>> If the sign is false, the shifted bits (mask) have to be 0. >> If the sign bit is true, the shifted bits (mask) have to be set. > > IIUC this logic handles sign bit + "shift - 1" bits. So if the last > shifted bit is different, the overflow is not detected.
Ah, right, because of the - 1ULL ... [...] >> This looks like some black magic :) > > Yeah, I felt this way too, but didn't come up with anything better and > just left a comment warning not to simplify. > I wonder if all we want is const uint64_t sign = 1ULL << 63; uint64_t mask = (-1ULL << (63 - shift)) & ~sign; For shift = * 0: 0000000...0b * 1: 0100000...0b * 2: 0110000...0b * 63: 0111111...1b Seems to survive your tests. -- Thanks, David / dhildenb
