Refer to derivative function

Question

I have a base class and a few derivative. I have to 'register' some static function from each of them. Here is the example:

class Base
{
   // Some interface...
};

class Der1 : Base
{
   static void Do();
};
class Der2 : Base
{
   static void Do();
};

void processStatic()
{
   SomeFunc(Der1::Do);
   SomeFunc(Der2::Do);
}

As you see, SomeFunc receives function pointer. I want to do that automatically with each new derivative class, is it possible? Maybe, predefine static function in Base class and register it there. But I think it's impossible, yes?

Maybe, this will be more easier to understand what do I want:

class Der1 : Base
{
   Der1() { SomeFunc(Der1::Do); }
   static void Do();
};
class Der2 : Base
{
   Der2() { SomeFunc(Der2::Do); }
   static void Do();
};

What does `SomeFunc` do? Do you want it to be called every construction, like your second example, or only once, like presumed in your first example? Does it matter when? Does order matter? — GManNickG, Apr 06 '11 at 20:05
Do you want to call the registration function once per class, or once per object? The first example implies the former; the second example implies the latter. — Robᵩ, Apr 06 '11 at 20:06
@rob-adams once per class. @gman that function put function pointer into some list which I use later. — Max Frai, Apr 06 '11 at 20:07
@gman The main idea is that each Derivative class will add it's static method into some map of function pointers. — Max Frai, Apr 06 '11 at 20:45
@Ockonal: I understand, but where is `Base` necessary? Or is that just noise with respect to the question? — GManNickG, Apr 06 '11 at 21:03
@gman I just thought how to do that with base class. But seems it's possible without it. — Max Frai, Apr 06 '11 at 21:04
@Ockonal: Okay, so you don't need `Base`? That was just an attempt? — GManNickG, Apr 06 '11 at 21:05
@gman yes :) Any way to automatically call SomeFunc with each new derived class. — Max Frai, Apr 07 '11 at 04:32

Robᵩ · Accepted Answer · 2011-04-06T20:23:08.097

EDIT: Completely replacing previous answer due to clarified requirements.

You could use the CRTP to declare a specialized base class that does nothing more than call your registration function:

#include <iostream>
void SomeFunc(void(*fp)()) {
  (*fp)();
};

template <class D>
struct ExtraBass {
  ExtraBass() {
    static bool once;
    if(!once)
      SomeFunc(D::Do);
    once = true;
  }
};

struct Bass {
};

struct Drive : Bass, ExtraBass<Drive>  {
  static  void Do() { std::cout << "Drive::Do\n"; }
};

struct Deride : Bass , ExtraBass<Deride> {
  static  void Do() { std::cout << "Deride::Do\n"; }
};

int main() {
  Drive d1;
  Deride d2;
  Deride d3;
}

Bass -> Base. :) Though extra bass can be sweet. – GManNickG Apr 06 '11 at 20:24 — GManNickG, Apr 06 '11 at 20:24

score 0 · Answer 2 · edited May 23 '17 at 11:55

0

This is not an easy thing to do in C++, but I'm not saying it's impossible. If all you need is a list of subclass names, these answers might help:

Somehow register my classes in a list

c++ List of classes without initializing them for use of static functions

Seems either macro magic or boost mpl is your tool of choice.

edited May 23 '17 at 11:55

Community

1
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answered Apr 06 '11 at 20:02

rubenvb

74,642
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Yes, I agree, it sounds impossible without some "macro magic" – satnhak Apr 06 '11 at 20:05

score 0 · Answer 3 · answered Apr 06 '11 at 20:40

0

I just wondering, if you did something like

void SomeFunc(void (*doFunc)())
{ 
   doFunc();
}

template <class T> int Register()
{
    SomeFunc(T::Do);
    return 0;
}

template <class T> class Base
{
   static int _i;
};

template <class T> int Base<T>::_i =  Register<T>();

class Derived : Base<Derived>
{
   public:
   static void Do() {  }
};

answered Apr 06 '11 at 20:40

satnhak

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`template int Base::_i = Register();` That is not automatic registering I was talking about. – Max Frai Apr 06 '11 at 20:50

Refer to derivative function

3 Answers3