C++: derived classes and virtual methods

Question

Possible Duplicates:
C++ : implications of making a method virtual
Why is 'virtual' optional for overridden methods in derived classes?

I wonder, what is documented behavior in the following case:

You have

class A
{
 virtual void A()
 {
   cout << "Virtual A"<<endl;
 }
 void test_A()
 {
   A();
 }
}

class B: public A
{
  void A()
  {
   cout << "Non-virtual A in derived class"<<endl;
  }

  void test_B()
  {
    A();
  }
}

A a; B b;
a.test_A();
b.test_A();
b.test_B();

What it supposed to do according to C++ standard and why? GCC works like B::A is also also virtual.

What shoudl happen in general when you override virtual method by non-virtual one in derived class?

I think this is also answered here: http://stackoverflow.com/questions/2963965/why-is-virtual-optional-for-overridden-methods-in-derived-classes — mkj, Nov 12 '10 at 11:18

Björn Pollex · Accepted Answer · 2010-11-12T11:24:48.540

3

The sub-class member function is implicitly virtual if a virtual base-class member function with the same name and signature exists.

edited Nov 12 '10 at 11:24

answered Nov 12 '10 at 11:12

Björn Pollex

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@Downvoter: Please explain, I want to learn. – Björn Pollex Nov 12 '10 at 11:18
You need to look at the code more carefully. BTW I haven't downlvoted. – Prasoon Saurav Nov 12 '10 at 11:19
1

The code is not working, I see that, but I answered the question that was asked in the last line. And with respect to that, my answer is correct. – Björn Pollex Nov 12 '10 at 11:21
I see. I have upvoted to counterbalance. BTW use the words `member function` instead of `method` in C++. C++ and Java are different. – Prasoon Saurav Nov 12 '10 at 11:24

score 2 · Answer 2 · answered Nov 12 '10 at 11:28

The code should not compile as you cannot name a method with the name of the class. But regarding what I understand that is your real question:

Will making a method virtual imply that the same method in all the derived classes is virtual even if the virtual keyword is not present?

The answer is yes. Once a method is declared virtual in a class, then all overrides of that method will be virtual, and the virtual keyword is optional in derived classes (even if I recommend typing it, if only for documentation purposes). Note that for a method in a derived class to be an override it has to have the same name and signature, with only potential difference being a covariant return type:

struct A {};
struct B : A {};
struct base {
   virtual A* foo();
   virtual A* bar();
};
struct derived : base {
   virtual B* foo();    // override, covariant return type
   virtual int bar();   // not override, return type is not covariant
   virtual A* bar(int); // not override, different argument list
};

Yes: I guess so. The last part of the question is exactly that. +1 — Chubsdad, Nov 12 '10 at 11:33

score 0 · Answer 3 · edited Nov 12 '10 at 11:27

0

According to standard it should be

A a; B b;
a.test_A();  //"Virtual A"
b.test_A(); //Non-virtual A in derived class
b.test_B(); //Non-virtual A in derived class

edited Nov 12 '10 at 11:27

Chubsdad

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answered Nov 12 '10 at 11:18

yozhik

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Chubsdad · Answer 4 · 2010-11-12T11:23:48.567

0

This code is ill-formed. A constructor cannot have a return type (as you have done for the constructor of 'A'). Also a constructor cannot be virtual.

After fixing A's constructor, class B is ill-formed as the constructor of A is private.

So, there are many problems with this code (including missing semicolons at the class definitions).

edited Nov 12 '10 at 11:23

answered Nov 12 '10 at 11:18

Chubsdad

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C++: derived classes and virtual methods

4 Answers4