over ridden method and super constructor

[nirjari patel]

public class SuperTest{

          public SuperTest(){}
          public SuperTest(int i){}
          public void testMethod(){
                    System.out.println("This is test method in super class");
          }
}

public class SubTest{
          public SubTest(){
                    super.SuperTest();
          }
          public void testMethod(){
                    System.out.println("This is test method in sub class");

          SuperTest sup = new SubTest();
          sup.testMethod();
          }
}

When I run this code it prints output "This is test method in sub class"

Why is it so ? SuperTest sup is parent class to SubTest. It has access to methods defined in super class. How come its accessing methos in a child class ? Accrding to my understanding it should print "This is test method in super class". PLease explain, why is a super class reference accessing method of a subclass ?

[saurav sarkar]

Hi Nirjari,

It is because the Instance you have created is of the sub class not the super class.
sup is just a reference of type super class which is actually pointing to the instance of the sub class.

Hope it helps .

cheers,
Saurav

[Unnar Björnsson]

You forgot to extend SuperTest

[Unnar Björnsson]

And how were you able to compile this code, there is no method SuperTest()?

[nirjari patel]

SuybTest declaration is as follows.

"public class SubTest extends SuperTest"

It is because the Instance you have created is of the sub class not the super class.
sup is just a reference of type super class which is actually pointing to the instance of the sub class.

I understand that instance created is of sub class. But reference type is of SuperTest which is super class. So this reference will have access to methods of super class only. In that case, it should access
testMethod() available in SuperTest.

Please explain.

[Unnar Björnsson]

When you override a method in a parent class you do not need to create an instance of the parent class and call it, you simply say

super.testMethod() // call to the parent method

to use the child method

testMethod() // call to child method

so

SuperTest sup = new SubTest();
sup.testMethod(); 

is unnecessary, all you really need is:

public class SuperTest
{
	public SuperTest(){}
	public SuperTest(int i){}
	public void testMethod()
	{
		System.out.println("This is test method in super class");
	}
}

public class SubTest extends SuperTest
{
	public SubTest()
	{
	}
	public void testMethod()
	{
		System.out.println("This is test method in sub class");
	}
}

[jishnu dasgupta]

Hi Nirjari,


Actually the method to be called at runtime is dependent on the object instance rather than an object references. So because you are using the subclass instance, the subclass method is being called, although at compile time, the superclass method is being checked for syntax. It is for the same reason that if your superclass method was written as:

public class SuperTest{   
  
          public SuperTest(){}   
          public SuperTest(int i){}   
          public void testMethod() throws Exception{   
                    try{
                           System.out.println("This is test method in super class");
                           throw new Exception();
                        }   
                   catch(Exception e)
                    {
                                   System.out.println(" i am being thrown by superclass");
                     }
          }   
}   

then the statement:

SuperTest sup = new SubTest();   
          sup.testMethod();   

would not compile, as the complier would be matching it to the superclass reference.

And guess what?? If you did include the proper try and catch statement, the ouput would be "This is test method in sub class"

as the sublass instance is being used to determine the overriden method to be run at runtime.!!!