CODEWITHSUNDEEP
javainheritance
user1265125
user1265125

Reputation: 2656

Method overriding in Java Inheritance

package P1;

public class Base {

   private void pri( ) { System.out.println("Base.pri()");  }

   void pac( ) {  System.out.println("Base.pac()");  }

   protected void pro( ) { System.out.println("Base.pro()"); }

   public void pub( ) { System.out.println("Base.pub()"); }

   public final void show( ) {
       pri();  
       pac();  
       pro();  
       pub(); 
   }    
}

and 

package P2;

import P1.Base;

public class Concrete1 extends Base {
   public void pri( ) { System.out.println("Concrete1.pri()"); }
   public void pac( ) { System.out.println("Concrete1.pac()"); }
   public void pro( ) { System.out.println("Concrete1.pro()"); }
   public void pub( ) { System.out.println("Concrete1.pub()"); }
}

And I'm executing

Concrete1 c1 = new Concrete1();
c1.show( );

Now, the output is shown to be

Base.pri()
Base.pac()
Concrete1.pro()
Concrete1.pub()

Can someone explain why this is so? From what I understood about inheritance, this should have happened:

1) P2.concrete1 inherits P1.Base
2) c1 object of concrete1 is created
3) c1.show() is called. Since P1.Base.show() is public, it can be called.
4) Now, in P2.concrete1 after inheritance, only it's own methods (pri, pac, pro, pub) and P1.Base's inheritable methods (pro, pub, show) are accessible.

Now WHY does it show Base.pri() and Base.pac() in the output when they're not even accessible?

It's clear that I've not got a very clear fundamental understanding of inheritance. Can someone explain this situation and how inheritance is actually 'structured'. I used to think that inheritable methods and fields of the superclass would just superimpose onto the subclass. But that line of reasoning is obviously wrong.

Thanks!

Upvotes: 8

Views: 5285

Answers (2)

Mr. 14
Mr. 14

Reputation: 9528

Private and no-modifier methods are by design not visible to its subclass. See here for detailed explanation.

Upvotes: 0

EdC
EdC

Reputation: 2349

The short answer is that you can only override methods that are visible. The first two methods, pri and pac are private and package protected respectively. Because nothing outside the class can see a private method, it can't be overridden. Similarly because Concrete1 is in a different package from Base it can't see Base.pac so cannot override it.

What this means is that while you define a pri and pac method in Concrete1, they're just methods that happen to have the same name as the methods in Base, not overrides. The other two methods pro and pub are protected and public respectively, so are visible to Concrete1. As a result the pro and pub methods in Concrete1 are overrides of the methods of the same name in Base.

Because show is defined in Base, it's compiled calling the 4 methods as defined in Base. When executed the JVM looks to see if any are overridden and if they are executes the overridden methods. As explained above, pri and pac are not overridden so the Base versions are executed where as pro and pub are so the Concrete1 versions are executed.

If you were to move the show method into Concrete1 instead of Base then it would execute the 4 methods as defined in Concrete1 as those would be the methods visible to show.

Upvotes: 12

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