Can i use C++ function pointers like a C#'s Action?

Question

In C ++, I first encountered function pointers.
I tried to use this to make it similar to Action and Delegate in C #.
However, when declaring a function pointer, it is necessary to specify the type of the class in which the function exists.
ex) void (A :: * F) ();
Can I use a function pointer that can store a member function of any class?

In general, function pointers are used as shown in the code below.

class A {
public:
    void AF() { cout << "A::F" << endl; }
};

class B {
public:
    void(A::*BF)();
};

int main()
{
    A a;
    B b;

    b.BF = &A::AF;
    (a.*b.BF)();

    return 0;
}

I want to use it like the code below.
is this possible?
Or is there something else to replace the function pointer?

class A {
public:
    void AF() { cout << "A::F" << endl; }
};

class B {
public:
    void(* BF)();
};

int main()
{
    A a;
    B b;

    b.BF = a.AF;

    return 0;
}

I solved the question through the answer.
Thanks!

#include <functional>
#include <iostream>

class A {
public:
    void AF() { std::cout << "A::F" << std::endl; }
};

class C {
public:
    void CF() { std::cout << "C::F" << std::endl; }
};

class B {
public:
    B(){}
    std::function<void()> BF;
};

int main() {
    A a;
    C c;
    B b;
    b.BF = std::bind(&A::AF, &a);
    b.BF();
    b.BF = std::bind(&C::CF, &c);
    b.BF();

    int i;
    std::cin >> i;
    return 0;
}

Why can't you just make `AF` a static method or a global function? You don't have to access any member of class `A` anyway. — user202729, Jun 01 '19 at 08:32
I can't read answer with code in this moment but You can read this answer https://stackoverflow.com/questions/9568150/what-is-a-c-delegate And from me, read about auto type ane std::variant from C++ 17 — Drock, Jun 01 '19 at 08:43
@Drock "read about auto type" - `auto` is not a type. It's just a keyword telling the compiler to deduce the correct type itself at initialization of the variable. The variable still has a static type and that type cannot change. For example; `int i = 42;` and `auto i = 42;` results in *exactly* the same. In both cases the stype of `i` is `int` and it will *stay* `int` forever. — Jesper Juhl, Jun 01 '19 at 08:54
@Drock Just out of interest, why is it worth reading up on variant? — George, Jun 01 '19 at 08:54
You can use the template `std::function` for member function pointers. The template type won't require the class. — Weak to Enuma Elish, Jun 01 '19 at 09:03

score 4 · Accepted Answer · edited Jun 01 '19 at 09:32

4

What you want to do is probably something like this. You can use std::function to hold a pointer to a member function bound to a specific instance.

#include <functional>
#include <iostream>

class A {
public:
    void AF() { std::cout << "A::F" << std::endl; }
};

class B {
public:
    B(const std::function<void()>& bf) : BF(bf) {}
    std::function<void()> BF;
};

int main() {
    A a;
    B b1(std::bind(&A::AF, &a)); // using std::bind
    B b2([&a] { a.AF(); });      // using a lambda

    b1.BF();
    b2.BF();

    return 0;
}

edited Jun 01 '19 at 09:32

Ted Lyngmo

93,841
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answered Jun 01 '19 at 09:14

snak

6,483
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3

`std::bind` is sooo old-school. https://stackoverflow.com/q/1930903/5945883 I'd upvote lamda-based answer. – R2RT Jun 01 '19 at 09:23
Well, with lambda, now it no longer answers the question, I think, because both capturing a lambda with `function<>` and calling a method in the lambda are both trivial and nothing special. – snak Jun 01 '19 at 09:32
@snak Also more efficient that `bind` since the resulting object is smaller and in practice will not require memory allocation in `std::function`. – Jun 01 '19 at 09:35
I have never used a lambda expression, so I made a little searching and understood. Thank you! – Lim Seungwook Jun 01 '19 at 09:47
@snak `B b2([&a] { a.AF(); });`, as already edited by Ted, answers question in 100% and saves OP headache when he decides to add any argument to delegated function. – R2RT Jun 01 '19 at 09:47

Avin Kavish · Answer 2 · 2019-06-01T10:14:18.993

1

Here's a C# style implementation of the accepted answer, It is memory efficient and flexible as you can construct and delegate at different points of execution which a C# developer might expect to do:

#include <iostream>
#include <functional>

using namespace std;

class A {
public:
    void AF() { cout << "A::F" << endl; }
    void BF() { cout << "B::F" << endl; }
};

class B {
public:
    std::function<void()> Delegate;
};

int main() {
    A a;
    B b;

    b.Delegate = std::bind(&A::AF, &a);
    b.Delegate();

    b.Delegate = [&a] { a.BF(); };
    b.Delegate();
    return 0;
}

edited Jun 01 '19 at 10:14

answered Jun 01 '19 at 10:07

Avin Kavish

8,317
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Technically, it is more efficient. Accepted answer's is forced to copy-initialize member `std::function` due to `const&` in constructor. In your case it will get initialized directly without any temporary objects. – R2RT Jun 01 '19 at 10:12
Oh okay. I'm a bit new to c++ too. I'll change accordingly. – Avin Kavish Jun 01 '19 at 10:13
I mean you don't have to change anything. – R2RT Jun 01 '19 at 10:14
The text at the top :) – Avin Kavish Jun 01 '19 at 10:14
I am familiar with C # style and this is a good answer for me. – Lim Seungwook Jun 01 '19 at 10:39

Can i use C++ function pointers like a C#'s Action?

2 Answers2