Base class pointer vs inherited class pointer?

Question

Suppose I have a class Dog that inherits from a class Animal. What is the difference between these two lines of code?

    Animal *a         


        

Answer 1

The difference is important when you try to call Dog's methods that are not Animal's method. In the first case (pointer to Animal) you have to cast the pointer to Dog first. Another difference is if you happen to overload non-virtual method. Then either Animal::non_virtual_method() (pointer to Animal) or Dog::non_virtual_method(pointer to Dog) will be called.

Answer 2

For all purposes of type-checking, the compiler treats a as if it could point to any Animal, even though you know it points to a Dog:

• You can't pass a to a function expecting a Dog*.
• You can't do a->fetchStick(), where fetchStick is a member function of Dog but not Animal.
• Dog *d2 = dynamic_cast<Dog*>(d) is probably just a pointer copy on your compiler. Dog *d3 = dynamic_cast<Dog*>(a) probably isn't (I'm speculating here, I'm not going to bother checking on any compiler. The point is: the compiler likely makes different assumptions about a and d when transforming code).
• etc.

You can call virtual functions (that is, the defined polymorphic interface) of Animal equally through either of them, with the same effect. Assuming Dog hasn't hidden them, anyway (good point, JaredPar).

For non-virtual functions which are defined in Animal, and also defined (overloaded) in Dog, calling that function via a is different from calling it via d.