java - 急切地条件#signal 是最佳实践吗？

Question

Condition JavaDoc具有以下代码示例:

class BoundedBuffer {
   final Lock lock = new ReentrantLock();
   final Condition notFull  = lock.newCondition(); 
   final Condition notEmpty = lock.newCondition(); 

   final Object[] items = new Object[100];
   int putptr, takeptr, count;

   public void put(Object x) throws InterruptedException {
     lock.lock();
     try {
       while (count == items.length)
         notFull.await();
       items[putptr] = x;
       if (++putptr == items.length) putptr = 0;
       ++count;
       notEmpty.signal();
     } finally {
       lock.unlock();
     }
   }

   public Object take() throws InterruptedException {
     lock.lock();
     try {
       while (count == 0)
         notEmpty.await();
       Object x = items[takeptr];
       if (++takeptr == items.length) takeptr = 0;
       --count;
       notFull.signal();
       return x;
     } finally {
       lock.unlock();
     }
   }
 }

如果我实现了 BoundedBuffer#take，我只会在 count == (items.length - 2) 时调用 notFull.signal() ) 除非必要，否则避免向其他线程发送信号。我还注意到 ArrayBlockingQueue#removeAt 急切地调用 notFull.signal();。

问题:我的支票会引入错误吗？ Java 并发编程的最佳实践是在条件为真时急切发出信号吗？我认为它会降低死锁的风险。它对性能有影响吗？

Answer 1

是的。您的检查引入了错误。用病理学的例子很容易证明。

假设 count = items.length:

Thread1: put(o1); // count == items.length. Waiting on notFull.
Thread2: put(o2); // Waiting on notFull.
Thread3: put(o3); // Waiting on notFull.
Thread4: take();  // count -> items.length - 1
                  // There's space in the buffer, but we never signalled notFull.
                  // Thread1, Thread2, Thread3 will still be waiting to put.
         take();  // count -> items.length - 2, signal notFull!
                  // But what happens if Thread4 manages to keep the lock?
         take();  // count -> items.length - 3
         ...
         take();  // count -> 0
Thread1: // Wakes up from Thread4's signal.
         put(o1); // count -> 1
Thread2: put(o2); // Never signalled, still waiting on notFull.
Thread3: put(o3); // Never signalled, still waiting on notFull.

这可以通过使用 signalAll 来缓解，但您仍然会遇到几个问题:

1. 由于即使只有一个空间被打开，您也会唤醒每个线程，因此会有更多的锁争用。
2. 当您的缓冲区中正好有一个空位时，您仍然会有线程等待put。根据实现/文档，我想这可能没问题，但在大多数情况下它仍然会令人惊讶。