Why is a vptr required when the derived class doesn't override the virtual function?

Question

class base {
public:
    void virtual fn(int i) {
        cout << "base" << endl;
    }
};

class der : public base{
    public:
    void  fn(char i) {
        cout << "der" << endl;
    }
};

int main() {

    base* p = new der;
    char i = 5;
    p->fn(i);
    cout << sizeof(base);
    return 0;
}

Here signature of function fn defined in base class is different from signature of function fn() defined in der class though function name is same. Therefore, function defined in der class hides base class function fn(). So class der version of fn cannot be called by p->fn(i) call; It is fine.

My point is then why sizeof class base or der is 4 if there is no use of VTABLE pointer? What is requirement of VTABLE pointer here?

Answer 1

Note that this is highly implementation dependent & might vary for each compiler.

The requirement for presence of vtable is that the Base class is meant for Inheritance and extension, and a class deriving from it might override the method.

The two classes Base and Derived might reside in different Translation Unit and the compiler while compiling the Base class won't really know if the method will be overidden or not. So, if it finds the keyword virtual it generates the vtable.

Answer 2

The vtable is usually not only used for virtual functions, but it is also used to identify the class type when you do some dynamic_cast or when the program accesses the type_info for the class.

If the compiler detects that no virtual functions are ever overridden and none of the other features are used, it just could remove the vtable pointer as an optimization.

Obviously the compiler writer hasn't found it worth the trouble of doing this. Probably because it wouldn't be used very often, and because you can do it yourself by removing the virtual from the base class.