一种高并发流控程序的简单轻量实现

实现一个流控程序，控制客户端每秒调用某个远程服务不超过N次，客户端是会多线程并发调用。

import java.util.Date;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Semaphore;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.atomic.AtomicInteger;

public class FlowConcurrentController {
	// 每秒并发访问控制数量
	final static int MAX_QPS = 10;
	// 并发控制信号量
	final static Semaphore semaphore = new Semaphore(MAX_QPS);
	// 监控每秒并发访问次数（理论上accessCount.get() <= 10）
	final static AtomicInteger accessCount = new AtomicInteger(0);

	// 模拟远程访问
	private static void remoteCall(int i, int j) {
		System.out.println(String.format("%s - %s: %d %d", new Date(), Thread.currentThread(), i, j));
	}

	private static void releaseWork() { // 每秒release一次
		// release semaphore thread
		Executors.newScheduledThreadPool(1).scheduleAtFixedRate(new Runnable() {
			@Override
			public void run() {
				semaphore.release(accessCount.get());
				accessCount.set(0);
			}
		}, 1000, 1000, TimeUnit.MILLISECONDS);
	}

	// 模拟并发访问控制
	private static void simulateAccess(final int m, final int n)
			throws Exception { // m : 线程数；n : 调用数
		ExecutorService pool = Executors.newFixedThreadPool(100);
		for (int i = m; i > 0; i--) {
			final int x = i;
			pool.submit(new Runnable() {
				@Override
				public void run() {
					for (int j = n; j > 0; j--) {
						try {
							Thread.sleep(5);
						} catch (InterruptedException e) {
							e.printStackTrace();
						}

						semaphore.acquireUninterruptibly(1);
						accessCount.incrementAndGet();
						remoteCall(x, j);
					}
				}
			});
		}

		pool.shutdown();
		pool.awaitTermination(1, TimeUnit.HOURS);
	}

	public static void main(String[] args) throws Exception {
		// 开启releaseWork
		releaseWork();
		
		// 开始模拟lots of concurrent calls: 100 * 1000
		simulateAccess(100, 1000);
	}

}

上面的代码中存在一个小问题，就是accessCount的释放后，存在负数的情况，也就是说高并发的情况下每秒会存在>MAX_QPS次的并发访问次数，还不能做到非常精确控制。

期待大家更加简单和轻量的方式。