C code to count the number of '1' bits in an unsigned char

Question

I need C code to return the number of 1's in an unsigned char in C. I need an explanation as to why it works if it's not obvious. I've found a lot of code for a 32-bit number but not much for an unsigned char.

Answer 1

const unsigned char oneBits[] = {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};

unsigned char CountOnes(unsigned char x)
{
    unsigned char results;
    results = oneBits[x&0x0f];
    results += oneBits[x>>4];
    return results
}

Have an array that knows the number of bits for 0 through 15. Add the results for each nibble.

Answer 2

The same code will work for an unsigned char. Loop over all bits testing them. See this.