java - 如果线程B希望看到线程A所做的更改，那么最后更改只能是volatile变量而不是全部变量吗？

Question

我看过this answer，它说明了如何:

Under the new memory model, when thread A writes to a volatile variable V, and thread B reads from V, any variable values that were visible to A at the time that V was written are guaranteed now to be visible to B.

因此，给出示例:

public class Main {
    static int value = -1;
    static volatile boolean read;

    public static void main(String[] args) {
        Thread a = new Thread(() -> {
            value = 1;
            read = true;
        });

        Thread b = new Thread(() -> {
            while (!read);

            System.out.println("Value: " + value);
        });

        a.start();
        b.start();
    }
}

尽管value不可变(仅value)，但是否仍对线程b看到read的更改(从-1到1)？

如果是这样的话，给定一系列更改是为了使另一个线程可见，那么除了最后一个可变变量之外，是否要对其他任何变量进行更改，是否有任何目的？

Answer 1

是的，保证对value的更改对于线程b是可见的。
JLS 17.4.4. Synchronization Order说:

• A write to a volatile variable v (§8.3.1.4) synchronizes-with all subsequent reads of v by any thread (where "subsequent" is defined according to the synchronization order).

JLS 17.4.5. Happens-before Order说:

Two actions can be ordered by a happens-before relationship. If one action happens-before another, then the first is visible to and ordered before the second.

If we have two actions x and y, we write hb(x, y) to indicate that x happens-before y.

• If x and y are actions of the same thread and x comes before y in program order, then hb(x, y).

• There is a happens-before edge from the end of a constructor of an object to the start of a finalizer (§12.6) for that object.

• If an action x synchronizes-with a following action y, then we also have hb(x, y).

• If hb(x, y) and hb(y, z), then hb(x, z).

项目符号1表示value = 1在read = true之前发生。
项目符号3说read = true在!read之前发生。
项目符号1表示!read在"Value: " + value之前发生。
项目符号4表示 value = 1在"Value: " + value 之前发生。