Are Rvalue members indeed Rvalues too? - c++

Question

They say that members of Rvalues are also Rvalues - which makes a lot of sense. So this is either a VC++-specific bug or a bug in my understanding of Rvalues.
Take this toy code:
#include <vector>
#include <iostream>
using namespace std;
struct MyTypeInner
{
MyTypeInner() {};
~MyTypeInner() { cout << "mt2 dtor" << endl; }
MyTypeInner(const MyTypeInner& other) { cout << "mt2 copy ctor" << endl; }
MyTypeInner(MyTypeInner&& other) { cout << "mt2 move ctor" << endl; }
const MyTypeInner& operator = (const MyTypeInner& other)
{
cout << "mt2 copy =" << endl; return *this;
}
const MyTypeInner& operator = (MyTypeInner&& other)
{
cout << "mt2 move =" << endl; return *this;
}
};
struct MyTypeOuter
{
MyTypeInner mt2;
MyTypeOuter() {};
~MyTypeOuter() { cout << "mt1 dtor" << endl; }
MyTypeOuter(const MyTypeOuter& other) { cout << "mt1 copy ctor" << endl; mt2 = other.mt2; }
MyTypeOuter(MyTypeOuter&& other) { cout << "mt1 move ctor" << endl; mt2 = other.mt2; }
const MyTypeOuter& operator = (const MyTypeOuter& other)
{
cout << "mt1 copy =" << endl; mt2 = other.mt2; return *this;
}
const MyTypeOuter& operator = (MyTypeOuter&& other)
{
cout << "mt1 move =" << endl; mt2 = other.mt2; return *this;
}
};
MyTypeOuter func() { MyTypeOuter mt; return mt; }
int _tmain()
{
MyTypeOuter mt = func();
return 0;
}
This code outputs:
mt1 move ctor
mt2 copy =
mt1 dtor
mt2 dtor
That is, MyTypeOuter's move ctor calls MyTypeInner's copy, not move. If I modify the code to:
MyTypeOuter(MyTypeOuter&& other)
{ cout << "mt1 move ctor" << endl; mt2 = std::move(other.mt2); }
The output is as expected:
mt1 move ctor
mt2 move =
mt1 dtor
mt2 dtor
It seems VC++ (both 2010 and 2013) do not respect this part of the standard. Or am I missing something?

Answer 1

Whether the member of an rvalue is an rvalue or not is not the issue here, because you are dealing with lvalues inside your assignment operator.
In this move assignment operator,
const MyTypeOuter& operator = (MyTypeOuter&& other)
{
cout << "mt1 move =" << endl;
mt2 = other.mt2;
return *this;
}
other is an lvalue (since it has a name), and by extension so is other.mt2. When you say mt2 = other.mt2 you can only invoke the standard assignment operator.
In order to invoke the move constructor, you need to make other.mt2 look like an rvalue, and this is what std::move achieves:
const MyTypeOuter& operator = (MyTypeOuter&& other)
{
cout << "mt1 move =" << endl;
mt2 = std::move(other.mt2);
return *this;
}
See this related question.