"return" statement inside try block

[Jonathan Nguyen]

Hi,
I'm new here. Just read this strange code block, and tried it.
But I can't explain why the result is...

public int test() {
  int r = 0;
  try {
    r = 1;
    return r;
  }
  finally {
    r = 2;
    // return r;
  }
}

As we knew, the flow still come to finally block and assign r = 2 even we have return r; in try block.

But the result that test() return is 1, not 2.
If we comment out "return r;" at finally block, the result will be 2.
Confused !

Anyone can explain it ? :-)
[ May 20, 2005: Message edited by: Jonathan Nguyen ]

[Tom Johnson]

  
public class Try {
        public static int test() {
               int r = 0;
                try {
                        r = 1;
                        System.out.println("try/return r : " + r);
                        return r;
                }
                finally {
                        r = 2;
                        System.out.println("finally r : " + r);
                        // return r;
                }
        }
 
        public static void main(String [] args) {
                Try t = new Try();
                System.out.println("return value : " + t.test());
        }
}

This prints
try/return r : 1
finally r : 2
return value : 1

It appears that the return statement in the try block takes the value of r as it is in the try block and assigning r=2 in the finally has no effect as return line has already read r, If the return line is uncommented in the finally block the output is

try/return r : 1
finally r : 2
return value : 2

The r=2 line in finally block updates r and the return r; in the finally block is the return statement which occurs and the value is then 2.
Didnt know that would happen till i ran it there tho

//Tom

[Animesh Shrivastava]

Hello Jonathan,
As u know that a finally block always gets to execute either after a try block or a catch block. So only after the full completion of a try block or a catch block(If exception gets thrwon), the finally block executes.

So in ur program what i think is that return r is executed before the finally starts its execution. Return value is already set to be returned. So even though the return value gets changed in the finally block, the return is executed with the value set before in the try block.

Hope i am making some sense

Regards,
Animesh

[fred rosenberger]

when the try block runs, you set the return value. the 'return r' line doesn't set r as what's returned, but the VALUE of r to be returned.

then you enter the finally block, and assign r to 2. we leave the try block, and now return that value we stored before, which is still 1.