Why does the compiler choose the copy ctor after calling move

Question

For the following code, I know that because I implement a copy constructor the default move is blocked by the compiler, what I don't understand is why the compiler choose the copy constructor, and not issuing an error?
Can a rvalue reference be the parameter for a reference function if no rvalue ref function exists?
If not, what do happens here?

#include <iostream>

using namespace std;

class A{
    public:

        A(){printf("constructor\n");}
        A(const A& a){printf("copy constructor");}
};

int main()
{ 
    A a;
    A b = std::move(a);
    return 1; 
}

The output is:

constructor
copy constructor

Thanks

Answer 1

The reason is that rvalue expressions can be bound to function parameters that are const lvalue references.

---

Also note that const rvalue expressions cannot be bound to rvalue references:

class A{
    public:    
        A(){printf("constructor\n");}
        A(const A& a){printf("copy constructor");}
        A(A&&) = default; // <-- has move ctor
}; 

int main()
{ 
    const A a; // <-- const!
    A b = std::move(a); // <-- copy not move!!!
    return 1; 
}

The code above still calls the copy constructor, even though there is a move constructor.