Comment by Liquid_Fire

> Sure you could do the check before the overflow but any way you slice it that calculation ultimately applies to something that is going to be UB so the compiler is free to optimise it out?

No, if you never do the calculation it's not going to be UB.

  int8_t x = some_input();
  if (x > 10) return bad_value;
  else x *= 10;

There is no UB here, because we never execute the multiplication in cases where it would have otherwise been UB. The compiler is not free to remove the check, because it can't prove that the value is not > 10.

> It has just rearranged code that it knows is UB

No - that's the problem. The compiler doesn't know that the code is UB, because this depends on the exact values at runtime, which the compiler doesn't know.

In some limited cases it could perform data flow analysis and know for sure that it will be UB, but those cases are very limited. In general there is no way for to know. So there are three things it could do:

A) Warn/error if there could possibly be UB. This would result in warnings in hundreds of thousands of pieces of legitimate code, where there are in fact guarantees about the value but the compiler can't prove or see it. It would require much more verbose code to work around these, or changing the language significantly. For example, you could represent this in the type system, or have annotations.

B) Insert runtime checks for the UB. This would have a significant performance overhead, as there are lots of "innocent" operations in the language that, in the right circumstances, lead to UB. So we would bloat the code with a lot of branches, 99.999% of which will never ever be taken, filling up the instruction cache and branch predictor. You get something more like (the runtime behaviour of) Python or JavaScript. Or even C if you enable UBSan.

C) Assume that the programmer has inserted these checks where they are needed, and omitted them where they are not. You get performance, but in exchange for that you are responsible for avoiding UB. This is what C chooses.

> C doesn't know whether your machine is 1s complement, 2s complement or 3s complement, it leaves it to the programmer to deal with the situation, if the programmer knows he's working on 2s complement machines that overflow predictably he can work on that assumption, the compiler isn't expected to know, but it should stay out of the way because the programmer does

This is mostly right, but with the caveat that you can't invoke UB. If you want to deal with whatever the underlying representation is, cast it to an unsigned type and then do whatever you want with it. The compiler will not mess with your unsigned arithmetic, because it's allowed to wrap around. But for signed types, you are promising to the compiler that you won't cause overflow. In exchange the compiler promises you fast signed arithmetic.

This promise is part of the language, not part of GCC. If you removed that promise, you would have to pay the price in reduced performance.

Could you have a C compiler that inserts these checks? Yes (see UBSan). But you would be throwing away performance - it would be slower than GCC/Clang/MSVC/etc. If you're writing performance-sensitive software, you are better off either ensuring you never trigger UB, or use another language like Rust. If performance is not so important, you are probably better off writing the thing in Go/JavaScript/whatever.