Why does the code behave so?

[Gitesh Ramchandani]

For the following code sample:

  
class tt{
 
public void method() {
 
try {
          throw new ArithmeticException();
          System.out.println("INSIDE");
 
   }catch(ArithmeticException re) {
System.out.println("in CATCH");
}
   finally {
System.out.println("in FINALLY");
}

System.out.println("OUTSIDE");
}
 
 
public static void main(String [] args){
 new tt().method();
}
 
}
 

The output is

  
unreachable statement
System.out.println("INSIDE");

But for the following code sample:

    
 
 
class tt{
 
public void method() {
 
try {
int i,j;
i=j=0;
int k=i/j;
System.out.println("INSIDE");
 
   }catch(ArithmeticException re) {
System.out.println("in CATCH");
}
   finally {
System.out.println("in FINALLY");
}

System.out.println("OUTSIDE");
}
 
 
public static void main(String [] args){
 new tt().method();
}
 
}

The output is:

    
in CATCH
in FINALLY
OUTSIDE

Why are the output different in the above two cases, even though in both an ArithmeticException is being thrown?

Thanks,
Gitesh
[ March 18, 2008: Message edited by: Gitesh Ramchandani ]

[Nadeem Khan]

Switch(Gitesh's code)
case 1:
The compiler knows that the code is definitely throwing an exception from the syntax "throw new ArithmeticException();" and hence it comes to know that the next statement within that block of code is UNREACHABLE during the phase of code optimization, & thus it gives you an error!


case 2:
Here compiler cannot deduce the logic just by parsing your syntax, so it doesnt know that this will throw an ArithmeticException! Hence no compiler error!

default:
Hope that helps

[Jelle Klap]

The first case does not compile, because an ArithmaticException is explicitly thrown before the System.out.println() statement, thereby making it unreachable code. The compiler can determine this.

The second case does compile, but throws the ArithmaticException at runtime, because the compiler can not pick-up on the divide-by-zero bug, and is therefor unable to determine that the System.out.prinln() statement is effectively unreachable code.

Edit: Late again...
[ March 18, 2008: Message edited by: Jelle Klap ]

[Gitesh Ramchandani]

Thanks for help, got the point