Is RVO (Return Value Optimization) on unnamed objects a universally guaranteed behavior?

Question

This question is in different aspect (also limited to gcc). My question is meant only for unnamed objects. Return Value Optimization is allowed to change the observable behavior of the resulting program. This seems to be mentioned in standard also.

However, this "allowed to" term is confusing. Does it mean that RVO is guaranteed to happen on every compiler. Due to RVO below code changes it's observable behavior:

#include<iostream>
int global = 0;
struct A {
  A(int *p) {}
  A(const A &obj) { ++ global; }
};

A foo () {  return A(0); }  // <--- RVO happens
int main () {
  A obj = foo(); 
  std::cout<<"global = "<<global<<"\n"; // prints 0 instead of 2
}

Is this program suppose to print global = 0 for all implementations irrespective of compiler optimizations and method size of foo ?

Answer 1

According to the standard, the program can print 0, 1 or 2. The specific paragraph in C++11 is 12.8p31 that starts with:

When certain criteria are met, an implementation is allowed to omit the copy/move construction of a class object, even if the copy/move constructor and/or destructor for the object have side effects.

Note that both copy elisions are not an optimization that falls in the as-if rule (which requires the behavior of the program to be consistent with the behavior of the same program as-if no optimization had taken place). The standard explicitly allows the implementation to generate different observable behaviors, and it is up to you the programmer to have your program not depend on that (or accept all three possible outcomes).

Note 2: 1 is not mentioned in any of the answers, but it is a possible outcome. There are two potential copies taking place, from the local variable in the function to the returned object to the object in main, the compiler can elide none, one or the two copies generating all three possible outputs.