#define FOO 17
void bar(int x, int y) {
if (x + y >= FOO) {
//do stuff
}
}
void baz(int x) {
bar(x, FOO);
}
the compiler can inline the call to bar in baz, and then optimise the condition to (x>=0)… because signed integer overflow is undefined, so can’t happen, so the two conditions are equivalent.The countless messages about optimisations like that would swamp ones about real dangerous optimisations.