 Do XOR between the numbers which then gives the difference bits as '1'
 Calculate those difference no.of bits by bitEND operation on preceding numbers until it is zero
...............
public int HammingDistance(int x, int y) {
int z = x ^ y;
int n = 0;
while(z > 0)
{
z = z & (z1);
++n;
}
return n;
}
...............
C#  Simple solution


// XOR returns no.of bits that are not same at corresponding positions
// 2 0010 and 6  0110. Both integers differ at positions (left to right) 3. 2 ^ 6 = 0010 ^ 0110 = 0100
// You may notice the 1 in the result.
// Now if we could calculate all those 1's from the result above we get the hamming distance// Any two consecutive numbers when written in base2 always differ by either
// only one bit or differ by all bits  Ex 10000 (16) and 1111(15) differ by all bits
// (ignore leading zeros) and 1111 (15) 1110(14) differ by 1.Doing & operation on the result
//continuously will exhaust all 1s and leads to zero. If we want to calculate number of bits in
//4...0100(4) & 0011(3) becomes zero by then n is incremented by 1 which is our answer

@krishnacs XOR does NOT return no.of bits that are not same at corresponding positions. Try it with 7 ^ 8

@Aevil1 Hi Aevil1,
XOR returns no.of bits that are not same.0 ^ 0  0
0 ^ 1  1
1 ^ 0  1
1 ^ 1  0Your case:
7 > 0 1 1 1 8 > 1 0 0 0  7^8 >1 1 1 1
You may notice 7 and 8 have 4 bits different. When you do AND operation on this result until it is zero, we get the count of set bits.

@krishnacs : what about difference between 96 (1100000) and 95 (1011111)? It has only 6 differences. can you help me understand it?