If the floating point trick worth it? When I was young, floating point operations were very slow and I still try to avoid them like the plague. I remembered th

If the floating point trick worth it? When I was young, floating point operations were very slow and I still try to avoid them like the plague. I remembered there was a trick to avoid popcount, but I didn't remember it. So I found https://stackoverflow.com/questions/51387998/count-bits-1-on-an-integer-as-fast-as-gcc-builtin-popcountint https://stackoverflow.com/questions/51387998/count-bits-1-on... That version counts, the 1 bit's but here we already know the number is not 0 and we want to know that the inverted number it has exactly 1 bit set, so instead of popcnt(n) == 1 we can use n & (n-1) == 0 Moreover, n is the inverted number n = 0b1111111111 - s so an alternative is to run the check in the original number, and I think that this does the trick: s | (s+1) == 0b1111111111