同步阻塞当我尝试CountDown [英] Synchronize Is Blocking When I Try To CountDown

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问题描述

我关注了我在此信息中发现的建议使用CountDownLatch和我遇到一个问题。

  private CountDownLatch lock = new CountDownLatch(1); 

@Test
public void testBlock(){
Runnable r = new Runnable(){
@Override
public void run(){
try
{
synchronized(this){
this.wait(50);
}
}
catch(InterruptedException e)
{
e.printStackTrace();
throw(new RuntimeException(e));
}
releaseLock();
}
};

线程t =新线程(r);
t.setDaemon(true);
t.start();

waitOnCallback();
}

private void releaseLock(){
synchronized(lock){// Thread t blocks here
lock.countDown();
}
}

private void waitOnCallback(){
synchronized(lock){
try
{
lock.await ();
}
catch(InterruptedException e)
{
throw new RuntimeException(e);
}
}
}

?CountDownLatch不是一个需要同步的对象(例如,删除 synchronized(),锁)块)。所有线程安全性都在对象本身内部处理。


I followed the advice I found in this post using CountDownLatch and i'm running into a problem. I wrote up this test and ran it and my thread I created blocks when i try to synchronize on lock.

  private CountDownLatch lock = new CountDownLatch(1);

  @Test
  public void testBlock() {
    Runnable r = new Runnable() {
      @Override
      public void run() {
        try
        {
          synchronized(this) {
            this.wait(50);
          }
        }
        catch (InterruptedException e)
        {
          e.printStackTrace();
          throw (new RuntimeException(e));
        }
        releaseLock();
      }
    };

    Thread t = new Thread(r);
    t.setDaemon(true);
    t.start();

    waitOnCallback();
  }

  private void releaseLock() {
    synchronized(lock) { // Thread t blocks here
      lock.countDown();
    }
  }

  private void waitOnCallback() {
    synchronized(lock) {
      try
      {
        lock.await();
      }
      catch (InterruptedException e)
      {
        throw new RuntimeException(e);
      }
    }
  }

Why isn't this working?

解决方案

A CountDownLatch is not an object on which you need to synchronize (i.e. remove the synchronized(lock) blocks). all thread-safety is handled internally to the object itself.

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