Can't understand the implementation of std::move

Question

Here is a possible implementation of std::move(). It's not fully conforming to the details of the standard, but it's very close:

template<class T>
typename std::remove_reference<T>::type&&
myMove( T&& Arg )
{   
    return ( ( typename std::remove_reference<T>::type&& )Arg );
}

I don't understand why it wouldn't work if we replace typename std::remove_reference<T>::type&& by the T&&, i.e.

template<class T>
typename std::remove_reference<T>::type&&
myMove( T&& Arg )
{   
    return ( (T&&) Arg );
}

`remove_reference` of course is important if `T` already is a reference. — Bo Persson, Mar 10 '16 at 07:56
This is where `type_name` earns its keep: http://stackoverflow.com/a/20170989/576911 :-) — Howard Hinnant, Mar 10 '16 at 15:06

score 6 · Accepted Answer · edited May 23 '17 at 11:59

The issue is not really about the move, but the reference collapsing that happens with T&&.

The are nice write-ups about this here and these Q&A here, and here.

I'll focus on the cast (T&&) and why that doesn't work.

Given the reference collapsing rules listed above;

T& & becomes T&
T& && becomes T&
T&& & becomes T&
T&& && becomes T&&

The problem arises when T is deduced to be an lvalue reference, then T&& becomes T& && which collapses to T&, given the cast, you merely cast it back to an lvalue reference and you have not gained the rvalue reference that enables moving to happen.

The std::remove_reference<T>::type is there to remove all the references and the && then adds back the rvalue ref and thus ensures the correct cast is made.

The typename is there to disambiguate the member type over a member variable.

Can't understand the implementation of std::move

1 Answers1