C++ function pointer (class member) to non-static member function

Question

class Foo {
public:
    Foo() { do_something = &Foo::func_x; }

    int (Foo::*do_something)(int);   // function pointer to class member function

    void setFunc(bool e) { do_something = e ? &Foo::func_x : &Foo::func_y; }

private:
    int func_x(int m) { return m *= 5; }
    int func_y(int n) { return n *= 6; }
};

int
main()
{
    Foo f;
    f.setFunc(false);
    return (f.*do_something)(5);  // <- Not ok. Compile error.
}

How can I get this to work?

Aren't `func_x` and `func_y` static functions even though they're not marked as such? — Sideshow Bob, Oct 10 '12 at 14:05

Nick Dandoulakis · Answer 1 · 2013-01-12T12:13:02.613

36

 class A{
    public:
        typedef int (A::*method)();

        method p;
        A(){
            p = &A::foo;
            (this->*p)(); // <- trick 1, inner call
        }

        int foo(){
            printf("foo\n");
            return 0;
        }
    };

    void main()
    {
        A a;
        (a.*a.p)(); // <- trick 2, outer call
    }

edited Jan 12 '13 at 12:13

answered Jun 13 '09 at 13:07

Nick Dandoulakis

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Tudok, after going through your post a second time I realized that you have given the right answer too (trick 2). Thanks. – Girish Jun 13 '09 at 18:24
Your code does not work. I copy/paste and try compile. test_function_pointer_2.C: In constructor 'A::A()': test_function_pointer_2.C:7: error: argument of type 'int (A::)()' does not match 'int (A::*)()' – user180574 Jan 12 '13 at 02:03
1

@user180574, I remember that it did work. Anyway, you're right. I tested it with gcc and got the same error. I fixed it. – Nick Dandoulakis Jan 12 '13 at 12:15

score 33 · Accepted Answer · answered Jun 13 '09 at 13:18

33

The line you want is

   return (f.*f.do_something)(5);

(That compiles -- I've tried it)

"*f.do_something" refers to the pointer itself --- "f" tells us where to get the do_something value from. But we still need to give an object that will be the this pointer when we call the function. That's why we need the "f." prefix.

answered Jun 13 '09 at 13:18

James Curran

101,701
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9

Technically, "f.do_something" returns the pointer, and the ".*" operator calls the pointer-to-member function on a class. – Todd Gardner Jun 13 '09 at 15:30
So I assume `return (f.*Foo::do_something)(5)` should also work. – ThomasMcLeod Jun 05 '13 at 20:00
1

@ThomasMcLeod It does not. – Joey Dumont Sep 29 '14 at 18:52
Ah, but this means that if you have `Foo g`, then this would also compile and run: `return (g.*f.do_something)(5);` Right? (even if there's no good reason to do that) – Andy Jun 14 '15 at 19:36
@ThomasMcLeod no, because `Foo::do_something` names the data member of the class. You need an instance to access a non-static data member. – Caleth Jan 15 '21 at 11:07
@Caleth, correct. This is sever years old but I think I misunderstood the question the first time. – ThomasMcLeod Jan 15 '21 at 15:15

Volkan Ozyilmaz · Answer 3 · 2012-11-06T14:30:50.943

class A {
    int var;
    int var2;
public:
    void setVar(int v);
    int getVar();
    void setVar2(int v);
    int getVar2();
    typedef int (A::*_fVar)();
    _fVar fvar;
    void setFvar(_fVar afvar) { fvar = afvar; }
    void insideCall() { (this->*fvar)(); }
};

void A::setVar(int v)
{
    var = v;
}

int A::getVar()
{
    std::cout << "A::getVar() is called. var = " << var << std::endl;
    return var;
}

void A::setVar2(int v2)
{
    var2 = v2;
}

int A::getVar2()
{
    std::cout << "A::getVar2() is called. var2 = " << var2 << std::endl;
    return var2;
}

int main()
{
    A a;
    a.setVar(3);
    a.setVar2(5);

//    a.fvar = &A::getVar;
    a.setFvar(&A::getVar);
    (a.*a.fvar)();

    a.setFvar(&A::getVar2);
    (a.*a.fvar)();

    a.setFvar(&A::getVar);
    a.insideCall();

    a.setFvar(&A::getVar2);
    a.insideCall();

    return 0;
}

I extended Nick Dandoulakis's answer. Thank you.

I added a function which set the member function pointer from outside of the class. I added another function which can be called from outside to show inner call of member function pointer.

Rather than only post a block of code, please *explain* why this code solves the problem posed. Without an explanation, this is not an answer. — Martijn Pieters, Nov 06 '12 at 12:56
@MartijnPieters I wrote that it is extended from Nick Dandoulakis answer. It was explained in there. I am new in stackoverflow maybe I need to do in a different way. — Volkan Ozyilmaz, Nov 06 '12 at 14:22
Well, you could explain what you changed, for example, and why. — Martijn Pieters, Nov 06 '12 at 14:23

score 0 · Answer 4 · answered May 11 '16 at 15:34

#include<iostream>
using namespace std;

class A {

public:
    void hello()
    {
        cout << "hello" << endl;
    };

    int x = 0;

};


void main(void)
{

    //pointer
    A * a = new A;
    void(A::*pfun)() = &A::hello;
    int  A::*v1 = &A::x;

    (a->*pfun)();
    a->*v1 = 100;
    cout << a->*v1 << endl << endl;

    //----------------------------- 
    A  b;
    void(A::*fun)() = &A::hello;
    int  A::*v2 = &A::x;

    (b.*fun)();
    b.*v2 = 200;
    cout << b.*v2 << endl;

}

score 0 · Answer 5 · answered Jun 13 '09 at 13:05

0

Try (f.*do_something)(5);

answered Jun 13 '09 at 13:05

SuPra

8,488
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score -4 · Answer 6 · answered Oct 22 '09 at 10:20

-4

I think calling a non static member of the class could also be done using a static member function.

answered Oct 22 '09 at 10:20

Pradip Bhojania

1

1

This is not possible. "When modifying a data member in a class declaration, the static keyword specifies that one copy of the member is shared by all instances of the class. When modifying a member function in a class declaration, **the static keyword specifies that the function accesses only static members**." See [MSDN reference](http://msdn.microsoft.com/en-us/library/s1sb61xd(v=vs.80).aspx) – ForceMagic Jul 19 '12 at 01:04
well, you could keep a list of all instances trough the constructor and then call it for all instances, but i guess that is somewhat of a whack idea :p – Willem D'Haeseleer Jul 30 '12 at 14:28

C++ function pointer (class member) to non-static member function

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