Java的状态比较不一致

Question

In Java Concurrency In Practice an example* is given to demonstrate the issue of visibility , where two different threads may not see the up-to-date state of any particular mutable object because of a lack of synchronisation.

public class Holder {
    private int n;
    public Holder(int n) { this.n = n; }

    public void assertSanity() {
        if (n != n){
            throw new AssertionError("This statement is false.");
        }
    }

In this particular example, the book states that should a Thread "A" first initialise and publish it via a Thread unsafe manner, like:

public Holder holder;
public void initialize() {
    holder = new Holder(42);
}

And then a Thread "B" calls holder.assertSanity() , that it is entirely possible that the AssertionError will be thrown, due to an inconsistent state.

Now, I understand the basic premise of the argument, where a change made to a mutable variable may never be observed by another thread. But what I am confused here is the fact that it is comparing the same (or so I think) reference n != n .

Doesn't this compare the value of the mutable primitive field private int n ? Regardless of the fact that n may now be of value 42 to Thread A, and of value 0 (default) to Thread B, shouldn't a direct call to check it's value in the same thread be consistent ? ie Calling assertSanity() in Thread A would check if 42 != 42 , and in Thread B, 0 != 0 ?

*Referenced from 3.5 Safe Publication, Listings 3.14 & 3.15 in the book.

Answer 1

The problem is that in the expression n != n the variable n will be loaded twice (assuming no optimization of the bytecode). Between these two loads, another thread could change the value.

Answer 2

During comparing in thread B n != n , B retrieve n two times. Meanwhile the constructor runs in thread A would modify the value of n, from default 0, to 42.