When the reference passed to f() is bound to an object that's defined const GCC could assume that the call to f() doesn't change it and emit better code but, unfortunately, it does not. As a result, even though GCC could avoid the test below, it emits it. void g (void) { const std::vector<int> v; int n = v.size (); f (v); if (v.size () != n) __builtin_abort (); } GCC does eliminate such tests in the simple cases when v is a fundamental type like int, but not when it's a struct (even a trivial one containing just an int).Hmm. if that is really true then I would be negatively surprised. I just tested it with the the latest g++ (v9) and it gives a compile error if one tries to modify the vector inside f(). And that is the expected behaviour, so I wonder which compiler & version you mean?Ok, you mean this: if (v.size () != n) __builtin_abort (); But the compliler cannot know that size() returns a vector-internal value, ie. it could also be an external value, therefore it can't optimize it away, IMO.
Jonathan already made it clear earlier that GCC can determine that the return value of vector::size() doesn't change between successive calls to the function because it's is trivially inline and doesn't modify any data, and therefore can eliminate all but the first call. For a reduced test case showing the missing optimization see bug 86318. Martin
