Right-shifting three bits would reduce the size of the lookup table to 32 slots I guess something like const int extra_bits = (sizeof(int) - 1) * 8; i

Right-shifting three bits would reduce the size of the lookup table to 32 slots I guess something like const int extra_bits = (sizeof(int) - 1) * 8; int x = __builtin_clz(~lead_byte); return (x == 0) + (x > 1 + extra_bits) * (x < 5 + extra_bits) * (x - extra_bits)); could work, although I've not tested it for all cases or checked if it's fast The idea there is to invert the bits, use a built in operation to count leading zeros (i.e. leading ones in the original byte) and then do some math to achieve the same semantics as the lookup table