try-catch and final variables

Question

I have a very silly question for you :)

For example, I have following code snippet:

class MyClass {

    public static void main (String[] args) {

        final String status;

        try {
            method1();
            method2();
            method3();
            status = "OK";
        } catch (Exception e) {
            status = "BAD"; // <-- why compiler complains about this line??
        }

    }

    public static void method1() throws Exception {
        // ...
    }

    public static void method2() throws Exception {
        // ...
    }

    public static void method3() throws Exception {
        // ...
    }

}

The question is inside: why compiler complains about this line?

IntelliJ IDEA says, that Variable 'status' might already have been assigned to.

But, as I can see, we don't ever reach line (where we set status = "OK") in case of exceptional situation. So the status variable will be BAD and everything should be ok. And if we don't have any exception, then we get OK. And we will set this variable only ONCE a time.

Any thoughts about this?

Thanks for your help!

compiler is not that smart, and if you are declaring that method could throw an exception compiler wont be struggling to check it, so for the sake of simplicity it does not allow such things to happen — Toumash, Jun 29 '15 at 18:47

score 11 · Accepted Answer · answered Jun 29 '15 at 18:47

11

The Java compiler doesn't see what you and I see -- that either status gets set to "OK" or it gets set to "BAD". It assumes that status can be set and an exception is thrown, in which case it gets assigned twice, and the compiler generates an error.

To workaround this, assign a temporary variable for the try-catch block, and assign the final variable once afterwards.

final String status;
String temp;

try {
    method1();
    method2();
    method3();
    temp = "OK";
} catch (Exception e) {
    temp = "BAD";
}

status = temp;

answered Jun 29 '15 at 18:47

rgettman

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I think it is still valid error, what if this thread gets interrupted and throws `InterruptedException` right after assignment – jmj Jun 29 '15 at 18:59
1

@JigarJoshi It can't throw `InterruptedException` (maybe something else, but not that, and probably not in this code). – Sotirios Delimanolis Jun 29 '15 at 19:02
if this was in separate thread and that got called on `interrupt()` – jmj Jun 29 '15 at 19:07
2

@JigarJoshi Then the interrupt flag would be set on the `Thread`. An `InterruptedException` will only happen if the thread is blocked on IO (or something similar: locks, etc.). – Sotirios Delimanolis Jun 29 '15 at 19:10
@Sotirios you are right - Thanks! – jmj Jun 29 '15 at 19:21

Vince · Answer 2 · 2015-06-29T18:58:29.567

2

What if the code that caused the exception occured after status = "OK"? The reason you get the error seems pretty obvious.

Take this for example:

final String status;

try {
    status = "OK":
    causeException();
}catch(Exception e) {
    status = "BAD";
}

void causeException() throws Exception() {
    throw new Exception();
}

This would result in reassigning the variable, which is why you get an error.

edited Jun 29 '15 at 18:58

answered Jun 29 '15 at 18:47

Vince

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2

The variable is final. Reassigning is forbidden. – swinkler Jun 29 '15 at 18:49
5

@swinkler That's the point I'm getting across. The error occurs because the code shown above is possible. Man, what's up with StackOverflow now-a-days.. – Vince Jun 29 '15 at 18:51

try-catch and final variables

2 Answers2