Doubts reg try block and return statements

hi
public int test()
          try
               System.out.println("hi");
               return 1;
          catch(Exception e)
          System.out.println("err");
          return 2;
          finally
               System.out.println("final");
               return 3;
          //System.out.println("after");
          //return 4;
when i call this function, it will printing 3... why ??
and also compilation error is coming when i put return statement after final block.. why is it so??

but if commented the return statements in catcch and finally then it iam not getting any compilation error.
public int test()
          try
               System.out.println("hi");
               return 1;
          catch(Exception e)
          System.out.println("err");
          // return 2;
          finally
               System.out.println("final");
          //     return 3;
          System.out.println("after");
          return 4;
and iam just learning ..... regarding return and try block

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Hi friends
Take a look in the code bellow
try
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finally
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statement will be achieved. In this case, what will
hapen ? The code in the finally block will be executed
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class sample{
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veldhanas wrote:
return abc;In your code there is no varible declared as abc. It is a value assigned in str.
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In this case, since there's no way the code in the given "try" block can throw an exception, it would have been better to not even have a try/catch block in the first place.
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Hi
Can anyone tell me which is the better practice - to put the return statement at the end of try block or after all the catch blocks ie at the end of method.
Eg
Method A()
String str;
try{
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OR
Method B()
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try{
str= [some code]
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System.out.println("Exception");
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I always try to work with only one exit point for each method.
For readability I always put the return statement as close to the end of the method as possible.
In this particular question,
I think you should put the return at the end of the method (for readability, since this is what you are
familiar with), but when I have a try-catch clause, I usually have an unrecoverable error and
should throw this further down the tree, so it usually becomes
try{
   return
}catch(...){
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}I think you should NEVER reflect the occurence of an error in the return-value of the
method when an unrecoverable exception occurs. Just rethrow this exception or throw another method.
Other methods look like
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This is possible in two versions:
With one return and a result-object
Object result = new ...
if(..)
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else if(..)
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return result;With every time a return
if(...)
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else if(...)
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but I find the second version (which I hated when I began programming)
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kind regards,

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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
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          try
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               System.out.println("Inside try block");
               catch(ArithmeticException e)
               System.out.println("hi in ame");
               //return;
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          System.out.println("inner finally");
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               System.out.println("in exception");
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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?
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Murthy.

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.

Doubts reg try block and return statements

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