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This explains it better, can be used for swapping block of bits :
graphics.stanford.edu/~seander/bithacks.html#SwappingBitsXOR

Swapping just two bits would be too simple as this can be done simply by checking if they are same or not. If not same then set 1 to 0 and 0 to 1 simply by xoring with ( 1<<position ).
Suppose positions are p and q and integer is n:

! (((n & (1 << p)) >> p) ^ ((n & (1 << q)) >> q)):
	bits are same, no need to swap
else
	n ^= ( 1<<p )
	n ^= ( 1<<q )

suppose,we are going to swap 2 bits at position m & n of num X . if bits of position m & n are same then there is no need two swap. if bit at m position is 0 and bit at n position is 1 then add 2^m-2^n in X. if bit at m position is 1 and bit at n position is 0 then add 2^n-2^m in X.

#include<stdio.h>
int main()
{

int n=5,i=1,j=3;
if( ((n&(1<<i))>>i) ^ (n&(1<<j))>>j ){
n^=(1<<i);
n^=(1<<j);
printf("bits are not same");
printf("the value after swapping the bits:%d",n);
}
else {
printf("bits are same");
}
}

How to swape pairs of bits in number ?

#include<stdio.h>
int main()
{

int n=5,i=1,j=3;
if( ((n&(1<<i))>>i) ^ (n&(1<<j))>>j ){
n^=(1<<i);
n^=(1<<j);
printf("bits are not same");
printf("the value after swapping the bits:%d",n);
}
else {
printf("bits are same");
}
}