views:

337

answers:

2

Hi. Probably asking this shows that I don't have the convenient grasp on OOP, but...

How can I call a derived function from a base class? I mean, being able to replace one function from the base to the derived class.

Ex.

class a
{
public:
   void f1();
   void f2();
};

void a::f1()
{
   this->f2();
}

/* here goes the a::f2() definition, etc */

class b : public a
{
public:
   void f2();
};

/* here goes the b::f2() definition, etc */    

void lala(int s)
{
  a *o; // I want this to be class 'a' in this specific example
  switch(s)
  {
   case 0: // type b
     o = new b();
   break;
   case 1: // type c
     o = new c(); // y'a know, ...
   break;
  }

  o->f1(); // I want this f1 to call f2 on the derived class
}

Maybe I'm taking a wrong approach. Any comment about different designs around would also be appreciated. Thanks!

+11  A: 

Declare f2() virtual in the base class.

class a
{
public:
       void f1();
       virtual void f2();
};

Then whenever a derived class overrides f2() the version from the most derived class will be called depending on the type of the actual object the pointer points to, not the type of the pointer.

sharptooth
Thx, now I know what the 'virtual' thing was... :D
huff
Actually, virtual is THE feature that makes real OOP possible in C++.
Thorsten79
Yeah, well... at least the inheritance part of it... imio (in my ignorant opinion), OOP would still be practical just using compositions...
huff
@huff: True. `virtual` isn't exactly the enabling idiom as far as OOP goes, precisely. `virtual` makes polymorphism possible, which is one part of OOP.
John Dibling
+3  A: 

Declare f2() as virtual.

Franci Penov