about finally

[dx wu]

confused about finally,for example:

public class ExcepTest
{
   public static void main(String[] args)
   {
      System.out.println(new ExcepTest().a());
   }
   
   public int a()
   {
      int a[] = new int[5];
      int i=0;
 
      try {
         System.out.println(a[5]);
         i++;
      }
      catch(Exception e)
      {
         e.printStackTrace();
         return i;
      }
      finally {
         i++;
      }
      return i;
   }
}

and before return ,the finally block will execute,and the value of i should be 1,and then return i,output will 1.but the result is 0.cloud someone tell me why?
thanks in advance

[ September 07, 2005: Message edited by: dx wu ]
[ September 07, 2005: Message edited by: Michael Ernest ]

[raghu babu]

This code doesn't even compile as a[5] within try block is not a statement.

Pl. provide correct code to clear your confusion.

[dx wu]

and have correct the code

[Michael Ernest]

The code compiled fine for me. I got this runtime result:

java.lang.ArrayIndexOutOfBoundsException
        at ExcepTest.a(ExcepTest.java:14)
        at ExcepTest.main(ExcepTest.java:5)
0

Here's a modification that may help you see what's happening:

public class ExcepTest
{
   public static void main(String[] args)
   {
      System.out.println(new ExcepTest().a());
   }
 
   public int a()
   {
      int a[] = new int[5];
      int i=0;
 
      try {
         System.out.println(a[5]);
         System.out.println("try");
 
         i++;
      }
      catch(Exception e)
      {
         e.printStackTrace();
         System.out.println("catch");
         return i;
      }
      finally {
         System.out.println("finally");
         i++;
      }
      System.out.println("below finally");
      return i;
   }
}

Here's the output I got:

java.lang.ArrayIndexOutOfBoundsException
        at ExcepTest.a(ExcepTest.java:14)
        at ExcepTest.main(ExcepTest.java:5)
catch
finally
0

[vjy chin]

Hi,
I have modified the code a little bit for more clarification.

 public class ExcepTest
{
   public static void main(String[] args)
   {
      System.out.println(new ExcepTest().a());
   }
 
   public int a()
   {
      int a[] = new int[5];
      int i=0;
 
      try {
         System.out.println(a[5]);
         System.out.println("Inside try");
 
         i++;
         System.out.println("I value in try block is: "+i);
      }
      catch(Exception e)
      {
         e.printStackTrace();
         System.out.println("catch");
         System.out.println("I value in catch block is: "+i);
         //return i; //If the value is returned, the value of i is returned, 
                     //whatever the c hanges made for i after this does not
                     //come into picture
      }
      finally {
         System.out.println("finally");
         i++;
         System.out.println("I value in finally block is: "+i);
         return i;//if the value is returned from here, the value is 1
      }
      //System.out.println("below finally");
      //System.out.println("I value in before returning is: "+i);
      //return i;
   } 

I have added my comments, but experts please correct me if I am wrong.

Thanks

[Todd Johnson]

This is not the result I expected either. After playing with the code a bit, I suspect here is what happens...

1)In the try block array element 5 is accessed generating an ArrayIndexOutOfBoundsException.

2)The exception is caught and the "return i" statement is executed in the catch block. The variable "i" at this point has a value of 0 which the JVM stores off as the return value for the function.

3) The finally block is executed. The variable "i" is incremented, but this has no effect on the value returned by the function, because the return statement has already executed in step #2.

4) The method exits with the return value that was saved off by the JVM in step #2.
[ September 08, 2005: Message edited by: Todd Johnson ]

[dx wu]

thanks everyone very much