java - 来自单线程执行器的 RejectedExecutionException

Question

下面是我的方法，其中我有单线程执行器在 run 方法中执行一些任务。

  private void trigger(final Packet packet) {

    // this line is throwing exception
    Executors.newSingleThreadExecutor().execute(new Runnable() {
      @Override
      public void run() {
        // some code here
      }
    });
  }

下面是我得到的异常，我不确定为什么？解决此问题的最佳方法是什么？

error= java.util.concurrent.RejectedExecutionException: Task com.abc.stuffProc$2@e033da0 rejected from java.util.concurrent.ThreadPoolExecutor@76c2da8f[Terminated, pool size = 0, active threads = 0, queued tasks = 0, completed tasks = 0]
    at java.util.concurrent.ThreadPoolExecutor$AbortPolicy.rejectedExecution(ThreadPoolExecutor.java:2048)
    at java.util.concurrent.ThreadPoolExecutor.reject(ThreadPoolExecutor.java:821)
    at java.util.concurrent.ThreadPoolExecutor.execute(ThreadPoolExecutor.java:1372)
    at java.util.concurrent.Executors$DelegatedExecutorService.execute(Executors.java:628)

如果多次调用我的 trigger 方法并且它仍在处理我之前线程的 run 方法，会发生什么情况？它会启动尽可能多的线程还是会等待一个线程完成然后启动另一个线程？

Answer 1

参见此处:What could be the cause of RejectedExecutionException

正如您从错误日志中看到的，您的 ThreadPoolExecutor 已终止。

也许这就是你想要的:

private void trigger(final Packet packet) {

    executor.execute(new Runnable() {
      @Override
      public void run() {
        // some code here
      }
    });
  }

private final ExecutorService executor = Executors.newFixedThreadPool(10);

编辑重现问题:

public static void main(String[] args) {
    final ExecutorTest et = new ExecutorTest();
    for (int i = 0; i < 50000; i++) {
        et.trigger(i);
    }
    System.out.println("Done");
}

private void trigger(int i) {

    try {
        Executors.newSingleThreadExecutor().execute(() -> {
            try {
                Thread.sleep(1000);
            } catch (final InterruptedException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
        });
    } catch (final Exception e) {
        System.out.println("Test " + i + " with " + Thread.activeCount());
        e.printStackTrace();
    }
}