Returning a value from a finally block

Question

I found this piece of code in a java certification website

   public class Test1{
        public static void main(String args[]){
            System.out.println(method());
        }
        public static int method(){
           try{
              return 1;
           }
           catch(Exception e){
              return 2;
           }
           finally{
              return 3;
           }
        }
    }

So the output of this piece of code is shown as 3. How is this possible..since it is returning 1 in the try block itself? The code will never reach finally right??

Answer 1

The code will never reach finally right? No, If there is a finally block control will go to finally after try or/and catch

Because in any case finally will always execute. Except as Mike Kobit said System.exit

Thanks Jason C, From JLS 14.17. The return Statement

It can be seen, then, that a return statement always completes abruptly.

The preceding descriptions say "attempts to transfer control" rather than just "transfers control" because if there are any try statements (§14.20) within the method or constructor whose try blocks or catch clauses contain the return statement, then any finally clauses of those try statements will be executed, in order, innermost to outermost, before control is transferred to the invoker of the method or constructor. Abrupt completion of a finally clause can disrupt the transfer of control initiated by a return statement.

From JLS 14.20.2. Execution of try-finally and try-catch-finally

If execution of the try block completes normally, then the finally block is executed, and then there is a choice:

1. If the finally block completes normally, then the try statement completes normally.
2. If the finally block completes abruptly for reason S, then the try statement completes abruptly for reason S.

So In you code try returns 1, then control goes to finally which returns 3. so return value of function will be 3.

Answer 2

As Sumit Singh mentioned in his answer, this happens because return causes the method to end abruptly.

You can observe the same behaviour, when you throw new Exceptions inside your try-catch-finally block, since throw is also abrupt.

See the following code:

public class FinallyTest {
    public static void main(String[] args) {
        abrupt();
    }

    @SuppressWarnings("finally")
    public static void abrupt() {
        try {
            throw new IllegalArgumentException("In Try");
        }
        catch(Exception x) {
            throw new IllegalArgumentException("In Catch");
        }
        finally {
            throw new NullPointerException("In Finally");
        }
    }
}

When running the code, the console shows

Exception in thread "main" java.lang.NullPointerException: In Finally
at FinallyTest.abrupt(FinallyTest.java:15)
at FinallyTest.main(FinallyTest.java:3)

Even though you throw a new IllegalArgumentException inside the try block, only the NullPointerException from the finally block is thrown, since this block ended abruptly.

Answer 3

Finally block will always get executed just before return statement get executed. (There are some exceptions like JVM itself crash or system.exit() is called)

Here is the reason:

We normally use finally to free up our resources or to cleanup. Now if developer will write return statement accidentally in try block then logically finally would never get executed. To overcome this problem, JVM by itself handle this scenario and execute finally block just before returning from try block.

Remember the finally block always executes when the try block exits. And above explained scenario is also applied to return, continue, or break.

Answer 4

Because if we think about priority finally has more than try or catch, to make it clear what if the exception haven't caught even then the finally executes and here it is working inside a method so logically when a method is suppose to return a single output int value then obviously it will always return values from finally.

Is it makes sense?