c++ - Why can't C++11 move a noncopyable functor to a std::function?

Question

//------------------------------------------------------------------------------
struct A
{
    A(){}
    A(A&&){}
    A& operator=(A&&){return *this;}
    void operator()(){}

private:
    A(const A&);
    A& operator=(const A&);

    int x;
};

//------------------------------------------------------------------------------
int main()
{
    A a;
    std::function<void()> func(std::move(a));
}

'A::A' : cannot access private member declared in class 'A'

It seems like when I capture something by reference or const I can make a non-copyable lambda. However when I do that it actually works to give it to a std::function.

Answer 1

The short answer is that the C++11 specification requires your A to be CopyConstructible to be used with std::function.

The long answer is this requirement exists because std::function erases the type of your functor within the constructor. To do this, std::function must access certain members of your functor via virtual functions. These include the call operator, the copy constructor and the destructor. And since these are accessed via a virtual call, they are "used" whether or not you actually use std::function's copy constructor, destructor or call operator.