Answer by cipilo for What is the fastest/most efficient way to find the position of the highest set bit (msb) in an integer in C?

For a 4 bit nibble n, the value can be computed in 3 instructions by using a constant 32bit unsigned value as a lookup table with 16 entries of two bits each.

(0xffffaa50 >> (n << 1)) & 3

This can be used to trim the end of the divide and conquer methods making them faster. The following functions have a constant dependency chain length of 16 for the 32bit version or 20 for the 64bit.

size_t blog32(uint32_t v) {    size_t r = 0, t;    t = (0 != (v >> 16)) << 4; v >>= t; r |= t;    t = (0 != (v >>  8)) << 3; v >>= t; r |= t;    t = (0 != (v >>  4)) << 2; v >>= t; r |= t;    return r + ((0xffffaa50 >> (v << 1)) & 3);}

For 64bit, add one more level:

size_t blog64(uint64_t v) {    size_t r = 0, t;    t = (0 != (v >> 32)) << 5; v >>= t; r |= t;    t = (0 != (v >> 16)) << 4; v >>= t; r |= t;    t = (0 != (v >>  8)) << 3; v >>= t; r |= t;    t = (0 != (v >>  4)) << 2; v >>= t; r |= t;    return r + ((0xffffaa50 >> (v << 1)) & 3);}

The same method can be used to process 4 bits at a time for the loop or the tree versions. The naive version for 32 bits becomes:

size_t logb(uint32_t v) {  // top 7 nibbles, from the top down  for(int i = 27; i > 0; i -= 4)  {    // nibble multiplied by 2    uint32_t n = (v >> i) & 30;    if (n)        return ++i | ((0xffffaa50 >> n) & 3);  }  // Bottom nibble, could add test for v == 0  return (0xffffaa50 >> (v << 1)) & 3);}

The function above will test very well with uniformly distributed inputs since it will return the first time through the loop in 15/16 of all the possible 32bit values.