try, catch question

[Jay Ashar]

public class test {
    public static void main(String args[]) { 
        int i=1, j=1;
        try {
            i++; 
            j--;
            if(i/j > 1)
                i++;
        }
        catch(ArithmeticException e) {
            System.out.println(0);
        }
        catch(ArrayIndexOutOfBoundsException e) {
            System.out.println(1);
        }
        catch(Exception e) {
            System.out.println(2);
        }
        finally {
            System.out.println(3);
        }
        System.out.println(4);
     }
}

A 0
B 1
C 2
D 3
E 4

what will be the output of this code... this mock exam answer says ADE
[ Jess added UBB [code] tags to preserve whitespace. Check 'em out! ]
[ October 31, 2002: Message edited by: Jessica Sant ]

[Jessica Sant]

Did you try running it??
The output of the program is 034, just like the mock exam answer.
during the first iteration of the code 'i' goes up to 2, 'j' goes down to 0
i/j ( -> 2/0) produces an ArithmeticException: Divide By Zero
which is caught by the catch (ArithmeticException e) and prints '0'
the finally is always executed and prints '3'
then the code exits the try/catch and prints '4'

[Ron Newman]

Did you try compiling and running it? That's the best way to understand the answers to questions like this.