question about exception

[Anonymous]

Hi guys
I am not clear about Exception part. Would you please take a look at the following code:

<font size=+0>
public class Test{
public static String output = "";
public static void foo( int i){
try{
if( i==1){
throw new Exception();
}
output +="1";
}
catch(Exception e){
output +="2";
return;
}
finally{
output +="3";
}
output +="4";
}
public static void main( String args[]){
foo(0);
foo(1);
System.out.println(output);
}
}

The output is 13423. I am not quite sure how the program goes
after it catch the exception. ( I mean why the result is not
like 134234? )
Thank you very much!


[This message has been edited by Marilyn deQueiroz (edited October 24, 2001).]

[Manish Hatwalne]

There's a *return* statement in catch block, when exception is thrown and caught, just before returning, "only" finally block is executed, and not the code after that. Hence 134 for foo(0), and 23 for foo(1); makes 13423.
HTH,
- Manish

[Marilyn de Queiroz]

Because of the "return" in the catch which executes before it gets to the last line (output +="4" ;).
[This message has been edited by Marilyn deQueiroz (edited October 24, 2001).]

[Fei Ng]

before the return on catch() the finally() get to run
and then the return in catch() returns back to the main() method.
but if it is System.exit() instead of return.
They fianlly() couldn't run. The program would just exit.

[Dave Vick]

calanthe
When you call foo the second time with the value of 1 as its argument it throws the exception. In the catch statement you add the '2' to the output and then before the return statement the finally block is executed which adds the '3' to the output. After the finally block is done then the rest of the catch is finished - namely, the return statement. So foo returns before the last line, that adds the '4', is executed.
hope that helped

------------------
Dave
Sun Certified Programmer for the Java� 2 Platform

[Jose Botella]

Hello wei
This example shows perfectly the way catch finally work:
After executing "output +="2";" a return is going to be executed, but stop, you know that finally must be executed under (almost) any circumstances. And if return were executed the method would return without doing it. So before return is executed the finally clause it is. Afer this the execution of finally return in the catch block is resumed so 4 is no added.
Think finally as a little subroutine that is called before any possible outing way of try/catch code, and after the completion of finally control is returned to the next instruction that was going to be executed.

[Jose Botella]

hey wei you can see how quick ranchers are...
5 anwsers in 6 minutes

[HanMing Low]

Hi,
I have some doubt on exception that I need some help.
Referring to the above problem, then in what kind of scenario will the output 4 be omitted if the return statement is not there.
What I mean is that when the return statement is commented out, the output+=4 will execute irregardless of whether the exception is thrown or not.
Then, in what case, will the finally excecute but not output+=4
Thanks for any advice.

Han Ming

[Manish Hatwalne]

Originally posted by HanMing Low:
I have some doubt on exception that I need some help.
Referring to the above problem, then in what kind of scenario will the output 4 be omitted if the return statement is not there.
What I mean is that when the return statement is commented out, the output+=4 will execute irregardless of whether the exception is thrown or not.
Then, in what case, will the finally excecute but not output+=4

If the exception thrown is unknown/unhandled, and not caught by the catch block, then finally block will execute but not the code after that.
HTH,
- Manish

[Nisheeth Kaushal]

hey friends,
I have a very small doubt, see
output is set to " ", then "i" gets the value 0, and then this 0(int) is converted to string when system calls its toString() method. That means output now have a value of "0" and then we add 1 to this
"0" + 1 so it shud be "0134" why it is "134" ?.
Please reply urgently.
Nisheeth.

[Manish Hatwalne]

Originally posted by Nisheeth Kaushal:
hey friends,
I have a very small doubt, see
output is set to " ", then "i" gets the value 0, and then this 0(int) is converted to string when system calls its toString() method. That means output now have a value of "0" and then we add 1 to this
"0" + 1 so it shud be "0134" why it is "134" ?.
Please reply urgently.
Nisheeth.

Have look at the code, carefully.
1) output is set to "", not " " (no space).
2) i is never getting appended to the String output.
Try running the code, that always helps.
HTH,
- Manish

[Nisheeth Kaushal]

I got it.