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Home/ Questions/Q 646067
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T21:34:17+00:00 2026-05-13T21:34:17+00:00

I have a problem with using a pointer to function in C++. Here is

  • 0

I have a problem with using a pointer to function in C++. Here is my example:

#include <iostream>

using namespace std;

class bar
{
public:
    void (*funcP)();
};

class foo
{
public:
    bar myBar;
    void hello(){cout << "hello" << endl;};
};

void byebye()
{
    cout << "bye" << endl;
}


int main()
{
    foo testFoo;

    testFoo.myBar.funcP = &byebye;         //OK
    testFoo.myBar.funcP = &testFoo.hello;  //ERROR
    return 0;
}

Compilator returns an error at testFoo.myBar.funcP = &testFoo.hello;:

ISO C++ forbids taking the address of a bound member function to form a
pointer to member function. Say
‘&foo::hello’

cannot convert ‘void (foo::)()’ to ‘void ()()’ in assignment

So i tried it like this:

class bar
{
public:
    void (*foo::funcP)();
};

But now the compilator adds one more:

‘foo’ has not been declared

Is there a way make it work?

Thanks in advance for suggestions

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  1. Editorial Team
    Editorial Team
    2026-05-13T21:34:17+00:00Added an answer on May 13, 2026 at 9:34 pm

    Taking everyone’s suggestions together, your final solution will look like:

    #include <iostream> 
    using std::cout;
    usind std::endl;
    
    class foo; // tell the compiler there's a foo out there.
    
    class bar 
    { 
    public: 
        // If you want to store a pointer to each type of function you'll
        // need two different pointers here:
        void (*freeFunctionPointer)();
        void (foo::*memberFunctionPointer)();
    }; 
    
    class foo 
    { 
    public: 
        bar myBar; 
        void hello(){ cout << "hello" << endl; }
    }; 
    
    void byebye() 
    { 
        cout << "bye" << endl; 
    } 
    
    
    int main() 
    { 
        foo testFoo; 
    
        testFoo.myBar.freeFunctionPointer = &byebye;
        testFoo.myBar.memberFunctionPointer = &foo::hello;
    
        ((testFoo).*(testFoo.myBar.memberFunctionPointer))(); // calls foo::hello()
        testFoo.myBar.freeFunctionPointer();   // calls byebye()
        return 0; 
    } 
    

    The C++ FAQ Lite has some guidance on how to simplify the syntax.

    Taking Chris’ idea and running with it, you could get yourself something like this:

    #include <iostream>
    using std::cout; using std::endl;
    
    class foo;
    typedef void (*FreeFn)();
    typedef void (foo::*MemberFn)();
    
    class bar
    {
    public:
      bar() : freeFn(NULL), memberFn(NULL) {}
      void operator()(foo* other)
      {
        if (freeFn != NULL) { freeFn(); }
        else if (memberFn != NULL) { ((other)->*(memberFn))(); }
        else { cout << "No function attached!" << endl; }
      }
    
      void setFreeFn(FreeFn value) { freeFn = value; memberFn = NULL; }
      void setMemberFn(MemberFn value) { memberFn = value; freeFn = NULL; }
    private:
      FreeFn freeFn;
      MemberFn memberFn;
    };
    
    class foo
    {
    public:
      bar myBar;
      void hello() { cout << "foo::hello()" << endl; }
      void operator()() { myBar(this); }
    };
    
    void bye() { cout << "bye()" << endl; }
    
    int main()
    {
      foo testFoo;
    
      testFoo();
    
      testFoo.myBar.setMemberFn(&foo::hello);
      testFoo();
    
      testFoo.myBar.setFreeFn(&bye);
      testFoo();
    
      return 0;
    }
    
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