Existence of VPTR in subobjects

Question

I have a class 'base' with a virtual destructor and thus a VTable and corresponding VTPR in it, and a class derived from it:

class base {
public:
    virtual ~base() {}
};

class der : base {};

main()
{

    int a = sizeof(base); // = 4 , fine !

    int b = sizeof(der);  // = 4 too ?
}

Now as derived class too is virtual, it'll have a VPTR of its own, but since it also has a subobject of the base class with a VPTR in it, shouldn't the size of the class 'der' be 8 bytes i.e. the size of the VPTR of the class 'der' + size of VPTR of the subobject of class 'base'? (when sizeof(void*) = 4 bytes ).

So basically my question is : When the subobject of class 'base' is made in 'der' does it have a seperate new VPTR ? And if it is so then why its size is not getting added while calculating the size of 'der'?

Can somebody please clarify this.

Answer 1

This is all implementation-specific. But in practice, there will only be one vptr in the derived class; there is no need for two. The whole point of the vptr is it's what gets used to dynamically call the correct override of the virtual function; der objects will simply have a different pointer value to base objects.

[Note: Your example is probably confused by the fact that you are (unintentionally?) using private inheritance, rather than the more typical public inheritance...]

Answer 2

I think you're confusing vtables and vptrs. Each class will have a vtable, and each object will store a pointer to its vtable as the vptr. The vtable is like a static global, it is shared between all instances of the class and thus doesn't take any space in the object.