method returns before executing body

[Marlene Miller]

In JLS section 8.1.3.4
http://java.sun.com/docs/books/jls/second_edition/html/classes.doc.html#36930
�This prevents method one and method two from being executed concurrently, and furthermore guarantees that the shared values of i and j are both updated before method one returns.�
Does this imply that without synchronization, method one could return before the expressions i++ and j++ are evaluated?
If so, would you please explain how this can happen?

[John Smith]

Does this imply that without synchronization, method one could return before the expressions i++ and j++ are evaluated?

I assume that you are refering to this code:

class Test {
    static int i = 0, j = 0;
    static void one() { i++; j++; }
    static void two() {
        System.out.println("i=" + i + " j=" + j);
    }
}

Since methods one() and two() are not synchronized, it could happen that method two() will print different values of i and j. It's not that method one() will return before its completion, it's that the thread that executes method one() is put in the wait() state by the thread scheduler right after the operation i++. At that time, another thread may execute method two() thus resulting in inconsistent output where i is less that j.
Eugene.

[Marlene Miller]

synchronization guarantees X
without synchronization, X is not guaranteed
without synchronization, not X is possible
X: values are updated before method returns
not X: values are not updated before method returns
not X: method returns and some value is not updated
without synchronization, when method returns some value is not updated

[Marlene Miller]

I figured out what was confusing me.
Before a method returns, the expressions will be evaluated and the variables in the working memory will be updated.
Without synchronization, when the method returns the variables in the shared memory might not be updated.