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.

Answer 1

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!

Answer 2

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.