java - 检查线程池是否有等待执行的任务？

Question

这就是我试图提出解决方案的场景:

You got a parking house with 2000 free slots. It has 4 entrances. When the parking house is full, there is only allowed 50 cars to wait in each entrance.

我最终做的是创建一个包含 2000 个线程的线程池，然后用 4 个线程代表入口。他们以随机的间隔创建汽车任务(它们基本上只是调用Thread.sleep())并将它们提交到线程池。

当线程池忙于执行 2000 个任务时，就会出现问题，而新提交的任务会被放入队列中等待执行。我该如何应对在入口处等待的情况？我似乎找不到一种方法来检查线程池是否“已满”或类似的情况。或者也许我的方法是完全错误的。

Answer 1

Now how do I deal with waiting in the entrances instead? I can't seem to find a way to check if the threadpool is "full" or anything like that.

嗯。使用 Java 并发类有多种方法可以做到这一点。一种方法是使用一个大小限制为 2000 的 BlockingQueue 来表示 parking 场槽位。然后有 4 个单线程执行器服务线程池，每个线程池都有一个 50 的队列，代表 4 个入口。一旦第 51 辆车尝试在入口处排队，就会抛出 RejectedExecutionException。

类似于:

final int NUM_SPACES_IN_LOT = 2000;
final int ENTRANCE_QUEUE_SIZE = 50;
final BlockingQueue<Car> parkingLot = new LinkedBlockingQueue<>(NUM_SPACES_IN_LOT);
...
ExecutorService entrance1 = new ThreadPoolExecutor(1, 1,
    0L, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<Runnable>(ENTRANCE_QUEUE_SIZE));
ExecutorService entrance1 = new ThreadPoolExecutor(1, 1,
    0L, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<Runnable>(ENTRANCE_QUEUE_SIZE));
ExecutorService entrance1 = new ThreadPoolExecutor(1, 1,
    0L, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<Runnable>(ENTRANCE_QUEUE_SIZE));
ExecutorService entrance1 = new ThreadPoolExecutor(1, 1,
    0L, TimeUnit.MILLISECONDS, new ArrayBlockingQueue<Runnable>(ENTRANCE_QUEUE_SIZE));
...
private class ParkCar implements Runnable {
    private final Car car;
    public ParkCar(Car car) {
        this.car = car;
    }
    public run() {
        parkingLot.put(car);
    }
}
...
Car car = new Car();
try {
   entrance1.submit(new ParkCar(car));
} catch (RejectedExecutionException jee) {
   // entrance is full
}
...
// remove a car from the lot, warning this is O(N)
parkingLot.remove(car);

您需要每个入口队列和每个入口线程来进行并发 parking 。如果您不希望入口已满，您可以调用 entranceX.setRejectedExecutionHandler(...) 设置处理程序，并且不希望它抛出异常。

我不喜欢这个实现的地方是，parkingLot.remove(...) 是O(N)，但如果不自己编写的话，就没有阻塞集。