c++ virtual function call without pointer or reference

Question

As far as I know, virtual function call usually requires pointer or reference. So I am very surprised by the following codes.

#include <iostream>
using namespace std;
class B{
public:
  void runB(){        call();  }
  virtual void call(){ cout<<"B\n"; };
};

class D: public B{
public:
  void runD(){       runB();   }
  void call(){       cout<<"D\n";  }
};

int main(){
 D d;
 d.runD();
}

The output is

D

Could someone please comment why this virtual function call works? Thanks。

To expand on chris's comment: Inside methods for every class there is always an implicit pointer of `*this`. So `call()` and `this->call()` is the same in the code you have. — Soren, Aug 13 '14 at 01:42
Virtual functions are just functions like any other function, the only difference is the key word "Virtual". Therefore all functions have the same attributes, you can define them as void or to return a statement. — Juniar, Aug 13 '14 at 01:43

score 14 · Answer 1 · answered Aug 13 '14 at 01:25

Within a member function, any references to other member functions or variables are implicitly resolved via the this pointer. So in the definition of runB(), the call() really means this->call(). The virtual function call is performed using the current object's virtual table.

AnT stands with Russia · Answer 2 · 2014-08-14T03:23:38.593

Firstly, virtual function call does not require a pointer or a reference. As far as the language is concerned, any call to virtual function is a virtual call, unless you explicitly suppress the virtual dispatch mechanism by using a qualified function name. For example, these

d.D::call(); // calls `D::call()` directly
d.B::call(); // calls `B::call()` directly

are calls which were explicitly forced to be non-virtual. However, this

d.call(); // calls `D::call()` virtually

is a virtual call. In this case it is immediately obvious to the compiler that the target function is D::call(), so the compiler normally optimizes this virtual call into a regular direct call. Yet, conceptually, d.call() is still a virtual call.

Secondly, the call to call() made inside B::runB() is made through a pointer. The pointer is present there implicitly. Writing call() inside B::runB() is just a shorthand for (*this).call(). this is a pointer. So that call is made through a pointer.

Thirdly, the key property of virtual call is that the target function is chosen in accordance with the dynamic type of the object used in the call. In your case, even when you are inside B::runB() the dynamic type of the object *this is D. Which is why it calls D::call(), as it should.

Fourthly, what you really need a pointer or a reference for is to observe the actual polymorphism. Polymorphism proper occurs when static type of the object expression used in the call is different from its dynamic type. For that you do indeed need a pointer or a reference. And that is exactly what you observe in that (*this).call() call inside B::runB(). Even though the static type of *this is B, its dynamic type is D and the call is dispatched to D::call().

score 5 · Answer 3 · answered Aug 13 '14 at 01:38

The difference between virtual to not virtual is:

not virtual - always goes by the caller object/reference/pointer type.

virtual - reference/pointer - goes by the created object type.

virtual - object - goes by the caller.

for example:

class A{
public:
    virtual void f(){
        cout <<"A\n";
    }
};
class B: public A{
public:
    virtual void f(){
        cout <<"B\n";
    }
};


B b;
A a,*pa=&b;
a.f(); //A: caller type = created type - same for not virtual
b.f(); //B: caller type = created type - same for not virtual
((A)b).f(); //A: object goes by the caller type - same for not virtual
pa->f(); // B: pointer goes by the created type - it would be A if it was not virtual!!

c++ virtual function call without pointer or reference

3 Answers3

Linked