Return statement in exception and consequences

Question

Hi,
Can any one give me the explanation for my doubt?
Case 1:
I am having some code throwing some exception lets say arithematic exception in my try block.
I caught it in the catch block and at last I am having one finally?
as per normal order try block next catch block and at last finally will execute.
I execute the above code by putting a return statement inside catch block.I got the same order of execution irrespective of return statement inside catch?
What is the significance of return inside catch?
Case 2:
Lets take the scenario.
class MyException
     public static void main(String[] args)
          int i=1,j=0;
          try
               try
               System.out.println("the value is: "+(i/j));
               System.out.println("Inside try block");
               catch(ArithmeticException e)
               System.out.println("hi in ame");
               //return;
          finally
          System.out.println("inner finally");
                                        System.out.println("outer try");
          catch(Exception e)
                         e.printStackTrace();
               System.out.println("in exception");
          finally
               System.out.println("plz wait");
If return statement is there inside the inner try and catch the code out of inside outer try not getting executed Why So?
Any clarifications?
Thanking you,
Murthy.

Answer 1

First, please format your code as per http://forum.java.sun.com/features.jsp#Formatting
I'm not sure what part you don't understand.
.    public static void main(String[] args) {
.        try {
.            try {
.                throw new ArithmeticException();
.                System.out.println("Inside try block"); // #1
.            catch(ArithmeticException e) {
.                System.out.println("hi in ame"); // #2
.                return;
.            finally {
.                System.out.println("inner finally"); // #3
.            System.out.println("outer try"); // #4
.        catch(Exception e) {
.            System.out.println("in exception"); // #5
.        finally {
.            System.out.println("plz wait"); // #6
.    }#1 -- You won't get here because you throw AME.
#2 -- You will get here because you caught the AME you just threw.
#3 -- You will get here because it's a finally
#4 -- You won't get here. I think that this is what you're asking about: Why don't we get here? Because we've already done 'return' and this line is NOT inside a finally. The only things that can get executed after a return are finally blocks.
#5 -- You won't get here because you already caught the AME and didn't rethrow it or any other exception.
#6 -- You will get here because it's a finally.
Once you do a return, the only code you can execute before exiting the method are finally blocks. First the one corresponding to the try/cathc block where your return is, then the finally for the try/catch that encloses that one, and so on outward. The code between one finally and the next (#4 here) is NOT executed.
If you still have a question, please try to clarify exactly what behavior you don't understand.
Don't return from inside t/c/f.