Question

Reading here:

JLS 8.3.1.4 volatile Fields

Without volatile it says

“then method two could occasionally print a value for j that is greater than the value of i, because the example includes no synchronization and”

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

With volatile is says

“It is possible, however, that any given invocation of method two might observe a value for j that is much greater than the value observed for i, because method one might be executed many times between the moment when method two fetches the value of i and the moment when method two fetches the value of j.”

In behaves ‘properly’ with synchornization, but I’m confused as to what benefit volatile brings here?

I thought volatile gaurantees the order is preserved, so I would have thought it SOME cases the value of i might be greater than j, but not the other way around since that implies the order of incrementing was changed.

Is that a typo in the doc? If not, please explain how j could be greater than i when using volatile.

Answer 1

It is saying that in the middle of method two method one could run several and that the value read for j would be higher than the value read for i.

read i
run method 1
run method 1
read j