Question

Let’s have a look at this code:

package rpg;

interface A {
    // An empty interface
}

public class WarriorClass extends CharacterClass implements A {

    final int bab;
    public WarriorClass(int a) {
        bab = a;
    }

    public static void main(String[] args) {
        A a = new WarriorClass(1);
        System.out.println(((WarriorClass) a).bab);
    }
}

First question: Why is it possible to use the interface A as the type of a?
Second question: Since A does not know about the iVar bab why does

System.out.println(((WarriorClass) a).bab);

print the correct value 1?

Answer 1

You do not store the class “in” the interface. You are storing a reference to the class in a variable of the interfice class.

Java only works with references. All have the same length (like a pointer in C). The references are typed so you must assign to it an object of an appropiate class.

As for the relationship, extending a class or implementing an interface means that the new class objects belong also to the original class. So all objects are instances of Object. Since by extending/implementing you affirm that the new class follows the contract (specification) of the superclass/interface, they are do belong to it.