多线程之Semaphore

多线程之Semaphore

Semaphore (信号量)

Counting semaphores are used to control the number of activities that can access a certain resource or perform a given action at the same time

A degenerate case of a counting semaphore is a binary semaphore, a Semaphore with an initial count of one. A binary semaphore can be used as a mutex with nonreentrant locking semantics; whoever holds the sole permit holds the mutex.

Semaphore管理着一组许可permit，许可的初始数量通过构造函数设定。

• 调用acquire()方法时，如果没有许可可用了，就将线程阻塞，等待有许可被归还了再执行。
• 当执行完业务功能后，需要通过release()方法将许可证归还，以便其他线程能够获得许可证继续执行。

如果初始化了一个许可为1的Semaphore，那么就相当于一个不可重入的互斥锁（Mutex）。

例子

车场仅有3个停车位，停车位就是有限的共享资源，许可数为3

public class SemaphoreTest {
    public static void main(String[] args) {
        Parking parking = new Parking(3);// 只能停3辆车的停车场

        for (int i = 0; i < 8; i++) {
            // 每一个线程表示一辆车到停车场停车
            new Thread() {
                public void run() {
                    parking.park();// 进入停车场
                };
            }.start();
        }
    }

    static class Parking {
        private Semaphore semaphore;// 信号量

        Parking(int count) {
            semaphore = new Semaphore(count);
        }

        public void park() {
            try {
                semaphore.acquire();// 获取许可
                long time = (long) (Math.random() * 10);
                System.out.println(Thread.currentThread().getName() + "进入停车场，停车" + time + "秒...");
                Thread.sleep(time);// 获取许可
                System.out.println(Thread.currentThread().getName() + "开出停车场...");
            } catch (InterruptedException e) {
                e.printStackTrace();
            } finally {
                semaphore.release();
            }
        }
    }
}

---

https://mp.weixin.qq.com/s?__biz=MzAxMjEwMzQ5MA==&mid=2448890836&idx=2&sn=6828a20fcb1c81684fc00fe2e3a100af&scene=21#wechat_redirect