////round-up

[chao-long liao]

An exception can be caught by a catch block with an argument type that is a superclass of the thrown exception??
Ans:True
What's the meaning of the above statement?
I can't get what it says.

[manasa teja]

Hope this code helps!!

 class Exc6 {
public static void main(String args[]) {
int d,a;
try {
    d =0;
a = 42/d;  // throws AirthmeticException
System.out.println ( "a=" +a);
}
catch (Exception e) { } // Exception is superclass of irthmeticException and is caught here as it does not find
//catch(ArithemeticException e) {} block here..
 
}
}
 

[ August 07, 2002: Message edited by: Murthy Ram ]

[Valentin Crettaz]

Murthy,
the intention of your code is good but it won't compile unless you remove the second catch block !!
The compile error is:

Exc6.java:12: exception java.lang.ArithmeticException has already been caught
catch(ArithmeticException e) {} //block here..
^
1 error

[manasa teja]

I commented out that out Val!! But, I dont know, why it came down to a new line. I think, I might clicked enter there... I will edit the post again!!