Question

I have to create a function bitParity(int x) that takes an integer and returns 1 if there is an odd number of 0‘s in the bit form of x, and 0 otherwise.

Ex: bitParity(5) = 0, bitParity(7) = 1

However, this is difficult as I can only use bit operators on this problem (! ˜ & ˆ | + << >> are the only legal ones). That means, no loops, if-then, or anything of the sort. Constants can be used.

So far, what I have doesn’t work, but I figured that I should shift the bits of the integer 16, 8, and 4 times and XOR the remaining integers.

Can anyone offer some advice? Thanks.

Answer 1

This is properly solved with a loop. But here is a way to do it without.

x = (x & 0x0000FFFF) ^ (x >> 16)
x = (x & 0x000000FF) ^ (x >> 8)
x = (x & 0x0000000F) ^ (x >> 4)
x = (x & 0x00000003) ^ (x >> 2)
x = (x & 0x00000001) ^ (x >> 1)

Edit: I don’t need the &. A better version:

x ^= x >> 16
x ^= x >> 8
x ^= x >> 4
x ^= x >> 2
x ^= x >> 1
x &= 1;