non-template std::reference_wrapper assignment operator and template constructor

Question

In the C++ 20 Standard the constructor of the class template std::reference_wrapper is a template.

template<class U>
constexpr reference_wrapper(U&&) noexcept(see below );

while the assignment operator is not a template

constexpr reference_wrapper& operator=(const reference_wrapper& x) noexcept;

What is the reason that this difference (template and non-template) between these special member functions exists?

On the other hand, I tried the following progarm using Visual C++ 2019.

#include <iostream>
#include <functional>

struct A
{
    void f() const { std::cout << "A::f()\n"; }
    virtual void g() const { std::cout << "A::g()\n"; }
};

struct B : A
{
    void f() const { std::cout << "B::f()\n"; }
    void g() const override { std::cout << "B::g()\n"; }
};

int main()
{
    B b;

    std::reference_wrapper<A> r( b );

    r.get().f();
    r.get().g();


    r = std::reference_wrapper<B>( b );
}

and the compiler did not issue an error message relative to the assignment operator.

Is it a bug of the compiler or have I missed something?

Brian Bi · Accepted Answer · 2020-05-18 23:33:42Z

4

If std::reference_wrapper<T> had a single constructor accepting T&, then it would lead to bugs like std::reference_wrapper<const T> being able to bind to temporaries of type T. So, originally there were two constructors (other than the copy constructor): one taking T&, and another taking T&&, which was defined as deleted, ensuring that a compile error would occur.

However, it was pointed out that this is not really what we want: it would be better if std::reference_wrapper<T> would have no constructor at all that accepts an rvalue of type T, as opposed to a deleted constructor. This is LWG 2993. (You'll notice that this issue is mentioned at the bottom of the cppreference page.) Thus, the constructor was changed to a template that is SFINAE-disabled as appropriate.

Once these issues are solved for the constructor, the assignment operator only needs to have a single overload taking reference_wrapper. The conversion issues will be handled by the constructor logic when the compiler forms the implicit conversion sequence.

answered May 18, 2020 at 23:33

Brian Bi

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4 Comments

Vlad from Moscow Over a year ago

But the standard does not allow accepting by the constructor an object of the type std::reference_wrapper.

Brian Bi Over a year ago

@VladfromMoscow there is a copy constructor, is there not?

Vlad from Moscow Over a year ago

The copy constructor is also a non-template constructor.

Brian Bi Over a year ago

@VladfromMoscow yes, there is a copy constructor and a template converting constructor, but I don't understand where you're going with this.

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