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method invocation problem

Ravi Bansal
Ranch Hand

Joined: Aug 18, 2008
Posts: 84
This below code is doing valid override of eat method . I just want to understand what is the problem with invoking the superclass version of method using the superclass object itself.

class Animal{
public void eat() throws Exception


public class dog extends Animal {
public void eat(){

public static void main(){

Animal d1=new Animal();; //this does not work ....why?

Animal d2= new dog();; //this does not work either.....why?


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marc weber

Joined: Aug 31, 2004
Posts: 11343

What is it about this code that doesn't work (aside from your main method lacking an argument and failing to declare or catch the possible exception)?

What are you expecting from these methods?

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Ganeshkumar cheekati
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Joined: Oct 13, 2008
Posts: 362
At first you must handle or declare the exception which throws by method in super class.


with out using String[] a in main method it compiles but JVM says that NoSuchMethodError.

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Ravi Bansal
Ranch Hand

Joined: Aug 18, 2008
Posts: 84
i m sorry ......i made a typo while writing main method(missed String args)........

another point....why should i handle an exception in the subclass ..if i m invoking the superclass version of method by using the superclass referece variable which is pointing to superclass object itself?
Ganeshkumar cheekati
Ranch Hand

Joined: Oct 13, 2008
Posts: 362
because super class method is throwing an exception so you must handle or declare it in main method from which you are calling super class method
[ November 07, 2008: Message edited by: Ganeshkumar cheekati ]
Ankit Garg

Joined: Aug 03, 2008
Posts: 9291

well you will have to handle it because you are using the super class reference to call the method. So at compile time the compiler will look at the method in the super class.

I have modified your code a bit

Now you might be saying that the method eat in class Animal actually doesn't throws any exception. But in it's throws clause it has a CHECKED exception (i.e. Exception). So it will force you to handle that exception whether it is actually being thrown or not. This is why call 1 will not work

Second thing about call 1. The actual method that will be called will be eat of dog class. And eat in dog doesn't thrown any exception. But the reference used to invoke eat is of type Animal. So the compiler will check the code of eat in Animal class and not dog class. Since it will find throws Exception there, it will force you to handle it.

call 2 will work as you are using dog reference to call it. Since the eat method in dog class doesn't throws any exception, so the call would be just fine and the compiler will not force you to handle any exception...

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