int res = (*fn[(n < 2)])(n); As stated by others, "n < 2" is a conditional or comparison or whatever (let's just say it's cheating :P) let's call it "m". So w

int res = (*fn[(n < 2)])(n); As stated by others, "n < 2" is a conditional or comparison or whatever (let's just say it's cheating :P) let's call it "m". So we choose our function pointer with: int res = (*fn[m])(n); where uint32_t m = n < 2?1:0; and let's assume n is non-negative... uint32_t top_bits = n & 0xFFFFFFFEu; uint32_t p = 0; for (int i = 0; i < 32; i++){ p |= (top_bits & (1u << i)) >> i; } uint32_t m = 1 - p; Edit: You can do the for loop to check if any bits are set in logarithmic time. For general comparisons see http://stackoverflow.com/questions/10096599/bitwise-operations-equivalent-of-greater-than-operator http://stackoverflow.com/questions/10096599/bitwise-operatio... Oh and the "i < 32" doesn't count as you could just unroll it...