将右值传递给函数

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) .