CODEWITHSUNDEEP
c++polymorphismvirtual
Vivek Ranga
Vivek Ranga

Reputation: 75

Virtual function call from a normal function

class base
{
public:
    void virtual func(){cout<<"base";}
    void check()
    {
        func();
    }
};
class derived: public base
{
public:
    void func(){cout<<"dervied";}
};
int main()
{
    base *obj = new derived();
    obj->check();
    return 0;
}

Above code prints derived on the console. Now, I understand the concept of virtual functions but I'm unable to apply it here. In my understanding whenever we call a virtual function, compiler modifies the call to "this->vptr->virtualfunc()" and that's how most heavily derived's class function gets invoked. But in this case, since check() is not a virtual function, how does the compiler determine that it needs to invoke func() of derived?

Upvotes: 2

Views: 251

Answers (2)

David Schwartz
David Schwartz

Reputation: 182753

Exactly the way you said, by using the vptr in the class member. It knows the function is virtual, therefore it knows it has to call it through the virtual function table.

Upvotes: 2

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

how does the compiler determine that it needs to invoke func() of derived?

In the same exat way - by invoking this->vptr->virtualfunc(). Recall that this belongs to the derived class even inside the base class, because each derived class is a base class as well, so the same way of accessing virtual functions works for it too.

Upvotes: 7

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