This is a solution that an interview candidate gave when I asked him to write a function to count the number of 1-bits set in a byte. The interview was for a position at Nimble Storage.

nbits.c

#include <stdio.h>
#include <stdlib.h>

int nbits(unsigned char value);

int testtab[] = {
   0x53,
   0xaa,
   0x55,
   0x01,
   0x00,
   0xf0,
   0xff,
   0xef,
   0x33,
   0x03
};
#define NTESTS (sizeof testtab / sizeof testtab[0])

int main(int argc, char **argv)
{
   int i;
   for (i = 0; i < NTESTS; i++) {
      printf("nbit(0x%0.2x) == %d\n", testtab[i], nbits(testtab[i]));
   }
   exit(0);
}

/*
 * Count the number of 1-bits set in a byte.
 */
int nbits(unsigned char value)
{
   value = ((value & 0xaa) >> 1) + (value & 0x55);
   value = ((value & 0xcc) >> 2) + (value & 0x33);
   value = ((value & 0xf0) >> 4) + (value & 0x0f);
   return value;
}