positive, maintain it. If it's negative, negate it".


What this means is that your logic and code should work like the following:

//If value is negative...
if ( value < 0 ) {
  //...negate it (make it a negative negative-value, thus a positive value).
  value = negate(value);
}

There are 2 ways you can negate a value:

  1. By, well, negating it's value: value = (-value);
  2. By multiplying it by "100% negative", or "-1": value = value * (-1);

Both are actually two sides of the same coin. It's just that you usually don't remember that value = (-value); is actually value = 1 * (-value);.


Well, as for how you actually do it in Java, it's very simple, because Java already provides a function for that, in the Math class: value = Math.abs(value);

Yes, doing it without Math.abs() is just a line of code with very simple math, but why make your code look ugly? Just use Java's provided Math.abs() function! They provide it for a reason!

If you absolutely need to skip the function, you can use value = (value < 0) ? (-value) : value;, which is simply a more compact version of the code I mentioned in the logic (3rd) section, using the Ternary operator (? :).


Additionally, there might be situations where you want to always represent loss or absence within a function that might receive both positive and negative values.

Instead of doing some complicated check, you can simply get the absolute value, and negate it: negativeValue = (-Math.abs(value));


With that in mind, and considering a case with a sum of multiple numbers such as yours, it would be a nice idea to implement a function:

int getSumOfAllAbsolutes(int[] values){
  int total = 0;
  for(int i=0; i<values.lenght; i++){
    total += Math.abs(values[i]);
  }
  return total;
}

Depending on the probability you might need related code again, it might also be a good idea to add them to your own "utils" library, splitting such functions into their core components first, and maintaining the final function simply as a nest of calls to the core components' now-split functions:

int[] makeAllAbsolute(int[] values){
  //@TIP: You can also make a reference-based version of this function, so that allocating 'absolutes[]' is not needed, thus optimizing.
  int[] absolutes = values.clone();
  for(int i=0; i<values.lenght; i++){
    absolutes[i] = Math.abs(values[i]);
  }
  return absolutes;
}

int getSumOfAllValues(int[] values){
  int total = 0;
  for(int i=0; i<values.lenght; i++){
    total += values[i];
  }
return total;
}

int getSumOfAllAbsolutes(int[] values){
  return getSumOfAllValues(makeAllAbsolute(values));
}

You want to wrap each number into Math.abs(). e.g.

System.out.println(Math.abs(-1));

prints out "1".

If you want to avoid writing the Math.-part, you can include the Math util statically. Just write

import static java.lang.Math.abs;

along with your imports, and you can refer to the abs()-function just by writing

System.out.println(abs(-1));

dont do this

number = (number < 0 ? -number : number);

or

if (number < 0) number = -number;

this will be an bug when you run find bug on your code it will report it as RV_NEGATING_RESULT_OF