为什么易失性发生 - 在指令重新排序之前失败？

Question

我hava遵循代码来测试volatile 。 bEnd和nCount定义为volatile。

nCount = 0, bEnd = false

Writer线程将设置

nCount = 100, bEnd = true

Reader线程读取这些变量并打印它们。 基于Java Happens-before命令，在我看来， 当bEnd = true时 ， volatile确保nCount = 100 。 但有时程序打印这个：

main thread done.
thread Reader running ...
thread Writer running ...
SharedData nCount = 0, bEnd = false
thread Writer bEnd = true
thread Reader nCount = 0, bEnd = true
thread Reader nCount = 100, bEnd = true
thread Reader nCount = 100, bEnd = true
thread Reader done.

Reader如何得到“nCount = 0，bEnd = true” ???

以下代码在windows10，jdk1.8.0_131上运行

public class HappensBeforeWithVolatile {

    public static void main(String[] args) {

        Thread threadWriter = new Thread(new Writer());
        Thread threadReader = new Thread(new Reader());
        threadWriter.start();
        threadReader.start();

        System.out.println("main thread done.");
    }
}

class Writer implements Runnable {

    @Override
    public void run() {
        System.out.println("thread Writer running ...");
        SharedData.nCount = 100;
//        System.out.println("thread Writer nCount = 100");
        SharedData.bEnd = true;
        System.out.println("thread Writer bEnd = true");
    }
}

class Reader implements Runnable {

    @Override
    public void run() {
        System.out.println("thread Reader running ...");
        System.out.println("thread Reader nCount = " + SharedData.nCount + ", bEnd = " + SharedData.bEnd);
        System.out.println("thread Reader nCount = " + SharedData.nCount + ", bEnd = " + SharedData.bEnd);
        if (SharedData.nCount == 0 && SharedData.bEnd) {
            System.out.println("thread Reader CODE REORDER !!!");
        }
        System.out.println("thread Reader nCount = " + SharedData.nCount + ", bEnd = " + SharedData.bEnd);
        System.out.println("thread Reader done.");
    }
}

class SharedData {
    volatile public static boolean bEnd = false;
    volatile public static int nCount = 0;

    static {
        System.out.println("SharedData nCount = " + nCount + ", bEnd = " + bEnd);
    }
}

Answer 1

当bEnd = true时，volatile确保nCount = 100

技术上，是的。 但是读者并没有原子地阅读它们。 因此它可能会打印nCount = 0 and bEnd = true 。

这是一个例子：

1. 读者读取nCount 0
2. Wirter写nCount = 100
3. Wirter写入bEnd bEnd = true
4. Writer打印thread Writer bEnd = true
5. 读者读取bEnd true

Answer 2

你的整个例子有点瑕疵。 要测试之前发生的情况，您需要测试后续操作，即使bEnd volatile和nCount不是volatile，然后将您的示例简化为：

static class Reader implements Runnable {
    @Override
    public void run() {

        while (true) {
            if (SharedData.bEnd) {
                System.out.println(SharedData.nCount);
                break;
            } else {
                System.out.println("Not yet seen as true");
                LockSupport.parkNanos(TimeUnit.MILLISECONDS.toNanos(100));
            }
        }
    }
}

static class Writer implements Runnable {
    @Override
    public void run() {
        LockSupport.parkNanos(TimeUnit.MILLISECONDS.toNanos(1000));
        SharedData.nCount = 100;
        SharedData.bEnd = true;

    }
}

这将输出100 ，总是（至少在这种情况下）。 正确的解释是，如果Reader Thread看到volatile变量的Write线程的更新，它将看到之前完成的所有事情，因此100 。