try catch finally blocks

[sankar]

I am confused with this concept pls.share knowledge
try{
}catch(subexcep se){System.out.println("subexcep");}
}catch(exce e){System.out.println("excep");}
finally{System.out.println("finally");}
System.out.println("after finally");}
1.what happens if subexcep occurs?
2.if excep occurs & caught does the statement afterfinally gets executed?
3.if some other eception occurs which is not handled what happens?
any good points to remember on this topic
sankar

[Anonymous]

Well Done. So this is just one more amguity that you have given me. There is a learning in your code.
In your code there is a method method1 that's overlaoded. So when you pass a byte variable to it, it is possible for this variable to be trapped by both the versions of the method due to arithmetic promotion.
Hence the ambiguity. To remove the ambiguity as to which varsion of the method you want to use just override it in the subclass.
Thanks again. Now I have 3 ways to create an amguity in Java.

[sankar]

But i didn't got any answer from ur explanation.May be i am stupid.Better explain step by step
thanks
sankar

[Anonymous]

When the code in a try block is executed,
follow these rules:
1) if exception occurs skip the code and jump straight to the specific exception, On its completion jump to the finally block and then to the statements that follow the finally block
This will not work if a System.exit(0) is seen in the catch block thus finally and the statements followed do not execute.
An example
class err
{
static String s=null;
static public void main(String[]p)
{
try
{
s.substring(5);
System.out.println(new err().s);
}
catch(NullPointerException e)
{
System.out.println("in");
}
catch(Exception r)
{
System.out.println("in ex");
r.printStackTrace();
}
finally
{
System.out.println("doing finally");
}
System.out.println("out");
}
}
gives the output
[b]
D:\>java err
in
doing finally
out
D:\>
[\b]

2) If there is no exception the each statement in the catch block is executed followed by everything in the finally block. Then the control is given to the lines below the finally block.
Moral finally is executed no matter whether an exception occurs or not except when a call to exit has been invoked

[sgwbutcher]

Hello sankar,
Here is a reworked version of the code you provided that should compile and run. Try using:
java Test 1
java Test 2
java Test 3
java Test 4
at the command line and see what happens. This should answer your questions and give you a better understanding.
Note that an uncaught checked exception can't be thrown because the compiler would require you to catch it or declare it in the throws clause of the test method...but you HAVE to handle it somewhere. That's why I throw an unchecked exception to show what happens if an uncaught exception occurs.
Best regards,
Steve Butcher
[e-mail]

// Test.java
public class Test {
 public static void main( String[] args ) {
   // check for arguments 
 if ( args == null ) {
 System.out.println( "Needs integer argument: 1, 2, 3, or 4" );
 System.exit( -1 );
 }
   // get an integer out of the first argument
 int i = 0;
 try {
 i = Integer.parseInt( args[ 0 ] );
 }
 catch ( NumberFormatException nfe ) {
 System.out.println( "Bad argument: enter an integer 1, 2, 3, or 4" );
 System.exit( -1 );
 }
 
  // based on the integer, throw a different exception
 try {
   new Test().test( i );
   }
   catch( SubExcep se ) {
   System.out.println( "SubExcep in main()" );
   }
 }
 
 public void test( int i) throws SubExcep {
 try{
  
  if ( i == 1 ) {
  throw new Excep( "" ); // (1)
  } else if ( i == 2 ) {
  throw new SubExcep( "" ); // (2)
  } else if ( i == 3 ) {
  throw new ArithmeticException(); // (3)
  }
  }
  catch (SubExcep se) {
    System.out.println("SubExcep in test()");
    throw se;
  }
  catch (Excep e) {
    System.out.println("Excep");
  }
  finally {
    System.out.println("finally");
  }
  
  System.out.println("after finally");
  }
}
class Excep extends Exception {

public Excep( String s ) {
super( s );
}
}
class SubExcep extends Excep {
public SubExcep( String s ) {
super( s );
}
}

Your output should look something like this (except for java Test 3 where the actual output will depend on your environment):

> java Test 1
Excep
finally
after finally
> java Test 2
SubExcep in test()
finally
SubExcep in main()
> java Test 3
finally
java.lang.ArithmeticException
java.lang.ArithmeticException
at Test.test(Test.java:33)
at Test.main(Test.java:18)
at org.jxe.tools.java.Java.main(Compiled Code)
at org.jxe.kernel.Tool$ToolThread.run(Tool.java:215)
> java Test 4
finally
after finally

[sankar]

Thanks Mr.Steve & Mr.lionel
your explanations & program helped me to clear my doubt.Shall i know from which place are u both???

sankar

[sankar]

Thanks Mr.Steve & Mr.lionel
your explanations & program helped me to clear my doubt.Shall i know from which place are u both???

sankar
if possible write to me
[e-mail]

[Herbert Maosa]

Lionel,
In your explanation you have this statement :
------------------------------------------------------------
2) If there is no exception the each statement in the catch block is executed followed by everything in the finally block. Then the control is given to the lines below the finally block.
----------------------------------------------------------------
I suppose you meant to say each statement in the TRY block and not in the catch block as it reads above.If there is no exception thrown, the catch block will not execute, so control will flow from the try block to the finally block and then the rest of the code following the finally block.
Regards,
Herbert.