Swap bits in a given number

reference: 

http://www.geeksforgeeks.org/swap-bits-in-a-given-number/

Problem Definition:

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.

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

Code:

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;
}