rvalue references and user defined conversion

Question

I am trying to figure what happens in the code below (it was inspired by shared_ptr(unique_ptr&&) constructor).

#include <iostream>
using namespace std;

class A {};

class B {
public:
    B(const A&) {
        cout << "ctor from A&\n";
    }
};

void f(B&&) { cout << "f(B&&)\n"; };
void f(const B&) { cout << "f(const B&)\n"; }

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

I am trying to understand how it works. Why is an rvalue a result of the conversion and f(B&&) called?

Answer 1

The function does not accept an object of type A. However, it does accept an object of type B, and A is convertible to B by virtue of the implicit converting constructor. The result of the conversion is a prvalue. The rvalue reference overload is a better match than the const lvalue reference when the argument is a prvalue.