C++ - calling a member function of another class using a pointer

Question

I'm testing, trying to call a member function being passed as a parameter, the member function has to be one of another class. this is an example, which gives an error:

"pointer-to-member selection class types are incompatible ("B" and "A")"

This is the code, what am I doing wrong?

#include <iostream>
using namespace std;
class A {
private:
public:
    void fA(int x) {
        cout << "hello" << endl;
    }
    void fB(int x) {
        cout << "good bye" << endl;
    }
    A() {
    }
};

class B {
private:
    void (A:: * f)(int) = NULL;
public:
    B(void (A:: * f)(int)) {
        this->f = f;
    }
    void call() {
        (this->*f)(10); //What's wrong here?
    }
};
A a = A();
B b = B(&(a.fA));
B b2 = B(&(a.fB));

int main(void) {
    b.call();
    b2.call();
}

Normal member functions are static. A pointer to them is not a pointer through the class instance (as with variables). Instead, they are passed a hidden 'this' pointer on class which indicates the instance. Therefore, &(a.fA) has no meaning. — Michael Chourdakis
– Michael Chourdakis, Commented Aug 14, 2021 at 10:51
Instead of using function pointers I suggest that you look into std::funtional: en.cppreference.com/w/cpp/utility/functional/function — Bart
– Bart, Commented Aug 14, 2021 at 11:52

Remy Lebeau · Accepted Answer · 2021-08-14 17:04:50Z

6

&(a.fA) is not legal C++ syntax. &A::fA is. As you can see, there is no object of type A anywhere of this syntax. &A::fA is just a pointer to a member function, not a pointer-to-member-together-with-an-object combo.

Now in order to call that pointer-to-member, you need an object of class A. In class B, you don't have any. You need to get one in there somehow, and call the function this way:

 (a->*f)(10);

where a is a pointer to that object of class A.

edited Aug 14, 2021 at 17:04

Remy Lebeau

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answered Aug 14, 2021 at 10:48

n. m. could be an AI

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