calculate number of bits set in byte

For 8-bit values, just use a 256-element lookup table.

For larger sized inputs, it's slightly less trivial. Sean Eron Anderson has several different functions for this on his Bit Twiddling Hacks page, all with different performance characteristics. There is not one be-all-end-all-fastest version, since it depends on the nature of your processor (pipeline depth, branch predictor, cache size, etc.) and the data you're using.

Why not just use the standard library? That way the optimal way should be determined by the implementation, and is likely better than any standards compliant code that you can actually write. For instance, if you're on an x86 this compiles to a single instruction but only if you're targeting CPUs that support it.

#include <bitset>
#include <iostream>

int main() {
  unsigned char bitfield = 17;
  std::cout << std::bitset<8>(bitfield).count() <<
    std::endl;
}

For one byte of data, the optimal way considering both speed and memory consumption:

uint8_t count_ones (uint8_t byte)
{
  static const uint8_t NIBBLE_LOOKUP [16] =
  {
    0, 1, 1, 2, 1, 2, 2, 3, 
    1, 2, 2, 3, 2, 3, 3, 4
  };


  return NIBBLE_LOOKUP[byte & 0x0F] + NIBBLE_LOOKUP[byte >> 4];
}

Calling this function from a for loop should yield quite an efficient program on most systems. And it is very generic.