java在可重入锁中等待条件

Question

以下代码取自JavaDoc of ^{}：

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();
    }
  } 
}

想象两个线程，消费者和生产者，一个使用take，一个put在BoundedBuffer的单个实例上

让我们假设消费者首先运行take()，他在其中锁定lock，现在在notEmpty.await();上循环

现在生产者如何通过锁定lock进入put()方法，而lock已经由消费者持有

我错过了什么？当线程等待某个条件时，lock是否“临时释放”？锁的重入到底意味着什么

Answer 1

我使用单监视器测试了以下代码，以下代码的性能总是更好——在2核机器上测试，条件性能平均低于10-15%

 final Object sync = new Object();
  AtomicInteger add=new AtomicInteger();
  AtomicInteger remove=new AtomicInteger();
   final Object[] items = new Object[1];
   int putptr, takeptr, count;        

 public void add(Object x) throws InterruptedException {
       add.incrementAndGet();

     synchronized (sync) {

       while (count == items.length)
         sync.wait();
       items[putptr] = x;
       if (++putptr == items.length) putptr = 0;
       ++count;
       sync.notify();
        }
   }

   public Object remove() throws InterruptedException {
       remove.incrementAndGet();

     synchronized (sync) {

       while (count == 0)
         sync.wait();
       Object x = items[takeptr];
       if (++takeptr == items.length) takeptr = 0;
        count;
       sync.notify();
       return x;

       }
   }


  public static void main(String[] args) {
    final BoundedBuffer bf=new BoundedBuffer();

    Thread put =new Thread(){
        public void run(){
        try {
            while(true)
            bf.add(new Object());
        } catch (InterruptedException e) {

        }
        }

    };
    put.start();

    Thread take= new Thread(){
        public void run(){
        try {
        while(true)
            bf.remove();
        } catch (InterruptedException e) {

        }
        }

    };
    take.start();

    try {
        Thread.sleep(1000L);
        put.interrupt();
        take.interrupt();
    } catch (InterruptedException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }


    System.out.println("add:"+bf.add);
    System.out.println("remove:"+bf.remove);

Answer 2

Lock和synchronized都允许一个线程在等待时放弃锁，而另一个线程可以获得锁。要停止等待，线程必须重新获取锁

注意：它们不会完全释放，如果进行堆栈跟踪，可能会有多个线程同时持有锁，但最多有一个线程正在运行（其余线程将被阻塞）

从Condition.await()

The lock associated with this Condition is atomically released and the current thread becomes disabled for thread scheduling purposes and lies dormant until one of four things happens:

• Some other thread invokes the signal() method for this Condition and the current thread happens to be chosen as the thread to be awakened; or
• Some other thread invokes the signalAll() method for this Condition; or
• Some other thread interrupts the current thread, and interruption of thread suspension is supported; or
• A "spurious wakeup" occurs.

In all cases, before this method can return the current thread must re-acquire the lock associated with this condition. When the thread returns it is guaranteed to hold this lock

Answer 3

就重入性而言，这意味着持有某个锁的线程可以再次获取同一个锁。如果不是这样，一个synchronized方法将无法调用同一对象的另一个synchronized方法

重新进入并不意味着理解你的问题