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Compute the integer absolute value (abs) without branching

We need not to do anything if a number is positive. We want to change only negative numbers. Since negative numbers are stored in 2’s complement form, to get the absolute value of a negative number we have to toggle bits of the number and add 1 to the result.

For example -2 in a 8 bit system is stored as follows 1 1 1 1 1 1 1 0 where leftmost bit is the sign bit. To get the absolute value of a negative number, we have to toggle all bits and add 1 to the toggled number i.e, 0 0 0 0 0 0 0 1 + 1 will give the absolute value of 1 1 1 1 1 1 1 0. Also remember, we need to do these operations only if the number is negative (sign bit is set).

Method 1
1) Set the mask as right shift of integer by 31 (assuming integers are stored using 32 bits).

 mask = n>>31 

2) For negative numbers, above step sets mask as 1 1 1 1 1 1 1 1 and 0 0 0 0 0 0 0 0 for positive numbers. Add the mask to the given number.

 mask + n 

3) XOR of mask +n and mask gives the absolute value.



 (mask + n)^mask 

Implementation:

C++

#include <bits/stdc++.h>
using namespace std;
#define CHARBIT 8 
  
/* This function will return absoulte value of n*/
unsigned int getAbs(int n) 
    int const mask = n >> (sizeof(int) * CHARBIT - 1); 
    return ((n + mask) ^ mask); 
  
/* Driver program to test above function */
int main() 
    int n = -6; 
    cout << "Absoute value of " << n << " is " << getAbs(n); 
    return 0; 
  
  
// This code is contributed by rathbhupendra

C

#include <stdio.h>
#define CHAR_BIT 8
  
/* This function will return absoulte value of n*/
unsigned int getAbs(int n)
{
  int const mask = n >> (sizeof(int) * CHAR_BIT - 1);
  return ((n + mask) ^ mask);
}
  
/* Driver program to test above function */
int main()
{
  int n = -6;
  printf("Absoute value of %d is %u", n, getAbs(n));
  
  getchar();
  return 0;
}

Java

// Java implementation of above approach
class GFG {
  
    static final int CHAR_BIT = 8;
    static final int SIZE_INT = 8;
  
    /* This function will return absoulte value of n*/
    static int getAbs(int n) 
    {
        int mask = n >> (SIZE_INT * CHAR_BIT - 1);
        return ((n + mask) ^ mask);
    }
  
    /* Driver code */
    public static void main(String[] args) 
    {
        int n = -6;
        System.out.print("Absoute value of " +
                            n + " is " + getAbs(n));
    }
}
  
//This code is contributed by Rajput-Ji



Method 2:
1) Set the mask as right shift of integer by 31 (assuming integers are stored using 32 bits).

 mask = n>>31 

2) XOR the mask with number

 mask ^ n 

3) Subtract mask from result of step 2 and return the result.

 (mask^n) - mask 

Implementation:

/* This function will return absoulte value of n*/
unsigned int getAbs(int n)
{
  int const mask = n >> (sizeof(int) * CHAR_BIT - 1);
  return ((n ^ mask) - mask);
}

On machines where branching is expensive, the above expression can be faster than the obvious approach, r = (v < 0) ? -(unsigned)v : v, even though the number of operations is the same.

Please see this for more details about the above two methods.

Please write comments if you find any of the above explanations/algorithms incorrect, or a better ways to solve the same problem.

References:
http://graphics.stanford.edu/~seander/bithacks.html#IntegerAbs



This article is attributed to GeeksforGeeks.org

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