Why do I have to call move on an rvalue reference?

Question

In the code below, why doesn\'t the first call mkme = mvme_rv dispatch to T& operator=(const T&&)?

#include 

        

Answer 1

As skypjack correctly comments, accessing an object through its name always results in an lvalue reference.

This is a safety feature and if you think it through you will realise that you are glad of it.

As you know, std::move simply casts an l-value reference to an r-value reference. If we use the returned r-value reference immediately (i.e. un-named) then it remains an r-value reference.

This means that the use of the r-value can only be at the point in the code where move(x) is mentioned. From a code-reader's perspective, it's now easy to see where x's state became undefined.

so:

 1: auto x = make_x();
 2: auto&& r = std::move(x);
 3: // lots of other stuff
35: // ...
54: // ...
55: take_my_x(r);

does not work. If it did, someone maintaining the code would have a hard time seeing (and remembering) that x (defined on line 1) enters an undefined state on line 55 through a reference taken on line 2.

This is a good deal more explicit:

 1: auto x = make_x();
 2: //
 3: // lots of other stuff
35: // ...
54: // ...
55: take_my_x(std::move(x));