Unreachable statement when using return in finally?

Question

This compiles:

class Ex1 {
    public int show() {
        try {
            int a=10/10;
            return 10;
        }   
        catch(ArithmeticException e) {
            System.out.println(e);
        }
        finally {
            System.out.println("Finally");
        }
        System.out.println("hello");
        return 20;
    }
}

on the other hand this doesn't:

class Ex15 {
    public int show() {
        try {
            int a=10/0;
            return 10;
        }
        catch(ArithmeticException e) {
            System.out.println(e);
        }
        finally {
            System.out.println("Finally");
            return 40;
        }

        System.out.println("hello");
        return 20;
    }
}

and gives unreachable statement System.out.println("hello"); error. why is it so?

What error do you get from the compiler? Just a warning that 10/0 is not a good idea? — bmargulies, Mar 23 '10 at 17:40
Probably a warning about unreachable code block because of the return in the finally. — Taylor Leese, Mar 23 '10 at 17:41
Please show us the compiler error, or it will be difficult to help. — FrustratedWithFormsDesigner, Mar 23 '10 at 17:42
@Taylor: http://meta.stackexchange.com/questions/11740/what-are-community-wiki-posts/11741#11741 — Jon Seigel, Mar 23 '10 at 18:15
@Pyrolistical: This new title makes it sound like OP already knows the answer to this question. :D — missingfaktor, Mar 23 '10 at 18:25

Taylor Leese · Accepted Answer · 2010-03-23T17:48:26.313

7

The finally has a return so you are probably getting an unreachable code block error.

finally
{   
    System.out.println("Finally"); 
    return 40;
}
System.out.println("hello"); // unreachable code
return 20;

This is actually a compile-time error in Java. See section 14.20.

It is a compile-time error if a statement cannot be executed because it is unreachable.

edited Mar 23 '10 at 17:48

answered Mar 23 '10 at 17:42

Taylor Leese

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Is that an error by default? I think I was able to change that to a warning in Eclipse. – FrustratedWithFormsDesigner Mar 23 '10 at 17:47
why only when finally's return is an error and not previously – abson Mar 23 '10 at 17:48
It's a compiler error. I don't see how you could configure Eclipse to suppress it. – Taylor Leese Mar 23 '10 at 17:49
3

@Abson - Every execution path will make it to the finally and return. There is no way the code that prints "hello" will execute. Java treats this as a compiler error. The first example works fine because there is no return in the finally block. – Taylor Leese Mar 23 '10 at 17:50
L: Eclipse has an option for "dead code" but I'm not sure it's the same thing (might check later today). – FrustratedWithFormsDesigner Mar 23 '10 at 18:04

score 7 · Answer 2 · edited May 23 '17 at 12:30

It's unreachable code. According to the compiler, System.out.println("hello"); can never be executed.

Beside that, DON'T EVER write return within a finally block. (see Hidden Features of Java for why you should not).

EDIT:

Yes, but what makes return in finally do this?

It's not because it is in a finally block or something. Even if you'd remove the finally keyword, you will still get the error.

   class ex15 {
        public int show() {
            int a = 10 / 0;
            return 40;
            System.out.println("hello");
            return 20;
        }
    }

Obviously, if you return 40, there is no way you can execute the next line. finally just means "do always, no matter what". So.

score 1 · Answer 3 · answered Mar 23 '10 at 17:47

When you put a "return" in the "finally" block, anything that comes after it will never be executed. The "return" statement ends the method right there.

You would get the same error if you put a System.out.println() in the first method, after the "return" statement in it.

score 0 · Answer 4 · answered Mar 23 '10 at 18:19

0

You have a return in the finally block. This makes any statements after that unreachable. Also you have a return in the try block and again in the finally block. This doesn't make sense.

answered Mar 23 '10 at 18:19

fastcodejava

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Unreachable statement when using return in finally?

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