Does this means that compilers cannot in general reorder non-side-effects operations across side effects, even if those operations wouldn't be globally visible if not for UB? Alternatively, is UB ordered with side effects? What's the ordering of UB with regard to other thread operations? Does it guarantee sequential consistency? I guess happens-before is guaranteed only if something else would guarantee happens before, but it means further constraint on ordering of, for example, potentially faulting operations across atomic operations.
Ie:
int ub(int idx, _Atomic int* x)) {
char y[1000];
int r = *x; // 2
r+= y[idx]; // 1
return r;
}
Statements 1 and 2 can't be reordered as any UB in accessing y[idx] is sequenced-after any side effect that happens-before 2, even if y is a purely local, non-escaping variable. This puts constraints on register allocation for example.
This opens a big can of worms.
edit: from a few quick tests GCC seems quite good at preserving ordering of potentially faulting instructions across side effects, even when reordering would be profitable (for example hoisting operations out of loops). It might be a posix requirement in practice because signals make them "visible".
edit2:
ok, this "fails":
extern volatile int x;
int ub(int d, int c) {
int r;
for (int i = 0; i < 100; ++i) {
r+= x; // 1
r+= d /c; // 2
}
return r;
}
GCC -O3 will hoists out of the looop the potentially faulting (and expensive) division at [2] before the volatile load at [1] (which is a side effect). Would you consider this a valid transformation or is it time-traveling UB?
Potentially this could be fixed by unrolling out the first iteration of the loop and preserving the first volatile access above the division, which can then be cached. This would also help with the variant where the volatile access is replaced by a function call that currently gcc doesn't optimize.