virtual function call from base class

Question

Say we have:

 Class Base {        virtual void f(){g();};     virtual void g(){//Do some Base related code;} };  Class Derived : public Base {        virtual void f(){Base::f();};     virtual void g(){//Do some Derived related code}; };  int main() {     Base *pBase = new Derived;     pBase->f();     return 0;   }  

Which g() will be called from Base::f()? Base::g() or Derived::g()?

Thanks...

Answer 1

The g of the derived class will be called. If you want to call the function in the base, call

Base::g(); 

instead. If you want to call the derived, but still want to have the base version be called, arrange that the derived version of g calls the base version in its first statement:

virtual void g() {     Base::g();     // some work related to derived } 

The fact that a function from the base can call a virtual method and control is transferred into the derived class is used in the template method design pattern. For C++, it's better known as Non-Virtual-Interface. It's widely used also in the C++ standard library (C++ stream buffers for example have functions pub... that call virtual functions that do the real work. For example pubseekoff calls the protected seekoff). I wrote an example of that in this answer: How do you validate an object’s internal state?

Answer 2

It is the Derived::g, unless you call g in Base's constructor. Because Base constructor is called before Derived object is constructed, Derived::g can not logically be called cause it might manipulate variables that has not been constructed yet, so Base::g will be called.