java - BlockingQueue.peekWait()

Question

我有一种情况需要 BlockingQueue 上的方法 peekWait。我会将此方法描述为

retrieves but not remove the head of the queue, waiting if necessary until an element becomes available

我在 BlockingQueue LinkedBlockingQueue 上看不到相应的方法。这是否存在或我可以以某种方式做到这一点？我考虑过执行 poll() + addFirst() 但队列可能会在中间填满，让我陷入困境。

Answer 1

将调用包装到 BlockingQueue#peek()在Callable ，执行它并等待 Future<T>.get(long, TimeUnit) :

final BlockingQueue<String> queue = new ArrayBlockingQueue<String>(10);
ExecutorService executor = Executors.newCachedThreadPool();
ScheduledExecutorService scheduler = Executors.newScheduledThreadPool(3);

scheduler.schedule(new Runnable()
{
  @Override
  public void run()
  {
    String e = UUID.randomUUID().toString();
    System.out.println("adding = " + e);
    queue.add(e);
  }
}, 2, TimeUnit.SECONDS);

Callable<String> task = new Callable<String>()
{
  @Override
  public String call() throws Exception
  {
    String result = null;
    while ((result = queue.peek()) == null)
    {
      Thread.sleep(100L);
    }
    return result;
  }
};

String peeked = null;
try
{
  peeked = executor.submit(task).get(1, TimeUnit.SECONDS);
  System.out.println("this should never be printed");
  queue.poll();
}
catch (TimeoutException e)
{
  System.out.println("null: peeked = " + peeked);
  e.printStackTrace();
}

try
{
  peeked = executor.submit(task).get(2, TimeUnit.SECONDS);
  System.out.println("not null: peeked = " + peeked);
}
catch (TimeoutException e)
{
  e.printStackTrace();
  System.out.println("should not throw an exception");
}