What exactly happens when returning const reference to a local object?

Question

struct A {
    A(int) : i(new int(783)) {
        std::cout << \"a ctor\" << std::endl;
    }

    A(const A& other) : i(new int(*(other.i))) {
              


        

Answer 1

You are misinterpeting "until function exit". If you really want to use a const reference to extend the life of an object beyond foo, use

A foo() {
    return A(32);
}
int main() {
    const A& a = foo();
}

You must return from foo by value, and then use a const reference to reference the return value, if you wish to extend things in the way you expect.

As @AndreyT has said, the object is destroyed in the function that has the const &. You want your object to survive beyond foo, and hence you should not have const & (or &) anywhere in foo or in the return type of foo. The first mention of const & should be in main, as that is the function that should keep the object alive.

You might think this return-by-value code is slow as there appear to be copies of A made in the return, but this is incorrect. In most cases, the compiler can construct A only once, in its final location (i.e. on the stack of the calling function), and then set up the relevant reference.