How copy constructor can be called with rvalue

Question

MWE:

struct A {
    A() {std::cout << "constructor" << std::endl; }
    A(const A& a) {std::cout << "copy constructor" << std::endl; }
    A(A&& a) {std::cout << "move constructor" << std::endl; }
};

int main() {
    A a1{};
    A a2{ a1 };
    A a3{ A{} };
    A a4{ std::move(a3) };
    return 0;
}

Output:

constructor
copy constructor
constructor
move constructor

fora2 copy elision is used which is an optimization of the compiler and everything seems to be fine. When I comment out the move constructor however, copy constructor is called in place of move constructor. How can an rvalue be converted to a const lvalue reference? Output:

constructor
copy constructor
constructor
copy constructor

The program is compiled in VS2017.

Answer 1

From en.cppreference.com:

If both copy and move constructors are provided and no other constructors are viable, overload resolution selects the move constructor if the argument is an rvalue of the same type (an xvalue such as the result of std::move or a prvalue such as a nameless temporary (until C++17)), and selects the copy constructor if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy constructor is provided, all argument categories select it (as long as it takes a reference to const, since rvalues can bind to const references), which makes copying the fallback for moving, when moving is unavailable.

This clearly says that rvalue can be bound to const lvalue reference.