Question

我正在为AtomicInteger和AtomicBoolean编写单元测试。它们将被用作参考测试，用于在objective-c中测试这些类的仿真，以用于翻译项目。

我认为AtomicInteger测试很好，主要是通过在大量for循环中执行可预测数量的递增，递减，加法和减法操作，每个循环都在自己的线程中运行（每个操作类型有很多线程）。实际操作使用CountDownLatch同时启动。

当所有线程都完成后，我通过比较原子整数和基于线程数，每个线程的迭代次数和每次迭代的预期增加/减少的预期整数值来断言。这个测试通过了。

但是如何测试AtomicBoolean？基本操作是get和set，因此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。我正在考虑的方向是使用两个应始终具有相反值的AtomicBooleans。像这样：

@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();
    }
}

这适用于一个线程，但失败了多个。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

我会专注于compareAndSet，这是AtomicBoolean和普通boolean之间的真正区别。

例如，使用compareAndSet(false, true)来控制关键区域。循环执行直到它返回false，然后进入临界区域。在关键区域，如果两个或多个线程同时运行，则执行极有可能失败的操作。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值应为t*iterations，其中t是主题数。

线程数应该与硬件可以同时运行的数量相似 - 这在多处理器上比在单个处理器上失败的可能性要大得多。失败的最高风险是在AtomicBoolean变为false之后。所有可用的处理器应该同时尝试获取它的独占访问权限，看它是假的，并自动将其更改为true。

Answer 2

我认为，正如你所指出的那样，测试这个比AtomicInteger更难。可能值的空间要小得多，因此可能出错的可能事物的空间要小得多。因为像这样的测试基本上归结为运气（有很多循环来增加你的机会），所以更难达到那个更小的目标。

我的建议是启动大量可以访问单个AtomicBoolean的线程。让他们每个人做CAS，只有如果他们成功，原子地增加AtomicInteger。当所有线程都完成后，您应该只看到AtomicInteger的一个增量。然后冲洗，起泡，重复。

Answer 3

这是四个原子操作。鉴于你只想让一个布尔值与另一个布尔值相反，只需要一个布尔值并继续切换它。您可以从此值计算另一个。

如何测试AtomicBoolean原子性？

3 个答案: