Lifetime extension, prvalues and xvalues

Question

Following the well accepted answer to this question Do rvalue references allow dangling references? It would seem that xvalues do not have their lifetime extended when assigned to a rvalue reference lvalue like in the question. However when I do this

#include <iostream>

using namespace std;

class Something {
public:
    Something() {
        cout << "Something()" << endl;
    }
    Something(const Something&) {
        cout << "Something(const Something&)" << endl;
    }
    Something(Something&&) {
        cout << "Something(Something&&)" << endl;
    }
    ~Something() {
        cout << "~Something()" << endl;
    }

    int a;
};

Something make_something() {
    return Something{};
}

int main() {
    auto&& something = make_something().a;

    return 0;
}

The lifetime of the object returned by a call to make_something is extended, even though make_something().a is an xvalue as per http://en.cppreference.com/w/cpp/language/value_category (the third bullet in the xvalues explanation lists the member access I have above as an xvalue,)

a.m, the member of object expression, where a is an rvalue and m is a non-static data member of non-reference type;

If value categories do not determine when the lifetime of an rvalue will be extended then what does? I am having a hard time understanding when the lifetime of an rvalue is extended in C++

Answer 1

Lifetime extension doesn't care about value categories. As stated by [class.temporary]/p6:

The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference

Emphasis added.

This says nothing here about the value category of the expression being referenced.

What determines whether a temporary is extended is exactly the above (and a few more rules).

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But that does not explain why adding an std::move() around the temporary to which the reference is being assigned does not extend the lifetime

std::move is not a magical, compiler-defined construct in C++. It is a function call, and therefore it behaves no differently from any other C++ function call.

So, if you have std::move(Type()), what does that mean? It means that you will create a temporary, bind it to the parameter of std::move, then call that function, which will return something.

Binding a temporary to a function parameter, as stated in [class.temporary]/p6, means that the lifetime of the temporary is fixed to be the lifetime of the full expression that created it (if not for that rule, then the temporary would have to be destroyed at the end of the function call, since that's the end of the reference's lifetime).

It doesn't matter what the function does, says, or implies. It doesn't matter if the compiler could perhaps inline things and determine that the return value is a reference to an argument that came from a temporary. The lifetime of that temporary is fixed to the expression, not extended.

Answer 2

If value categories do not determine when the lifetime of an rvalue will be extended then what does? I am having a hard time understanding when the lifetime of an rvalue is extended in C++

Note that value categories describe expressions not objects. Value categories ( xvalue, prvalue, or whatever ) won't be extended in any way. Only objects can have a lifetime.

From the n4296 standard draft:

• §12.2.1

  Temporaries of class type are created in various contexts: binding a reference to a prvalue (8.5.3), returning a prvalue (6.6.3), a conversion that creates a prvalue (4.1, 5.2.9, 5.2.11, 5.4), throwing an exception (15.1), and in some initializations (8.5).

and

• §12.2.4

  There are two contexts in which temporaries are destroyed at a different point than the end of the full- expression. [...]
  The second context is when a reference is bound to a temporary. The temporary to which the reference is bound or the temporary that is the complete object of a subobject to which the reference is bound persists for the lifetime of the reference

Note: I didn't quote the first context since it has minor revelance regarding the questions. Italic emphasis added by me.

Thus the value category of the function expression makesomething() is a prvalue creating a temporary of class type, according to the first paragraph cited above.
makesomething().a accesses a temporary, complete subobject. Binding this temporary to a reference leads, according to the second context quoated above, to an extended lifetime.

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The lifetime of the subobject a is coupled to the lifetime of the previous created temporary making it an expiring value (xvalue). Without extending its lifetime by binding it to a reference it would be destroyed together with the temporary class object. Thus, in this case, after the ;.