Move constructor on derived object

Question

When you have a derived object with a move constructor, and the base object also has move semantics, what is the proper way to call the base object move constructor from the derived object move constructor?

I tried the most obvious thing first:

 Derived(Derived&& rval) : Base(rval)  { } 

However, this seems to end up calling the Base object's copy constructor. Then I tried explicitly using std::move here, like this:

 Derived(Derived&& rval) : Base(std::move(rval))  { } 

This worked, but I'm confused why it's necessary. I thought std::move merely returns an rvalue reference. But since in this example rval is already an rvalue reference, the call to std::move should be superfluous. But if I don't use std::move here, it just calls the copy constructor. So why is the call to std::move necessary?

Answer 1

rval is not a Rvalue. It is an Lvalue inside the body of the move constructor. That's why we have to explicitly invoke std::move.

Refer this. The important note is

Note above that the argument x is treated as an lvalue internal to the move functions, even though it is declared as an rvalue reference parameter. That's why it is necessary to say move(x) instead of just x when passing down to the base class. This is a key safety feature of move semantics designed to prevent accidently moving twice from some named variable. All moves occur only from rvalues, or with an explicit cast to rvalue such as using std::move. If you have a name for the variable, it is an lvalue.

Answer 2

Named R-value references are treated as L-value.

So we need std::move to convert it to R-Value.