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

问题

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

    A(const A& other) : i(new int(*(other.i))) {
        std::cout << "a copy ctor" << std::endl;
    }

    ~A() {
        std::cout << "a dtor" << std::endl;
        delete i;
    }

    void get() {
        std::cout << *i << std::endl;
    }

private:
    int* i;
};

const A& foo() {
    return A(32);
}

const A& foo_2() {
    return 6;
}

int main()
{
    A a = foo();
    a.get();
}

I know, returning references to local values is bad. But, on the other hand, const reference should extend a temporary object lifetime.

This code produce an UB output. So no life extention.

Why? I mean can someone explain whats happening step by step?

Where is fault in my reasoning chain?

foo():

A(32) - ctor
return A(32) - a const reference to local object is created and is returned
A a = foo(); - a is initialized by foo() returned value, returned value goes out of scope(out of expression) and is destroyed, but a is already initialized;

(But actually destructor is called before copy constructor)

foo_2():

return 6 - temp object of type A is created implicitly,a const reference to this object is created(extending its life) and is returned
A a = foo(); - a is initialized by foo() returned value, returned value goes out of scope(out of expression) and is destroyed, but a is already initialized;

(But actually destructor is called before copy constructor)

回答1:

Rules of temporary lifetime extension for each specific context are explicitly spelled out in the language specification. And it says that

12.2 Temporary objects

5 The second context is when a reference is bound to a temporary. [...] A temporary bound to the returned value in a function return statement (6.6.3) persists until the function exits. [...]

Your temporary object is destroyed at the moment of function exit. That happens before the initialization of the recipient object begins.

You seem to assume that your temporary should somehow live longer than that. Apparently you are trying to apply the rule that says that the temporary should survive until the end of the full expression. But that rule does not apply to temporaries created inside functions. Such temporaries' lifetimes are governed by their own, dedicated rules.

Both your foo and your foo_2 produce undefined behavior, if someone attempts to use the returned reference.

回答2:

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.

来源：https://stackoverflow.com/questions/11747652/what-exactly-happens-when-returning-const-reference-to-a-local-object

标签

c++

const-reference