Question

我正在开展一个项目，我需要做以下事情：

将特定套接字上的某些数据发送到另一个系统。我必须在给定的套接字上发送一个特定的字节数组。每个字节数组都有一个唯一的长地址。
然后继续使用以下实施的RetryStrategy重试发送相同的数据。
启动后台轮询线程，告知您是否在其他系统上接收了您发送的数据。如果收到它，那么我们将它从pending队列中删除，以便它不会被重试，如果由于某种原因没有收到它，那么我们将重新使用RetryStrategy重新发送相同的数据我们使用。

例如：如果我们发送了byteArrayA，其唯一的长地址为addressA，并且如果它是在其他系统上收到的，那么我的轮询线程会将此addressA恢复为确认意味着已收到，所以现在我们可以从待处理队列中删除此地址，以便它不会再次重试。

我有两个RetryStrategy已实施ConstantBackoff和ExponentialBackoff。所以我想出了下面的模拟器，它模拟了上面的流程。

public class Experimental {
  /** Return the desired backoff delay in millis for the given retry number, which is 1-based. */
  interface RetryStrategy {
    long getDelayMs(int retry);
  }

  public enum ConstantBackoff implements RetryStrategy {
    INSTANCE;
    @Override
    public long getDelayMs(int retry) {
      return 1000L;
    }
  }

  public enum ExponentialBackoff implements RetryStrategy {
    INSTANCE;
    @Override
    public long getDelayMs(int retry) {
      return 100 + (1L << retry);
    }
  }

  /** A container that sends messages with retries. */    
  private static class Sender {
    private final ScheduledExecutorService executorService = Executors.newScheduledThreadPool(20);
    private final ConcurrentMap<Long, Retrier> pending = new ConcurrentHashMap<>();

    /** Send the given (simulated) data with given address on the given socket. */
    void sendTo(long addr, byte[] data, int socket) {
      System.err.println("Sending " + Arrays.toString(data) + "@" + addr + " on " + socket);
    }

    /** The state of a message that's being retried. */
    private class Retrier implements Runnable {
      private final RetryStrategy retryStrategy;
      private final long addr;
      private final byte[] data;
      private final int socket;
      private int retry;
      private Future<?> future;

      Retrier(RetryStrategy retryStrategy, long addr, byte[] data, int socket) {
        this.retryStrategy = retryStrategy;
        this.addr = addr;
        this.data = data;
        this.socket = socket;
        this.retry = 0;
      }

      private synchronized void start() {
        if (future == null) {
          future = executorService.submit(this);
          pending.put(addr, this);
        }
      }

      private synchronized void cancel() {
        if (future != null) {
          future.cancel(true);
          future = null;
        }
      }

      private synchronized void reschedule() {
        if (future != null) {
          future = executorService.schedule(this, retryStrategy.getDelayMs(++retry), MILLISECONDS);
        }
      }

      @Override
      synchronized public void run() {
        sendTo(addr, data, socket);
        reschedule();
      }
    }

   /** 
    * Get a (simulated) verified message address. Just picks a pending 
    * one. Returns zero if none left.
    */      
    long getVerifiedAddr() {
      System.err.println("Pending messages: " + pending.size());
      Iterator<Long> i = pending.keySet().iterator();
      long addr = i.hasNext() ? i.next() : 0;
      return addr;
    }

    /** A polling loop that cancels retries of (simulated) verified messages. */        
    class CancellationPoller implements Runnable {
      @Override
      public void run() {
        while (!Thread.currentThread().isInterrupted()) {
          try {
            Thread.sleep(1000);
          } catch (InterruptedException ex) {
            Thread.currentThread().interrupt();
          }
          long addr = getVerifiedAddr();
          if (addr == 0) {
            continue;
          }
          System.err.println("Verified message (to be cancelled) " + addr);
          Retrier retrier = pending.remove(addr);
          if (retrier != null) {
            retrier.cancel();
          }
        }
      }
    }

    private Sender initialize() {
      executorService.submit(new CancellationPoller());
      return this;
    }

    private void sendWithRetriesTo(RetryStrategy retryStrategy, long addr, byte[] data, int socket) {
      new Retrier(retryStrategy, addr, data, socket).start();
    }
  }

  public static void main(String[] args) {
    Sender sender = new Sender().initialize();
    for (long i = 1; i <= 10; i++) {
      sender.sendWithRetriesTo(ConstantBackoff.INSTANCE, i, null, 42);
    }
    for (long i = -1; i >= -10; i--) {
      sender.sendWithRetriesTo(ExponentialBackoff.INSTANCE, i, null, 37);
    }
  }
}

我想看看上面的代码中是否有任何竞争条件或任何线程安全问题？因为在多线程中获得正确的东西是很困难的。

让我知道是否有更好或更有效的方法来做同样的事情。

Answer 1

如果您的代码中没有任何针对第三方库的内容，则可以查看此https://github.com/nurkiewicz/async-retry。假设我正确地理解了你的问题，这个lib允许你在不重新发明轮子的情况下处理这类问题。它写得非常清楚，并且有很好的文档记录，所以我希望你能找到自己的方向。：）

如果未收到数据，则发送和重试数据？

1 个答案: