move semantics std::move

Question

I don\'t understand very well the std::move function

template 
typename remove_reference::type&&
move(T&& a)         


        

Answer 1

Because rvalue reference to lvalue reference would decay to lvalue reference, and returing lvalue reference would have different semantics from those you would expect from move.

Edit: Huh, why the downvote? Check out this code:

template < typename T > T&& func(T&& x) { return x; }

int main()
{
        int x;

        int &y = func(x);
}

Further reading: http://www.justsoftwaresolutions.co.uk/cplusplus/rvalue_references_and_perfect_forwarding.html

Answer 2

Think about what happens if T is an lvalue reference, for example MyClass &. In that case, T && would become MyClass & &&, and due to reference collapsing rules, this would be transformed into MyClass & again. To achieve the right result, typename remove_reference<MyClass&>::type&& first removes any reference decorations from the type, so MyClass & is mapped to MyClass, and then the rvalue reference is applied to it, yielding MyClass &&.