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
– Sideshow Bob, Commented Oct 10, 2012 at 14:05

Nick Dandoulakis · Accepted Answer · 2013-01-12 12:13:02Z

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, 2013 at 12:13

answered Jun 13, 2009 at 13:07

Nick Dandoulakis

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3 Comments

Girish Over a year ago

Tudok, after going through your post a second time I realized that you have given the right answer too (trick 2). Thanks.

user180574 Over a year ago

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::*)()'

Nick Dandoulakis Over a year ago

@user180574, I remember that it did work. Anyway, you're right. I tested it with gcc and got the same error. I fixed it.

James Curran · Accepted Answer · 2009-06-13 13:18:41Z

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, 2009 at 13:18

James Curran

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6 Comments

Todd Gardner Over a year ago

Technically, "f.do_something" returns the pointer, and the ".*" operator calls the pointer-to-member function on a class.

ThomasMcLeod Over a year ago

So I assume return (f.*Foo::do_something)(5) should also work.

Joey Dumont Over a year ago

@ThomasMcLeod It does not.

Andy Over a year ago

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)

Caleth Over a year ago

@ThomasMcLeod no, because Foo::do_something names the data member of the class. You need an instance to access a non-static data member.

|

Volkan Ozyilmaz · Accepted Answer · 2012-11-06 14:30:50Z

3

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.

edited Nov 6, 2012 at 14:30

answered Nov 6, 2012 at 12:33

Volkan Ozyilmaz

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3 Comments

Martijn Pieters Over a year ago

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.

Volkan Ozyilmaz Over a year ago

@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.

Martijn Pieters Over a year ago

Well, you could explain what you changed, for example, and why.

S.P · Accepted Answer · 2009-06-13 13:05:12Z

0

Try (f.*do_something)(5);

answered Jun 13, 2009 at 13:05

S.P

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Comments

SaeidMo7 · Accepted Answer · 2016-05-11 15:34:32Z

0

#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;

}

answered May 11, 2016 at 15:34

SaeidMo7

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Comments

Pradip Bhojania · Accepted Answer · 2009-10-22 10:20:45Z

-4

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

answered Oct 22, 2009 at 10:20

Pradip Bhojania

1

2 Comments

ForceMagic Over a year ago

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

Willem D'Haeseleer Over a year ago

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

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