question about polymorphism

[zhu weitao]

class Parent{

private void method1(){

System.out.println("Parent's method1");
}

public void method2(){

System.out.println("Parent's method2");
    method1();
}
}
 
class Child extends Parent{

public void method1(){

System.out.println("Child's method1");
}

public static void main(String [] args){

Parent p = new Child();

p.method2();
}
}

The result is

Parent's method2
Parent's method1

but if change the method method1 in class Parent to default protected or public ,the result is

Parent's method2
Child's method1

why???

[Vijitha Kumara]

As your code stands method1() is not inherited to Child class so it's not an override in the Child class. So Parent class methods called. But if you make the modifier default/public/protected of method1() in Parent then the method inherited so overriden by the Child class. Hence method1() of Child class called.

[Sagar Rohankar]

Originally posted by Vijitha Kumara:
As your code stands method1() is not inherited to Child class so it's not an override in the Child class.

As its not inherited to child class , and we had defined a method1(), in Child class, so the code must executes its own method1(), printing

Child's method1

I`m confused with above flow ? Pl elaborate it more !

[zhu weitao]

I know method1() is not inherited to Child class , but the class Child has the method1() method itself , why doesn't it invoke it's method1() method

[Rob Spoor]

Because you are not overriding method1. Overriding and polymorphism are not applicable for private methods. So instead of overriding method1, you are creating a new method that just happens to have the same signature as the private method.

If you would make method1 protected in Parent you would get what you expect.

[Sagar Rohankar]

Originally posted by Rob Prime:
So instead of overriding method1, you are creating a new method that just happens to have the same signature as the private method.

Ok, I agree !

Now, this what compiler saw ,

 class Parent{

private void method1(){

System.out.println("Parent's method1");
}

public void method2(){

System.out.println("Parent's method2");
    method1();
}
}
 
class Child extends Parent{

public void method1(){

System.out.println("Child's method1");
}
        /////////////////////////////////
        // inherited from Parent
         public void method2(){ // 1

System.out.println("Parent's method2");
    method1(); // 2
}
        ////////////////////////////////

public static void main(String [] args){

Parent p = new Child();

p.method2();
}
}   

Now, here object 'p' is of type 'Child' , so it gonna call method2 at line 1.. which in turn calls method1 at line 2. Now we have method1 body in Child class , which prints "Child's method1"..

This is how I conceive it ?

Where I`m wrong ?
[ September 17, 2008: Message edited by: Sagar Rohankar ]

[Rob Spoor]

Because you are overriding method2 as well it will work. That's because Child's method2 does not know anything about Parent's method1 and call Child's method1.

[lahiru dharmasena]

hi Rob Prime,

why is that doesn't happen when parents method1 is private?

[Sagar Rohankar]

Still confused !

Actually , I want to say like The compiler insert this code in Child class, which is inherited from Parent class..

  class Parent{

private void method1(){

System.out.println("Parent's method1");
}

public void method2(){

System.out.println("Parent's method2");
    method1();
}
}
 
class Child extends Parent{

public void method1(){

System.out.println("Child's method1");
}
        /////////////////////////////////
        // This code is compiler generated, I'm not inserting it , explicitly !
        // inherited from Parent
         public void method2(){ // 1

System.out.println("Parent's method2");
    method1(); // 2
}
        ////////////////////////////////

public static void main(String [] args){

Parent p = new Child();

p.method2();
}
}      

Now this method2 code is inserted by compiler with call to method1() in it and in Child class we have our own implementation of method1(), which prints "Child's method1"..

So How the original code works ?

[Campbell Ritchie]

In that case there is no access to a private method from the subclass. If you have a method of the same name, the compiler has to ignore the previous method.

Remember a private method is entirely for "internal consumption;" it is only for use inside its own class. When you create a new method with the same name, the compiler sees that as something completely new and independent.

[Vijitha Kumara]

Sagar Rohankar:
Now this method2 code is inserted by compiler with call to method1() in it

I'm not sure about that the compiler inserts that part in to child code or at the runtime it goes for the Parent object and execute method2() when it sees that child doesn't override that method. In that case Parent's method1() will be executed since it can't access the private method of Child from there (This is what Campbell Ritchie means i think). Correct me if I'm wrong.

[lahiru dharmasena]

In that case Parent's method1() will be executed since it can't access the private method of Child from there

so, your saying look up is comes from parent to child. first go to the parent and look if the method can override the child's method. if it can override, use child's method if it can't, use parent's method. mmm... according to my knowledge it's other way around. (correct me if i'm wrong)

first, look on child's class and if not found go to the parent.

[Vijitha Kumara]

No, What I said ,it goes for the Parent object and execute method2() when it sees that child doesn't override that method, Not it comes back again to child class.

[lahiru dharmasena]

ok. i got it now. thanks

[Sagar Rohankar]

At runtime , when I checked thy type of object 'p', using Object#getClass(), which returns class Child..

Now , Child class must contain definition of 'method2()', in order to execute ! But there is no method defined as such , but it is inherited from Parent class !!.

Now this code contains the call to method1(), the object is of type 'Child' , so the this code must get execute ..

  class Child extends Parent{

public void method1(){ [B]// This method must get execute , but it not ![/B]

System.out.println("Child's method1");
}

public static void main(String [] args){

Parent p = new Child();

p.method2();
}
}  

To simplify , I rewrite the code as :

 class Parent{

private void method1(){

System.out.println("Parent's method1");
}

public void method2(){
[I]System.out.println("Im executing method2" + this.getClass());[/I] 
System.out.println("Parent's method2");
    [B]this.[/B]method1();
}
}
 
class Child1 extends Parent{

public void method1(){

System.out.println("Child's method1");
}

public static void main(String [] args){

Parent p = new Child1();
[I]System.out.println("Type of p : " + p.getClass());[/I]
p.method2();
}
}

Output :

Type of p : class Child
I'm executing method2 class Child
Parent's method2
Parent's method1 // I'm expecting " Child's method1"

Where I`m wrong ?

May be I'm missing something silly or any simplest specification !

[Sagar Rohankar]

Originally posted by Vijitha Kumara:
No, What I said ,it goes for the Parent object and execute method2() when it sees that child doesn't override that method, Not it comes back again to child class.

If that's How overridden methods call gets executed , then I got it !