Polymorphism

[Tanatep Pichetvasin]

From the below code, the class B is the subclass of class A. After the instance a is created, the method go is called. Java run this method of the class B. And the value of the attribute x is changed within the method go.
Then I display the value of the attribute x of a. The initial value of x in class A is shown. My question is why doesn't Java show the lastest value of x after the method go finish ?

import java.io.*;
 
public class TestJava {
   
   public static void main(String[] args) throws IOException
   {
      A a = new B();
      a.go();
      System.out.println("x = " + a.x); /* Question */
      System.out.println("----------------------------");
 
      B b = new B();
      b.go();
      System.out.println("x = " + b.x);
   }
   
}
 
class A {
 
   public int x = 10;
 
   void go() throws IOException {
      x *= 10;
      System.out.println("A.go() : x = " + x);
   }
}
 
class B extends A {
 
   public int x = 20;
 
   void go() throws EOFException {
      x *= 30;
      System.out.println("B.go() : x = " + x);
   }
   void gone() {}
}

output >>
B.go() : x = 600
x = 10
----------------------------
B.go() : x = 600
x = 600

[edit]Add code tags. CR[/edit]
[ August 27, 2008: Message edited by: Campbell Ritchie ]

[Seetharaman Venkatasamy]

Originally posted by Tanatep Pichetvasin:
import java.io.*;


class A {

public int x = 10;

void go() throws IOException {
x *= 10;
System.out.println("A.go() : x = " + x);
}
}

class B extends A {

public int x = 20;

void go() throws EOFException {
x *= 30;
System.out.println("B.go() : x = " + x);
}
void gone() {}
}

the overridding method should not throw new checked exception.it should give compilation error.

Hope This Helps

[Vijitha Kumara]

the overridding method should not throw new checked exception.it should give compilation error.

EOFExeption extends IOException, so no compiler error. Instance variables accessed by the reference type. That's why a.X output 10.

hope this helps

[Prashanti Mukund]

In the code you are declaring A's instance of type B so when you call a.go the control is passed to B's go method as a result A's x value is not changed
try creating A's instance as

A a1=new A();

and then call

 a.go()

method now a will return 100 not 10

hope you understood

[Prashanti Mukund]

import java.io.*;
 
public class TestJava {
 
public static void main(String[] args) throws IOException
{
A a = new A();
a.go();
System.out.println("x = " + a.x); /* Question */
System.out.println("----------------------------");
 
A a1 = new B();
a1.go();
System.out.println("x = " + a1.x); /* Question */
System.out.println("----------------------------");
 
B b = new B();
b.go();
 
System.out.println("x = " + b.x);
 
B c = new B();
c.go();
 
System.out.println("x = " + c.x);
 
}
 
}
 
class A {
 
 int x = 10;
 
void go() throws IOException {
x *= 10;
System.out.println("A.go() : x = " + x);
}
}
 
class B extends A {
 
int x = 20;
 
void go() throws EOFException {
System.out.println("Initial value of x="+x);
x *= 30;
System.out.println("B.go() : x = " + x);
 
}
void gone() {}
}

[Campbell Ritchie]

Morning, and welcome to JavaRanch both of you

Tanatep Pichetvasin: please find the CODE button for posts in future; I have edited your post to add code tags, and you can see how they make the text easier to read.

[Seetharaman Venkatasamy]

Thanks , i got it

[Tanatep Pichetvasin]

Thank for your explaination
But why doesn't Java override the attribute ? Why override for only method ?

[Ilja Preuss]

The simple answer is: because that's the way it is specified.

Now, why has it been specified that way?

I guess there are mainly two reasons:

- in a good OO design, data should be private, anyway, in which case polymorphism isn't needed, and

- making fields polymorphic would make the JVM more complex and could lead to decreased performance.

As an aside, i don't know of any OO language where fields are polymorphic. So another reason could be tradition...

[Tanatep Pichetvasin]

Ilja, Thank you your answer

[Garrett Rowe]

Originally posted by Ilja Preuss:
As an aside, i don't know of any OO language where fields are polymorphic. So another reason could be tradition...

Fields in Scala are polymorphic. Actually, from the clients perspective there's no difference between fields and methods. Fields in a subclass can override no-argument method declarations in a parent class or trait. But Scala is a multi-paradigm language so maybe this doesn't come from the OO side of it's family.
[ August 28, 2008: Message edited by: Garrett Rowe ]

[Ilja Preuss]

Garrett, thanks for the interesting tidbit!