Doubt in exception handling????

[Amit Das]

hi all,

plz help me by providing suitable reasoning for the behavoir exhibited by this code:

class Amit {
void method() throws InterruptedException {
System.out.println("reachable");
throw new InterruptedException("Amit");
//int i = 0;//unreachable
}
}
public class Question46 extends Amit {
public static void main(String[] args) throws Exception {
try{
new Question46().method();
System.out.println("unreachable but compiler not knows");
} catch(InterruptedException ae) {
System.out.println("in InterruptedException handler"); //1)
throw new Exception("thrown from first catch");
} catch(Exception e){
System.out.println("in Exception handler");
throw new Exception("thrown from second catch");
} finally {
System.out.println("in finally"); //2)
return;
}
//System.out.println("Shouldn't come here");//unreachable
}
}
 
/******************************
Goes from 1) to 2)
o/p:
reachable
in InterruptedException handler
in finally
*******************************/

why does controll goes directly to 2) from 1)???

thanx,
amit
[ May 11, 2005: Message edited by: Mark Spritzler ]

[Mark Spritzler]

1) First, your throw exception in the catch throws it up to the JVM and not to the try catch, so the catch of Exception will not be run.
2) Finally is always run. So after the catch of the InterruptedException, then it runs the finally code.

Mark

[Amol Fuke]

Since its a exception of type InteruptedException it will be caught at line 1.Had the exception be of Exception other than InterruptedException then it will be caught in Exception clause.