Guava 基于键的信号量与 ConcurrentHashMap 中的信号量

Question

我需要一个基于 Semaphore mechanism in my application and stumbled upon Striped.semaphore(int, int) Guava 的密钥。但是，它的行为并不像预期的那样。

使用以下代码，有时会获取 returns null。这两种方法都由不同的线程访问。我希望调用 fetch 的线程等到地图中有 Blubb 可用。

private final Striped<Semaphore> semaphores = Striped.semaphore(64, 0);

private final Map<String, Blubb> blubbs = Collections.synchronizedMap(new HashMap<String, Blubb>());

private Semaphore getSemaphore(final String key) {
    return semaphores.get(key);
}

@Override
public void put(String key, Blubb blubb)  {
    blubb.put(key, blubb);
    final Semaphore semaphore = getSemaphore(toUser);
    semaphore.release();
}

@Override
public blubb fetch(final String key) {
    try {
        final Semaphore semaphore = getSemaphore(key);
        final boolean acquired = semaphore.tryAcquire(30, TimeUnit.SECONDS);
        return blubbs.get(key);
    } catch (final InterruptedException e) {
        e.printStackTrace();
    }

    return null;
}

如果我使用以下代码切换回基本 Java，一切都会按预期进行。

private final Map<String, Semaphore> semaphoresMap = new ConcurrentHashMap<String, Semaphore>();

private Semaphore getSemaphore(final String key) {
    Semaphore semaphore = semaphoresMap.get(key);
    if (semaphore == null) {
        semaphore = new Semaphore(0);
        semaphoresMap.put(key, semaphore);
    }
    return semaphore;
}

我在这里错过了什么？谢谢

Answer 1

Guava 的 Striped 指定多个键可能映射到同一个信号量。来自 Javadoc：

The guarantee provided by this class is that equal keys lead to the same lock (or semaphore), i.e. if (key1.equals(key2)) then striped.get(key1) == striped.get(key2) (assuming Object.hashCode() is correctly implemented for the keys). Note that if key1 is not equal to key2, it is not guaranteed that striped.get(key1) != striped.get(key2); the elements might nevertheless be mapped to the same lock. The lower the number of stripes, the higher the probability of this happening.

您的代码中的基本假设似乎是，如果与特定对象关联的信号量具有许可，则该对象在映射中有一个条目，但事实并非如此——如果在恰好与同一 Semaphore 相关联的另一个对象的地图，那么该许可可能会被 fetch 在完全不同的对象上获取，该对象实际上在地图中没有条目。

Answer 2

'basic java' 示例有一个潜在的竞争条件，computeIfAbsent 是一个原子操作并解决了这个问题：

private final Map<String, Semaphore> semaphoresMap = new ConcurrentHashMap<String, Semaphore>();

private Semaphore getSemaphore(final String key) {
    return semaphoresMap.computeIfAbsent(key, (String absentKey) -> new Semaphore(0));
}

Guava 基于键的信号量与 ConcurrentHashMap 中的信号量

Key based Semaphores with Guava vs Semaphores in a ConcurrentHashMap

java

concurrency

guava

semaphore