Why does std::reference_wrapper explicitly define a copy assignment operator?

Question

A std::reference_wrapper explicitly declares the copy assignment operator. Why? This should prevent move construction/assignment from being implicitly defined so my assumption is that they do not want you to move a std::reference_wrapper, but I am not sure why that would matter.

If you look at the "possible implementation", they just default the operator: https://en.cppreference.com/w/cpp/utility/functional/reference_wrapper/operator%3D

Just looking for some history on why the committee felt the need for this.

There is no "possible implementation" on this page. Is it the right page? — user253751, Mar 05 '20 at 12:48
"possible implementation" is here: https://en.cppreference.com/w/cpp/utility/functional/reference_wrapper — user253751, Mar 05 '20 at 12:49

Maxim Egorushkin · Answer 1 · 2020-03-05T12:30:35.910

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This should prevent move construction/assignment from being implicitly defined

std::reference_wrapper is a non-owning reference (a pointer, really), move operation is no different from copy. Don't confuse it with std::shared_ptr.

edited Mar 05 '20 at 12:30

answered Mar 05 '20 at 12:24

Maxim Egorushkin

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I'm not, really looking more for why move would not have been added. My guess is it is because the reference_wrapper is just a pointer, so a move would be the same as a copy. – Rian Quinn Mar 05 '20 at 12:54

score 5 · Accepted Answer · answered Mar 05 '20 at 12:30

They declare it because it has work to do.

Yes, this inhibits an automatically-generated move constructor. No, that doesn't matter. The authors would have had to define their own anyway (for the same reason they need to create a copy constructor), except that there is no use in "moving" something that doesn't own a resource. It would be the same as a copy.

That doesn't mean that a std::reference_wrapper isn't movable, it just means that "moving" one is the same as copying it. And there's no point in implementing that twice!

Honestly I think this is best explanation I can come up with as well. In the end, the move would be the same as a copy, so there is not need to add it. — Rian Quinn, Mar 05 '20 at 12:58

score 3 · Answer 3 · answered Mar 05 '20 at 12:26

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std::reference_wrapper is an object with reference-like semantics. But unlike built in references, a reference_wrapper can be re-bound, copied and moved. This allows it it to be used in standard containers. So I do not understand your confusion, by its design a reference_wrapper is supposed to be copyable and movable.

answered Mar 05 '20 at 12:26

bolov

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The question is why it is not movable. – user253751 Mar 05 '20 at 12:49
Although you can use a move, it is not actually moved since the implicit move is not possible. – Rian Quinn Mar 05 '20 at 12:56

score 2 · Answer 4 · answered Mar 05 '20 at 12:47

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Consider what state you would expect the moved-from std::reference_wrapper to be in. You don't want to have a reference wrapper referring to something that doesn't exist if you can help it

answered Mar 05 '20 at 12:47

wreckgar23

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Why does std::reference_wrapper explicitly define a copy assignment operator?

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