What is ->* operator in C++?

Question

C++ continues to surprise me. Today i found out about the ->* operator. It is overloadable but i have no idea how to invoke it. I manage to overload it in my class but i hav

Answer 1

Just like .*, ->* is used with pointers to members. There's an entire section on C++ FAQ LITE dedicated to pointers-to-members.

#include <iostream>

struct foo {
    void bar(void) { std::cout << "foo::bar" << std::endl; }
    void baz(void) { std::cout << "foo::baz" << std::endl; }
};

int main(void) {
    foo *obj = new foo;
    void (foo::*ptr)(void);

    ptr = &foo::bar;
    (obj->*ptr)();
    ptr = &foo::baz;
    (obj->*ptr)();
    return 0;
}

Answer 2

Like any other opperator, you can also call it explicitly:

a.operator->*(2);

Answer 3

The overloaded ->* operator is a binary operator (while .* is not overloadable). It is interpreted as an ordinary binary operator, so in you original case in order to call that operator you have to do something like

A a;
B* p = a->*2; // calls A::operator->*(int)

What you read in the Piotr's answer applies to the built-in operators, not to your overloaded one. What you call in your added example is also the built-in operator, not your overloaded one. In order to call the overloaded operator you have to do what I do in my example above.