Why destructor is not called when object goes out of scope

Question

Following program C++ compiled using g++ -O0 a.cpp -std=c++98:

class A
{
public:
        A(const A& a)  { cout << "A copy" << endl; }
        A() { cout << "A" << endl; }
        ~A() {cout << "~A" << endl; }
};

A f()
{
        cout << "f1" << endl;
        A a1;
        cout << "f2" << endl;
        return a1;
}

int main()
{
        cout << "m1" << endl;
        A a2 = f();
        cout << "m2" << endl;
}

Produces output:

m1
f1
A
f2
m2
~A

What bothers me is that after f2 the destructor and copy constructors are not called. From my understanding a1 goes out of scope and its destructor must be called. It looks like the object is moved from local a1 to a2 without copying. I understand that there is return value optimization but it is optimization and optimization must not change the logic of code and there is missing output saying ~A and A copy.

That will not compile. You declare a variable `a1` in the function `f()` but return `a` which is not declared. — Cory Kramer, Jul 21 '14 at 17:50
@chris but optimization must not change of application behaviour. If such optimization is not applied by compiler output would be different which means application behaviour is changed. — Trismegistos, Jul 21 '14 at 17:51
@Cyber This was a typo when preparing an example. It is corrected now. — Trismegistos, Jul 21 '14 at 17:52
@Trismegistos: You refer to the as-if rule. Return value optimization (and more generally, copy elision) is an exception to the as-if rule, explicitly mentioned in the standard. — Benjamin Lindley, Jul 21 '14 at 17:54
@BenjaminLindley Thanks, that explains the situation. Are there some other optimizations that can ignore equivalence of behaviour? — Trismegistos, Jul 21 '14 at 17:58

Why destructor is not called when object goes out of scope

0 Answers0