java - notifyAll() 不起作用

Question

在下面的代码中调用了 notifyAll() 但没有重新激活其他线程。我得到的输出是

• beta 等待通知时间:1441870698303，activeWriters:1

• alpha 等待通知时间:1441870698303，activeWriters:1

• 增量通知时间:1441870698403，activeWriters:0

  公共(public)类 Waiter 实现 Runnable{

      private static int activeWriters;
  
      public Waiter(Message msg){
      }
  
      @Override
      public void run() {
          beforeWrite();
          try {
              Thread.sleep(100);
          } catch (InterruptedException e) {
              // TODO Auto-generated catch block
              e.printStackTrace();
          }
          afterWrite();
      }
  
      protected synchronized void beforeWrite(){
          while (activeWriters > 0 ) {
              try {
                  System.out.println(Thread.currentThread().getName() +" waiting to get notified at time: "+System.currentTimeMillis()+ ", activeWriters: " + activeWriters);
                  wait();
                  System.out.println(Thread.currentThread().getName() +" waiting got notified at time: "+System.currentTimeMillis()+ ", activeWriters: " + activeWriters);
              } catch (InterruptedException e) {
                  e.printStackTrace();
              }
          }
          ++activeWriters;
      }
      protected synchronized void afterWrite(){
          --activeWriters;
          System.out.println(Thread.currentThread().getName() +" notify all at time: "+System.currentTimeMillis() + ", activeWriters: " + activeWriters);
          notifyAll();
      }
  
  }
  
  public class WaitNotifyTest {
  
      public static void main(String[] args) {
          Message msg = new Message("process it");
          Waiter waiter1 = new Waiter(msg);
          Waiter waiter2 = new Waiter(msg);
          Waiter waiter3 = new Waiter(msg);
          new Thread(waiter1,"alpha").start();
          new Thread(waiter2, "beta").start();
          new Thread(waiter3, "delta").start();
  
      }
  
  }
  

Answer 1

调用 wait() 和 notify*() 作用于指定的对象，这意味着 notify*() 唤醒线程在同一个对象上调用了 wait()。

在您的情况下，您在 3 个未连接的不同对象上调用了 wait() 和 notifyAll()，因此这是行不通的。

你可以添加一个静态互斥量:

private static final Object mutex = new Object();

然后对该对象调用wait() 和notify*()。请记住首先在互斥量上同步:

synchronized (mutex) {
    ...
    mutex.wait();
    ...
}

和:

synchronized (mutex) {
    ...
    mutex.notifyAll();
    ...
}

所有对 activeWriters 的访问都必须在这些 synchronized block 中进行，原因有两个。目前对它的访问实际上是不同步的，因为您在 3 个不同的对象上同步。除此之外，activeWriters 是您的条件变量，并且您希望 notify*() 其他线程发生更改。为此，变量的更改和 notify*() 调用必须在同一个 synchronized block 中。