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.

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Examples:

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:

#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);
 
    /* Moce 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;
}
 
/* Drier program to test above function*/
int main()
{
    int res =  swapBits(28, 0, 3, 2);
    printf("\nResult = %d ", res);
    return 0;
}

Output:

Result = 7

Following is a shorter implementation of the same logic