exam watch question

[jayalakshmi charugundla]

Hi all

This question is from K&B book in the exam watch topic.

The below code won't compile. Here I will get unreported exception MyException; must be caught or declared to be thrown.

I dont understand what actually the code is telling. As I know doStuff() declaring the checked exception which is thrown by us and this method is not handling that exception

Am I right? If I am wrong please explain me what is going on in the program?

class MyException  extends Exception{
void someMethod(){
doStuff()();
}
void doStuff() throws MyException{
try{
throw new MyException();
}
catch(MyException me){
throw me;
}
}
}

Thanks in advance,
Jaya

[Hiruka Sandev]

jayalakshmi charugundla wrote:
As I know doStuff() declaring the checked exception which is thrown by us and this method is not handling that exception

That is correct. 'someMethod' needs to explicitly handle the 'MyException' exception which is thrown by the 'doStuff' method.

[S Bhanage]

Hi,

Since MyException extends "EXCEPTION", it can be handled as:
1) Using try-catch
2) Declare in throws clause.

Your someMethod() should use either of the above step.
OR
You can extend "RuntimeException" and no need to change
any code further.

It compiles with this minor change.
Hope this is helpful to you.