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。