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

user_5606784 0
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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.

c++
4 Answers

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.

user_4120800

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.

user_28053050

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!

user_43862780

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

user_3905570
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