Why the exception is being caught - Suggestion is required.

hi,
Acording to the knowledge i have both theoretical and practial
when a method declares that it throws an ExceptionType then that method never handles that exception at all even though there exists a catch(Exception e) block in the try block.
The above scenario and ethics is violated when i run the following program.
I want to knw why the Exception block is catching the IllegalArgumentException ?
public class TestClass {
private void testMethod() throws IllegalArgumentException {
try {
int i = 0;
throw new IllegalArgumentException("IA");
} catch (Exception e) {
System.out.println("Test message 2");
e.printStackTrace();
public static void main(String[] args) {
try {
new TestClass().testMethod();
} catch (IllegalArgumentException iae) {
System.out.println("Exception caught in main method");
When I run the above program the exception being thrown is caught in the testMethod only and the message Test message 2" is being printed
Why is it happening ?
Thanx in advance
Mahesh

Hi,
What ever u said is not real. When a method declares that it throws an exception it will never be caught by the generic Exception method.
This what is happened.
Generally in my programming i throw a BusinnesException when input data violates some business exception. ( I declare that my method throws BusinessException ).
I also catch Exception as generic to avoid abnormal termination due to other exception causes.
What i knw is that if a method declares that it is throws an exception ti will never handle that even though it has a catch block to catch parent exception class
mahesh

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