Is there a way to call a move constructor from a constructor that takes an rvalue reference?

Suppose I have a class A with a move constructor:

class A {
  public:
    A(A &&other) = default;
}

Suppose I then have a class with an A member:

class B {
    A a;
}

The first part of the question: How can I write a constructor that takes an rvalue reference to A and calls A(A &&)?

If I write this:

class B {
    A a;
  public:
    B(A &&a) : a(a) {}
}

I get the following error:

move_constructor_call.cc:9:16: error: call to implicitly-deleted copy constructor of 'A'
    B(A &&a) : a(a) {}
               ^ ~
move_constructor_call.cc:3:5: note: copy constructor is implicitly deleted because 'A' has a user-declared move constructor
    A(A &&other) = default;
    ^

I assume this is because a is considered an lvalue so the move constructor is not called. But I am not sure why it is not called even though a should have type A &&. Is it a requirement of a move constructor that the value constructed from is an rvalue, as it is not here? I had been under the impression an lvalue of an rvalue reference would suffice.

This is the second part of the question: Why does it not call A(A &&) here?

>Solution :

You can change the initializer list to a(std::move(pa)) as shown below.

This is because the expression a inside the parenthesis in your example has the value category of an lvalue expression.

class B {
    A a;
  public:
    //------------vvvvvvvvv----------->explicitly use std::move
    B(A &&pa) : a(std::move(pa)) {} 
};