Re: Difference between compile time and runtime reference/objects

From:

Wojtek <nowhere@a.com>

Newsgroups:

comp.lang.java.programmer

Date:

Mon, 10 Mar 2008 18:47:20 GMT

Message-ID:

<mn.52c37d8339725db0.70216@a.com>

lielar wrote :

Hi

I have the following code
-------------------------------------------------
interface I {
        int i = 0;
}

class A implements I {
        int i = I.i + 1;

}

class B extends A {
    int i = I.i + 2;

    static void printAll(A obj) {
           System.out.println(obj.i);
    }

The signature for the printAll method specifies the 'A' class. Since
'B' extends 'A', 'B' can be passed in, however it will be treated as an
'A'.

    public static void main(String [] args) {
           B b = new B();
           A a = new B();
           printAll(a);
           printAll(b);
      }

}

So printAll(a) is a 'B', which is then treated as an 'A'.
printAll(b) is also treated as an 'A'

-----------------------------------------------------

I get 1, 1. As the second object I passed is of reference and object
type B , why is it that the answer it prints? Without overloading the
method, can I make it print '2'?

No. You must have:

static void printAll(B obj) {
System.out.println(obj.i);

The compiler will then choose the proper method for the class you are
passing in.

Side note: Always use properly named variables, classes. There is some
confusion with:

B b = new B();
A a = new B();

--
Wojtek :-)