Semaphore

信号量玩的也是aqs的state。

package com.hust.grid.leesf.semaphore;
import java.util.concurrent.Semaphore;
class MyThread extends Thread {
private Semaphore semaphore;
public MyThread(String name, Semaphore semaphore) {
    super(name);
    this.semaphore = semaphore;
}
public void run() {
    int count = 3;
    System.out.println(Thread.currentThread().getName() + " trying to acquire");
    try {
        semaphore.acquire(count);
        System.out.println(Thread.currentThread().getName() + " acquire successfully");
        Thread.sleep(1000);
    } catch (InterruptedException e) {
        e.printStackTrace();
    } finally {
        semaphore.release(count);
        System.out.println(Thread.currentThread().getName() + " release successfully");
    }
}
}
public class SemaphoreDemo {
public final static int SEM_SIZE = 10;
public static void main(String[] args) {
    Semaphore semaphore = new Semaphore(SEM_SIZE);
    MyThread t1 = new MyThread("t1", semaphore);
    MyThread t2 = new MyThread("t2", semaphore);
    t1.start();
    t2.start();
    int permits = 5;
    System.out.println(Thread.currentThread().getName() + " trying to acquire");
    try {
        semaphore.acquire(permits);
        System.out.println(Thread.currentThread().getName() + " acquire successfully");
        Thread.sleep(1000);
    } catch (InterruptedException e) {
        e.printStackTrace();
    } finally {
        semaphore.release();
        System.out.println(Thread.currentThread().getName() + " release successfully");
    }
}
}

说以说白了没有什么难度。

首先分配10个信号量。

第一个分走了5个。

第二个分走了3个，也还能跑。

第三个不够数了，所以只能照例park了，然后自己躺在待解救的aqs sync队列中。

上面有人爽完了，释放了几个坑，第三个也就解脱了。

最终，所有人都爽完了，aqs状态变回10结束。