Semaphore

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1. CountDownLatch
2. CyclicBarriar--循环屏障
3. Semaphore--信号量

使用场景：常用于使用有限的资源，限制线程并发的最大数量。

1. 设定信号量的最大个数：Semaphore semaphore=new Semaphore(3);
2. 获取信号量：
   • semaphore.acquire(); //获取信号量
   • semaphore.acquire(3); //获取多个许可
   • semaphore.tryAcquire(3); //尝试获取多个许可
   • semaphore.tryAcquire(5, TimeUnit.SECONDS); //给定时间内获取许可
3. 释放信号量：semaphore.release();

公平性：

默认情况下，信号量是非公平性的（先等待先执行为公平。类似于买东西的时候大家排队付款，先来的先付款是公平的。但是这时候有人插队，那就是非公平的）

image.png

• 信号量构造器。可选择的

image.png

示例代码：

• 一次拿一个许可

class SemaphoreExample1 {
    private static final int total=20;

    public static void main(String[] args) {
        ExecutorService executorService = Executors.newCachedThreadPool();
        Semaphore semaphore=new Semaphore(3);
        for (int i=0;i<total;i++){
            final int threadNum=i;
            executorService.execute(()-> {
                try {
                    semaphore.acquire();//获取许可
                    test(threadNum);
                    semaphore.release();//释放许可
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            });
        }

        executorService.shutdown();
    }

    private static void test(int time){
        System.out.println(time+" --- ");
    }

• 一次拿多个许可

class SemaphoreExample2 {
    private static final int total=20;

    public static void main(String[] args) {
        ExecutorService executorService = Executors.newCachedThreadPool();
        Semaphore semaphore=new Semaphore(3);
        for (int i=0;i<total;i++){
            final int threadNum=i;
            executorService.execute(()-> {
                try {
                    semaphore.acquire(3);//获取多个许可
                    test(threadNum);
                    semaphore.release(3);//释放多个许可
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            });
        }

        executorService.shutdown();
    }

    private static void test(int time){
        System.out.println(time+" --- ");
    }
}

• 尝试拿许可，拿到返回true，拿不到返回false

class SemaphoreExample3 {
    private static final int total=20;

    public static void main(String[] args) {
        ExecutorService executorService = Executors.newCachedThreadPool();
        Semaphore semaphore=new Semaphore(5);
        for (int i=0;i<total;i++){
            final int threadNum=i;
            executorService.execute(()-> {
                try {
                    //尝试同时拿多个许可。拿到返回true，拿不到返回false
                    //第一次拿3个，第2次只拿到了2个，不满足3个，
                    if (semaphore.tryAcquire(3)){
                        test(threadNum);
                        semaphore.release(3);//释放多个许可
                    }
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            });
        }

        executorService.shutdown();
    }

    private static void test(int time) throws InterruptedException {
        System.out.println(time+" --- ");
        Thread.sleep(1000);
    }
}

• 给定时间内拿到许可返回true，拿不到返回false

class SemaphoreExample4 {
    private static final int total=20;

    public static void main(String[] args) {
        ExecutorService executorService = Executors.newCachedThreadPool();
        Semaphore semaphore=new Semaphore(3);
        for (int i=0;i<total;i++){
            final int threadNum=i;
            executorService.execute(()-> {
                try {
                    if (semaphore.tryAcquire(5, TimeUnit.SECONDS)){//尝试拿许可，时间超过5秒，返回false
                        test(threadNum);
                        semaphore.release();//释放多个许可
                    }
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            });
        }

        executorService.shutdown();
    }

    private static void test(int time) throws InterruptedException {
        System.out.println(time+" --- ");
        Thread.sleep(1000);
    }
}