Computing the Parity

Question

I don\'t fully understand this algorithm of calculating the parity bit. Can someone please explain in detail?

The following code is taken from the \'Hacker\'s Deligh

Answer 1

If x had only 1 bit, clearly ((x ^ (x >> 1)) & 1 would calculate the parity (just xor the bits with each other).

This pattern can be extended to more bits.

If you have 4 bits, you get (at least, this is one way to do it)

y = x ^ (x >> 1);
y = y ^ (y >> 2);
return y & 1;

where the bits do this:

x = a     b     c     d
y = a   a^b   b^c    c^d
y = a  a^b  a^b^c  a^b^c^d

If you extend the pattern all the way to 32 bits you get the code you showed in the question.