When is the move constructor called in the `std::move()` function?

问题

The function std::move() is defined as

template<typename T>
typename std::remove_reference<T>::type&& move(T && t)
{ 
    return static_cast<typename std::remove_reference<T>::type&&>( t ); 
}

There are four places where I can imagine the move constructor to be called:

1. When the parameter is passed.
2. When the cast is performed.
3. When the result is returned.
4. Not in the std::move() function itself but possibly at the place where the returned reference ultimately arrives.

I would bet for number 4, but I'm not 100% sure, so please explain your answer.

回答1:

There is no move construction going on. std::move() accepts a reference and returns a reference. std::move() is basically just a cast.

Your guess 4. is the right one (assuming that you are actually calling a move constructor in the end).

回答2:

std::move is just a type cast, it tells the compiler that the type is an rvalue.

来源：https://stackoverflow.com/questions/14749587/when-is-the-move-constructor-called-in-the-stdmove-function