4 ms·
Umm, signed integers are UB on overflow; unsigned is always fine.
by tekne 1y ago
Umm, signed integers are UB on overflow; unsigned is always fine.
- zero_shift 1y agoSorry, you are correct. I don't think unsigned overflow behaviour was defined until C99 though. Anyway, in answer to the question, I would guess the reason was because of signed / unsigned type promotion.