Passing an rvalue to a function

Question

I read a lot of information about rvalue links, I understood everything, but I met this example:

template<class T, class Arg>
T* make_raw_ptr(Arg &&arg)
{
    return new T(arg);
}; 

If you pass rvalue to the make_raw_ptr function without using the my_forward function, when using new T, there will be a copy constructor, not a move constructor. I understand that arg will be lvalue, despite the fact that it is an rvalue reference, but I have one question. Why make a static_cast arg to an rvalue link when it is already an rvalue link, while using a static_cast<A&&> to arg, the move constructor will be called?

Answer 1

arg itself is an expression that represents an object referred to by arg and its value category is lvalue. It doesn't matter what the type of arg actually is. A name of a variable / function parameter itself is always an lvalue expression.

static_cast<A&&>(arg) is an expression that represents the very same object as arg, but its category is rvalue (and xvalue). When you use this expression as the constructor argument, the move constructor will be preferred. The same effect is with std::move(arg).