Passing member function pointer to member object in c++

Question

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

Bill · Accepted Answer · 2013-12-02T22:23:39.743

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;
}

Nice, but how to run the function that memberFunctionPointer points at? testFoo.myBar.*memberFuncionPointer() results error — Moomin, Mar 03 '10 at 21:23
@moomin: I added a correct sample of calling the functions this time. — Bill, Mar 03 '10 at 21:33
Just for completeness, you might want to add a default constructor to `bar` to set both pointers to `0`. You could set up the class so that it always only has one member set (i.e. setting one member `0`s the other) and then add an `operator()` function that calls the set member. Though it'll be more complicated than that in the end. — Chris Lutz, Mar 03 '10 at 21:40

score 12 · Answer 2 · answered Mar 03 '10 at 21:02

As the error says, methods belong to the class, not to individual instances. For this reason pointers to free functions and pointers to non-static methods are completely different things. You'll also need an instance to call the method on.

//declaring and taking the address of a foo's method 
void (foo::*method)() = &foo::hello; //as the compiler nicely suggests

//calling a function through pointer
free_func();

//calling a method through pointer
foo instance;
(instance.*method)();

You can use libraries like Boost.Bind and Boost.Function (also in std::tr1 I think) to abstract away the difference and also bind an instance to the method:

#include <iostream>
#include <boost/bind.hpp>
#include <boost/function.hpp>

using namespace std;

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

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


int main()
{
    foo testFoo;

    boost::function<void()> helloFunc(boost::bind(&foo::hello, testFoo));
    boost::function<void()> byeFunc(byebye);

    helloFunc();
    byeFunc();
    return 0;
}

Could please explain how to run the "hello()" from testFoo.myBar? I did 'testFoo.myBar.funcP = &foo::hello;' but '(testFoo.myBar.*funcP)();' returns 'funcP was not declared...' during compilation — Moomin, Mar 03 '10 at 21:17

score 1 · Answer 3 · answered Mar 03 '10 at 20:52

1

To make your second option work, declare foo so the compiler knows that it is a class.

Also note that your function pointer syntax is incorrect. The * comes just before the name of the variable:

class foo;

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

answered Mar 03 '10 at 20:52

R Samuel Klatchko

74,869
16
134
187

score 0 · Answer 4 · answered Mar 03 '10 at 20:54

0

forward foo's declaration in front of bar:

class foo;

answered Mar 03 '10 at 20:54

Andrey

59,039
12
119
163

Passing member function pointer to member object in c++

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