The relation between the declared type and created type

Question

I have an question about the following code (Is this call dynamic binding?). I feel confused about 3 point.

First, what is the mean of the variable pq? Does pd still be the data type of P or be the Q?

Second, when I invoke the pq.m(pp) method, why the result become Q::P but not P::Q?

Finally, what is this mean ((P) qq).m(qq);? I hope somebody could solve my problem.

The result of the following code will be
P::Q, Q::P, Q::Q, R::P, Q::P, Q::Q, Q::Q

    class Test {
    public static void main(String[] args) {
        P pp  = new P();     
        Q qq  = new Q();
        R rr = new R();
        P pq  = qq;
        pp.m(qq);    
        pq.m(pp);  
        pq.m(qq);         
        rr.m(pp); 
        qq.m(pq);   
        qq.m(qq); 
        ((P) qq).m(qq);      
    }
}
class P {
    public void m(P p){System.out.println("P::P"); } 
    public void m(Q p){System.out.println("P::Q"); }
    public void m(R c){System.out.println("P::R"); }
}
class Q extends P {
    public void m(P p){System.out.println("Q::P"); } 
    public void m(Q p){System.out.println("Q::Q"); }
    public void m(R c){System.out.println("Q::R"); }
}
class R extends Q {
      public void m(P p){System.out.println("R::P"); } 
      public void m(Q p){System.out.println("R::Q"); }
    public void m(R c){System.out.println("R::R"); }
}

There is no variable `pd`, so it's difficult to answer your first question... — Oliver Charlesworth, Apr 19 '12 at 12:51
But to answer all your questions, you should read about [inheritance in Java](http://docs.oracle.com/javase/tutorial/java/IandI/subclasses.html). — Oliver Charlesworth, Apr 19 '12 at 12:52

score 0 · Answer 1 · answered Apr 19 '12 at 12:59

Starting, I think you mean pq, not pd. Since Q extends P, Q is also a P type. It's like saying that an apple is a fruit. So you take the apple (Q) and says: It's a fruit (P). When you call the pq methods, they will call the methods from the Q class, since pq is still a Q object. In the last part, when you do ((P) qq).m(qq);, is the same as doing the following:

P p = (P) qq;
q.m(qq);

So as said above, the code will still call the method from the Q class, printing "Q::Q"

score 0 · Accepted Answer · answered Apr 19 '12 at 13:05

P pq = qq; means that pq is known to the rest of the program as a type P. But as the creator, you know it's really of type Q. So this means that when you call pq.m(), it's really calling the implementation from class Q.

It's called overriding a method. So when you call pq.m(pp), you are really calling: public void m(P p){System.out.println("Q::P"); because that is the method from class Q.

If Q did not have a m(P) method, then it would automatically call the superclass method, the one from P.

((P) qq).m(qq); is the same as doing:

P pqq = (P)qq;   // pqq is known as P type, but it's instance is still the original Q type
pqq.m(qq); // Again, since pqq is truly an instance of Q, it calls Q.m(Q)

You should really read about inheritance. This is a bigger subject than can be explained here.

All this being said, your example doesn't illustrate its power well. But for example, if class Q had an extra method, public void sayHello();, then

Q q = new Q();
P p = new Q();
q.sayHello(); // This would be legal
p.sayHello(); // This would be illegal because the compiler knows p as a declared instance of P, even though you know it's truly a Q.
((Q)p).sayHello(); // This would be legal because you told the compiler to look at p as an instance of Q. It's called a cast.

I hope this all helps. Be sure to go read up on object orientation.

score 0 · Answer 3 · answered Apr 19 '12 at 13:09

Dynamic binding, and therefore polymorphism, only works for the object to the left of the dot in a method call (o.m(x) -- only for o). The arguments types are resolved statically, at compile time. Take a more well-known situation:

class A { 
  public boolean equals(A other) { 
    System.out.println("A.equals called"); return true;
  }
}

A a1 = new A(), a2 = new A();
Object o = a1;
o.equals(a1); // doesn't print anything
a1.equals(o); // doesn't print anything
a1.equals(a2); // prints "A.equals called"

The point here is that class A doesn't override Object.equals(Object), but instead just adds another overloaded method A.equals(A) -- and that one gets called only when the argument is of a declared type A.

sorry...i un-downvoted. i found your explanation very confusing because i thought it sounded like you were describing c++ and not java. the last paragraph you added improves your answer. — stevevls, Apr 19 '12 at 13:13

The relation between the declared type and created type

3 Answers3