Comment by rocqua
1 year ago
Notably, unsigned integers also have defined behavior for overflow. This means compilers can do less optimization on unsigned integers. For example, they can't assume that. x + 1 > x for unsigned ints, but are free to assume that for standard ints.
That is just another reason to stick with signed ints unless there is a very specific behavior you rely on.
> For example, they can't assume that. x + 1 > x for unsigned ints, but are free to assume that for standard ints.
No they ain't:
Integers are integers, and can roll over regardless of whether or not they're signed. Avoiding rollover is not a reason to stick with signed integers; indeed, rollover is a very good reason to avoid using signed integers unless you're specifically prepared to handle unexpectedly-negative values.
It depends on the language. I linked a set of c++ guidelines and for c++, they are correct: it is undefined behaviour to do signed integer overflow. Some languages do specify it, eg rust, and even in c++ it might appear to work, but even then it is still undefined and should be strongly avoided.
That's what I'm saying, though: rollovers can happen regardless of whether the integer is signed or unsigned. x + 1 > x is never a safe assumption for integers of the same fixed width, no matter if they're i16 or u16. Whether it's specifically acknowledged as defined or undefined behavior doesn't really change that fundamental property of fixed-width integer addition.
(As an aside: I'm personally fond of languages that let you specify what to do if an integer arithmetic result doesn't fit. Zig, for example, has separate operators for rollover v. saturation v. panicking/UB, which is handy. Pretty sure C++ has equivalents in its standard library.)