throw/catch exception

[Moya Green]

http://www.jiris.com/
mockexam 2 Question 27:
---------------------------------
import java.io.*;
public class Test027
{
public static void main(String args[]) {
System.out.println(method());
}
static int method() {
try {
throw new IOException();
}
catch (IOException ioe) {
System.out.println("0");
return 0;
}
finally {
System.out.println("1");
return 1;
}
return 2;
}
}
------------------------------------
During compiling, error occurred with "unreachable statement, return 2". I thought new IOException was thrown in try block, then catch block caught it and returned "0". Apparently it did not work like that. Can any one explain this? Thank you very much!
Moya

[Thomas Paul]

return 2 is unreachable because the return in the finally will always be executed.

[Moya Green]

Can you briefly explain how the code compile step by step? I did not get your points. Thanks again. I appreciate it.
Moya

[Anthony Villanueva]

Order inside method() is:
1. code inside try block
2. code inside catch block
3. code inside finally block
4. code after //2 if there was no return statement inside finally block
Please note both return statements execute (i++ is used to check this), although the method actually returns on // 1

import java.io.*;
public class Test77
{
static int i = 0;
public static void main(String args[]) 
{
System.out.println(method());
}
static int method() 
{
try 
{ 
throw new IOException();
}
catch (IOException ioe) 
{
System.out.println("inside catch");
return ++i;
}
finally 
{
System.out.println("inside finally");
return ++i; // 1
} 
// 2
//System.out.println("end of method()");
//return -1;
}
}

[Billy Talton]

Moya,
I completely disagree with Thomas.. okay sort of. What I think should be noted/expounded is that when you purposedly throw an exception, have a <B>return</B> a value in the catch <I>and</I> then a finally clause, you effectively have a if/else logic for handling the exception (and exiting he method) which guarantees that anything after the finally will not be reached. Try this... I removed the <B>return</B> from the catch and finally. I will now compile.
try
{
throw new IOException();
} catch (IOException ioe)
{
System.out.println("0");
//return 0;
} finally
{
System.out.println("1");
//return 1;
}
return 2;
______
Billy Talton
[email protected]

[Barkat Mardhani]

Hi Billy:
I guess you and Thomas are saying same thing. The mere fact that finally has a return statement suffices the fact that any statements after finally will never be reached....
[ September 26, 2002: Message edited by: Barkat Mardhani ]

[Moya Green]

Clear.
Thank you very much, everyone! You ar ethe best!
Moya

[Thomas Paul]

Originally posted by Billy Talton:
Moya,
I completely disagree with Thomas.. okay sort of. What I think should be noted/expounded is that when you purposedly throw an exception, have a <B>return</B> a value in the catch <I>and</I> then a finally clause, you effectively have a if/else logic for handling the exception (and exiting he method) which guarantees that anything after the finally will not be reached.

You are mistaken. The return in the finally will ALWAYS be executed. The only time it isn't is if the catch has a System.exit().
Try this:

import java.io.*;
public class Test027
{ 
public static void main(String args[]) {
System.out.println(method());
}

static int method() {
try { 
throw new IOException();
}
catch (IOException ioe) {
return 0;
}
finally {
return 1;
} 
}
}

The method will return 1, the value from the finally and not 0, the value from the catch.