Swap bits in a given number

Given a number x and two positions (from right side) in binary representation of x, write a function that swaps n bits at given two positions and returns the result. It is also given that the two sets of bits do not overlap.



Let p1 and p2 be the two given positions.

Example 1
Input:
x = 47 (00101111)
p1 = 1 (Start from second bit from right side)
p2 = 5 (Start from 6th bit from right side)
n = 3 (No of bits to be swapped)
Output:
227 (11100011)
The 3 bits starting from the second bit (from right side) are
swapped with 3 bits starting from 6th position (from right side)

Example 2
Input:
x = 28 (11100)
p1 = 0 (Start from first bit from right side)
p2 = 3 (Start from 4th bit from right side)
n = 2 (No of bits to be swapped)
Output:
7 (00111)
The 2 bits starting from 0th postion (from right side) are
swapped with 2 bits starting from 4th position (from right side)

Solution
We need to swap two sets of bits. XOR can be used in a similar way as it is used to swap 2 numbers. Following is the algorithm.

1) Move all bits of first set to rightmost side
   set1 =  (x >> p1) & ((1U << n) - 1)
Here the expression (1U << n) - 1 gives a number that 
contains last n bits set and other bits as 0. We do & 
with this expression so that bits other than the last 
n bits become 0.
2) Move all bits of second set to rightmost side
   set2 =  (x >> p2) & ((1U << n) - 1)
3) XOR the two sets of bits
   xor = (set1 ^ set2) 
4) Put the xor bits back to their original positions. 
   xor = (xor << p1) | (xor << p2)
5) Finally, XOR the xor with original number so 
   that the two sets are swapped.
   result = x ^ xor

Implementation:

C

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// C Program to swap bits 
// in a given number
#include<stdio.h>
  
int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n)
{
    /* Move all bits of first set to rightmost side */
    unsigned int set1 =  (x >> p1) & ((1U << n) - 1);
  
    /* Move all bits of second set to rightmost side */
    unsigned int set2 =  (x >> p2) & ((1U << n) - 1);
  
    /* XOR the two sets */
    unsigned int xor = (set1 ^ set2);
  
    /* Put the xor bits back to their original positions */
    xor = (xor << p1) | (xor << p2);
  
    /* XOR the 'xor' with the original number so that the 
       two sets are swapped */
    unsigned int result = x ^ xor;
  
    return result;
}
  
/* Driver program to test above function*/
int main()
{
    int res =  swapBits(28, 0, 3, 2);
    printf("\nResult = %d ", res);
    return 0;
}

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Java

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//Java Program to swap bits 
// in a given number
  
class GFG {
      
    static int swapBits(int x, int p1, int p2, int n)
    {
        // Move all bits of first set
        // to rightmost side 
        int set1 = (x >> p1) & ((1 << n) - 1);
      
        // Move all bits of second set 
        //to rightmost side 
        int set2 = (x >> p2) & ((1 << n) - 1);
      
        // XOR the two sets 
        int xor = (set1 ^ set2);
      
        // Put the xor bits back to 
        // their original positions 
        xor = (xor << p1) | (xor << p2);
      
        // XOR the 'xor' with the original number 
        // so that the  two sets are swapped 
        int result = x ^ xor;
      
        return result;
    }
      
    // Driver program 
    public static void main(String[] args)
    {
        int res = swapBits(28, 0, 3, 2);
        System.out.println("Result = " + res);
    }
}
  
// This code is contributed by prerna saini.

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Python3

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# Python program to
# swap bits in a given number
  
def swapBits(x,p1,p2,n):
  
    # Move all bits of first
    # set to rightmost side 
    set1 =  (x >> p1) & ((1<< n) - 1)
   
    # Moce all bits of second
    # set to rightmost side 
    set2 =  (x >> p2) & ((1 << n) - 1)
   
    # XOR the two sets 
    xor = (set1 ^ set2)
   
    # Put the xor bits back
    # to their original positions 
    xor = (xor << p1) | (xor << p2)
   
      # XOR the 'xor' with the
      # original number so that the 
      # two sets are swapped
    result = x ^ xor
   
    return result
      
# Driver code
  
res =swapBits(28, 0, 3, 2)
print("Result =",res)
  
# This code is contributed
# by Anant Agarwal.

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C#

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// C# Program to swap bits 
// in a given number
using System;
  
class GFG {
      
    static int swapBits(int x, int p1, int p2, int n)
    {
        // Move all bits of first 
        //set to rightmost side 
        int set1 = (x >> p1) & ((1 << n) - 1);
      
        // Move all bits of second set 
        // set to rightmost side 
        int set2 = (x >> p2) & ((1 << n) - 1);
      
        // XOR the two sets 
        int xor = (set1 ^ set2);
      
        // Put the xor bits back to 
        // their original positions 
        xor = (xor << p1) | (xor << p2);
      
        // XOR the 'xor' with the original number 
        // so that the two sets are swapped 
        int result = x ^ xor;
      
        return result;
    }
      
    // Driver program 
    public static void Main()
    {
        int res = swapBits(28, 0, 3, 2);
        Console.WriteLine("Result = " + res);
    }
}
  
// This code is contributed by vt_m.

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PHP

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<?php
// PHP Program to swap bits 
// in a given number
  
// function returns
// the swapped bits
function swapBits($x, $p1, $p2, $n)
{
      
    // Move all bits of first 
    // set to rightmost side 
    $set1 = ($x >> $p1) & 
            ((1 << $n) - 1);
  
    // Move all bits of second
    // set to rightmost side 
    $set2 = ($x >> $p2) & 
            ((1 << $n) - 1);
  
    // XOR the two sets 
    $xor = ($set1 ^ $set2);
  
    // Put the xor bits back to 
    // their original positions 
    $xor = ($xor << $p1) | 
           ($xor << $p2);
  
    // XOR the 'xor' with the 
    // original number so that
    // the two sets are swapped 
    $result = $x ^ $xor;
  
    return $result;
}
  
    // Driver Code
    $res = swapBits(28, 0, 3, 2);
    echo "\nResult = ", $res;
      
// This code is contributed by anuj_67.
?>

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Output:

 Result = 7

Following is a shorter implementation of the same logic

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int swapBits(unsigned int x, unsigned int p1, unsigned int p2, unsigned int n)
{
    /* xor contains xor of two sets */
    unsigned int xor = ((x >> p1) ^ (x >> p2)) & ((1U << n) - 1);
  
    /* To swap two sets, we need to again XOR the xor with original sets */
    return x ^ ((xor << p1) | (xor << p2));
}

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References:
Swapping individual bits with XOR

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Improved By : vt_m



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