C++: Is "Virtual" inherited to all descendants

Question

Assume the following simple case (notice the location of virtual)

class A {
    virtual void func();
};

class B : public A {
    void func();
};

class C : public B {
    void func();
};

Would the following call call B::func() or C::func()?

B* ptr_b = new C();
ptr_b->func();

Interlining

yes, declaring a method as virtual makes all its descendants virtual – Corazoncorban

Answer 1

Examples using pointers as well as reference.

• Using pointer

  B *pB = new C();
  pB->func(); //calls C::func()
  
  A *pA = new C();
  pA->func(); //calls C::func()
  

• Using reference. Note the last call: the most important call.

  C c;
  B & b = c;
  b.func(); //calls C::func() 
  
  //IMPORTANT - using reference!
  A & a = b;
  a.func(); //calls C::func(), not B::func()
  

Online Demo : http://ideone.com/fdpU7

Investigate

Answer 2

1. Your code is invalid C++. What are the parentheses in class definition?
2. It depends on the dynamic type of the object that is pointed to by pointer_to_b_type.

3. If I understand what you really want to ask, then 'Yes'. This calls C::func:

   C c;
   B* p = &c;
   p->func();
   

Fist

@Amir: he IS asking yes/no question. – Investigate

@Amir: "Is “Virtual” inherited to all descendants" sounds like a yes/no question to me. – Bratwurst

@Amir: really? I've always thought that questions of the form "Is ..." and "Would ..." are yes/no questions. – Fist

Answer 3

It calls the function in the class that you're referring to. It works it's way up if it doesn't exist, however.

Try the following code:

#include <iostream>
using namespace std;

class A {
    public:
    virtual void func() { cout << "Hi from A!" << endl; }
};

class B : public A {
    public:
    void func()  { cout << "Hi from B!" << endl; }
};

class C : public B {
    public:
    void func()  { cout << "Hi from C!" << endl; }
};

int main() {
  B* b_object = new C;
  b_object->func();
  return 0;
}

Hope this helps

Sunflower

Apologies for that - I completely missed that in his post. Amended. My answer still stands though ;) – Sunflower