Question

Say a given class A is defined with a public copy constructor and a private move constructor. If a function f returns an object of type A, and f is used to initialize a local instance of variable of type A, then by default (since the value returned is a rvalue) the compiler will try to use the move constructor. I believed it is sensible to expect the compiler to use the copy constructor once it detects that the move constructor is private, however to my surprise I received a compiler error stating that the move constructor is private. My question is as follows, given the following code:

 #include<iostream>

 using namespace std;

 class A
 {

    friend A f();

 public:
    A(const A&) { cout << "copy\n"; }


 private:
    A() {}
    A(A&&) { cout << "move\n"; }
 };

 A f()
 {
    A a;
    return a;
 }

 int main()
 {
    A a = f();
 }

How can I change the code (without changing A or f) so that I could initialize the variable in main using the copy constructor?

Answer 1

I would change the class since it’s not sensible.

Alternatively derive from the class or wrap it.

If you just want a quick hack you can do

template< class Type >
Type& tempref( Type&& t ) { return t; }

then do

A a = tempref( f() )

Disclaimer: code not touched by compiler’s hands.