CODEWITHSUNDEEP
c++11
Hei
Hei

Reputation: 1894

std::move() is just casting?

I have read a few (pesudo) implementions of std::move(). And all are just casting away the reference of the parameter and then returning it as a rvalue reference.

It doesn't do anything more than that.

However, I am curious: 1. whether it does more than that. 2. whether standard explicitly states that so that the caller should be aware the side effect.

In the book The C++ Programming Language 4th edition, it states "move(x) marks x for destruction so that move(x) should be used with care".

Does it mean that there is a side effect imposed by the standard and so compiler can do certain optimization?

Thanks in advance.

Upvotes: 3

Views: 1671

Answers (2)

RiaD
RiaD

Reputation: 47658

Yes, that's exactly as it is described in the standard. In N4659 (which is last draft I found)

it says in §23.2.5

template <class T> constexpr remove_reference_t<T>&& move(T&& t) noexcept;

Returns: static_cast<remove_reference_t<T>&&>(t)

It doesn't mark anything for destruction, and it doesn't change the object but object may be changed in function that accepts rvalue (such as move constructor, move assignment operator)

Upvotes: 5

kmdreko
kmdreko

Reputation: 60662

Yes, std::move is a bit of a misnomer as it doesn't actually move anything. It is used to indicate that an object may be "moved from".

It does this by casting the object to a T&&. cppreference states the return value is static_cast<typename std::remove_reference<T>::type&&>(t). (btw, that is exactly what VS2017 does)

I don't know precisely what the standard says on the matter.

Upvotes: 1

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