About Question enthuware.ocajp.i.v7.2.1260 :

[R2-D2-]

Hi Paul,

I tried to play around with the code and there's one thing I don't understand. If I have

Code: Select all

public class TestClass {
    public static void main(String[] args) {
        int j = 1;
        int i;
        try {
            i = doIt() / (j = 2);
            System.out.println(i);
        } catch (Exception e) {
            System.out.println(" j = " + j);

        }
    }

    public static int doIt() throws Exception {
        throw new Exception("FORGET IT");
    }
}

then the code will compile. (Notice that I have moved the declaration of variable i before the try block and added a println in the try).

However, if I add a

Code: Select all

System.out.println(i);

in the catch clause, or right after the catch clause, it will say the variable might not have been initialized. My question is: why does the compiler not complain in the first case?

My guess: in the first case, the compiler knows that i might not be intialized and throw an Exception, but will not complain because the println will not be executed in that case.

Could you please point out if there's a flaw in my reasoning or give a little more info on how this works?

Thanks a lot!

[admin]

Your reasoning is correct. Complier will complain only if it notices a possible execution path in which a variable will be accessed without prior initialization.


hth,
Paul.

[wangchit]

I run the code and get the result j=1.
But the answer is "it will not compile". Why?

[admin]

I see that option 1 i.e. It will print j =1 is indeed set as the correct option.

HTH,
Paul.