如何测试AtomicBoolean原子性？

Question

我正在为AtomicInteger和AtomicBoolean编写单元测试。它们将被用作在objective中测试这些类的仿真的参考测试，用于翻译的项目。

我认为AtomicInteger测试的效果很好，主要是通过在大量for循环中执行可预测的增量、减少、加和减操作，每个循环在自己的线程中运行(每个操作类型有多个线程)。实际操作使用CountDownLatch同时启动。

当完成所有线程时，我根据线程数、每个线程的迭代次数和每次迭代的预期增减值，通过比较原子整数和预期整数值来断言。这个测试通过了。

，但是如何测试AtomicBoolean?基本操作是get和set，所以在许多线程中多次调用并期望最终结果为true或false似乎没有意义。我想的方向是使用两个AtomicBooleans，它们应该总是有相反的值。如下所示：

@Test
public void testAtomicity() throws Exception {

    // ====  SETUP  ====
    final AtomicBoolean booleanA = new AtomicBoolean(true);
    final AtomicBoolean booleanB = new AtomicBoolean(false);

    final int threadCount = 50;

    final int iterationsPerThread = 5000;

    final CountDownLatch startSignalLatch = new CountDownLatch(1);
    final CountDownLatch threadsFinishedLatch = new CountDownLatch(threadCount);

    final AtomicBoolean assertFailed = new AtomicBoolean(false);

    // ====  EXECUTE: start all threads ====
    for (int i = 0; i < threadCount; i++) {

        // ====  Create the thread  =====
        AtomicOperationsThread thread;
        thread = new AtomicOperationsThread("Thread #" + i, booleanA, booleanB, startSignalLatch, threadsFinishedLatch, iterationsPerThread, assertFailed);
        System.out.println("Creating Thread #" + i);

        // ====  Start the thread (each thread will wait until the startSignalLatch is triggered)  =====
        thread.start();
    }

    startSignalLatch.countDown();

    // ====  VERIFY: that the AtomicInteger has the expected value after all threads have finished  ====
    final boolean allThreadsFinished;
    allThreadsFinished = threadsFinishedLatch.await(60, TimeUnit.SECONDS);

    assertTrue("Not all threads have finished before reaching the timeout", allThreadsFinished);
    assertFalse(assertFailed.get());

}

private static class AtomicOperationsThread extends Thread {

    // #####  Instance variables  #####

    private final CountDownLatch startSignalLatch;
    private final CountDownLatch threadsFinishedLatch;

    private final int iterations;

    private final AtomicBoolean booleanA, booleanB;

    private final AtomicBoolean assertFailed;

    // #####  Constructor  #####

    private AtomicOperationsThread(final String name, final AtomicBoolean booleanA, final AtomicBoolean booleanB, final CountDownLatch startSignalLatch, final CountDownLatch threadsFinishedLatch, final int iterations, final AtomicBoolean assertFailed) {

        super(name);
        this.booleanA = booleanA;
        this.booleanB = booleanB;
        this.startSignalLatch = startSignalLatch;
        this.threadsFinishedLatch = threadsFinishedLatch;
        this.iterations = iterations;
        this.assertFailed = assertFailed;
    }

    // #####  Thread implementation  #####

    @Override
    public void run() {

        super.run();

        // ====  Wait for the signal to start (so all threads are executed simultaneously)  =====
        try {
            System.out.println(this.getName() + " has started. Awaiting startSignal.");
            startSignalLatch.await();  /* Awaiting start signal */
        } catch (InterruptedException e) {
            throw new RuntimeException("The startSignalLatch got interrupted.", e);
        }

        // ====  Perform the atomic operations  =====
        for (int i = 0; i < iterations; i++) {

            final boolean booleanAChanged;
            booleanAChanged = booleanA.compareAndSet(!booleanB.get(), booleanB.getAndSet(booleanA.get()));  /* Set A to the current value of B if A is currently the opposite of B, then set B to the current value of A */

            if (!booleanAChanged){
                assertFailed.set(true);
                System.out.println("Assert failed in thread: " + this.getName());
            }
        }

        // ====  Mark this thread as finished  =====
        threadsFinishedLatch.countDown();
    }
}

此操作只适用于一个线程，但对于多个线程则失败。我想这是因为booleanAChanged = booleanA.compareAndSet(!booleanB.get(), booleanB.getAndSet(booleanA.get()));不是一个原子操作。

有什么建议吗？

Answer 1

我会集中讨论compareAndSet，这是AtomicBoolean和普通boolean的真正区别。

例如，使用compareAndSet(false, true)控制关键区域。循环执行，直到返回false，然后输入关键区域。在关键区域，如果同时运行两个或多个线程，则执行非常可能失败的操作。例如，在读取旧值和写入新值之间增加一个具有短睡眠时间的计数器。在关键区域的末尾，将AtomicBoolean设置为false。

在启动线程之前，将AtomicBoolean初始化为false，将globalCounter初始化为零。

for(int i=0; i<iterations; i++) {
  while (!AtomicBooleanTest.atomic.compareAndSet(false, true));
  int oldValue = AtomicBooleanTest.globalCounter;
  Thread.sleep(1);
  AtomicBooleanTest.globalCounter = oldValue + 1;
  AtomicBooleanTest.atomic.set(false);
}

最后，globalCounter值应该是t*iterations，其中t是线程数。

线程的数量应该与硬件能够同时运行的数目相似--这在多处理器上比在单个处理器上更有可能失败。失败的最高风险是在AtomicBoolean变为假之后。所有可用的处理器都应该同时尝试获得对它的独占访问，将其视为false，并将其原子化地更改为true。

Answer 2

正如您所指出的，我认为比AtomicInteger更难测试这一点。可能的价值空间要小得多，因此可能出错的东西的空间要小得多。因为像这样的测试基本上取决于运气(通过大量的循环来增加你的机会)，实现这个较小的目标将更加困难。

我的建议是启动许多可以访问单个AtomicBoolean的线程。让他们每个人做一个CAS，并且只有当他们成功的时候，原子地增加一个AtomicInteger。当所有线程都完成时，您应该只看到该AtomicInteger的一个增量。然后冲洗，泡沫，重复。

Answer 3

这是四个原子操作。假设你只想要一个布尔值与另一个布尔值相反，那么就有一个布尔值，然后继续切换。您可以从这个值计算另一个值。