Java中的 Semaphores

今天药忘吃喽~ 2022-09-12 11:51 269阅读 0赞

# 1.介绍 #

探索 Java 中信号量和互斥体的基础知识。

# 2. Semaphore #

从 java.util.concurrent.Semaphore 开始，可以使用信号量来限制访问特定资源的并发线程数。

在下面的例子中，将实现一个简单的登录队列来限制系统中的用户数量：

public class LoginQueueUsingSemaphore { 
        private Semaphore semaphore;
    
        public LoginQueueUsingSemaphore(int slotLimit) { 
            semaphore = new Semaphore(slotLimit);
        }
    
      public  boolean tryLogin() { 
            return semaphore.tryAcquire();
        }
    
       public void logout() { 
            semaphore.release();
        }
    
       public int availableSlots() { 
            return semaphore.availablePermits();
        }
    }

请注意如何使用以下方法：

*  tryAcquire () – 如果许可立即可用则返回真，否则返回假，但 acquire() 获取许可并阻塞直到一个可用
 *  release () – 释放许可证
 *  availablePermits () – 返回当前可用许可证的数量

为了测试登录队列，首先尝试达到限制并检查下一次登录尝试是否会被阻止：

@Test
        public void test1() throws InterruptedException { 
            int slots = 10;
            ExecutorService executorService = Executors.newFixedThreadPool(slots);
            LoginQueueUsingSemaphore loginQueue = new LoginQueueUsingSemaphore(slots);
            IntStream.range(0, slots)
                    .forEach(user -> executorService.execute(loginQueue::tryLogin));
            executorService.shutdown();
            executorService.awaitTermination(10, TimeUnit.SECONDS);
            assertEquals(0, loginQueue.availableSlots());
            assertFalse(loginQueue.tryLogin());
        }

查看注销后是否有可用的许可

@Test
        public void test2() throws InterruptedException { 
            final int slots = 10;
            final ExecutorService executorService = Executors.newFixedThreadPool(slots);
            final LoginQueueUsingSemaphore loginQueue = new LoginQueueUsingSemaphore(slots);
            IntStream.range(0, slots)
                    .forEach(user -> executorService.execute(loginQueue::tryLogin));
            executorService.shutdown();
            executorService.awaitTermination(10, TimeUnit.SECONDS);
    
            assertEquals(0, loginQueue.availableSlots());
            loginQueue.logout();
            assertTrue(loginQueue.availableSlots() > 0);
            assertTrue(loginQueue.tryLogin());
        }

# 3.定时信号量 #

接下来，讨论 Apache Commons TimedSemaphore. TimedSemaphore 允许多个许可作为一个简单的信号量，但在给定的时间段内，在此时间段之后时间重置并释放所有许可。

public class DelayQueueUsingTimedSemaphore { 
    
        private TimedSemaphore semaphore;
    
        public DelayQueueUsingTimedSemaphore(long period, int slotLimit) { 
            semaphore = new TimedSemaphore(period, TimeUnit.SECONDS, slotLimit);
        }
    
        public boolean tryAdd() { 
            return semaphore.tryAcquire();
        }
    
        public int availableSlots() { 
            return semaphore.getAvailablePermits();
        }
    }

当使用以2秒为时间段的延迟队列并且在2秒内使用所有插槽后，应该没有可用的：

@Test
        public void test1() throws InterruptedException { 
            final int slots = 50;
            final ExecutorService executorService = Executors.newFixedThreadPool(slots);
            //2s 之后信号量许可自动重置
            final DelayQueueUsingTimedSemaphore delayQueue = new DelayQueueUsingTimedSemaphore(2, slots);
            IntStream.range(0, slots)
                    .forEach(user -> executorService.execute(delayQueue::tryAdd));
            executorService.shutdown();
            //睡眠1.5s
            Thread.sleep(1500);
            //获取不到可用的槽
            assertEquals(0, delayQueue.availableSlots());
            assertFalse(delayQueue.tryAdd());
        }

@Test
        public void test2() throws InterruptedException { 
            final int slots = 50;
            final ExecutorService executorService = Executors.newFixedThreadPool(slots);
            //2s 之后信号量许可自动重置
            final DelayQueueUsingTimedSemaphore delayQueue = new DelayQueueUsingTimedSemaphore(2, slots);
            IntStream.range(0, slots)
                    .forEach(user -> executorService.execute(delayQueue::tryAdd));
            executorService.shutdown();
            //睡眠2.1s
            Thread.sleep(2100);
            //可用的槽
            assertTrue(delayQueue.availableSlots() > 0);
            assertTrue(delayQueue.tryAdd());
        }

# 4. Semaphore vs. Mutex #

Mutex 的作用类似于二进制信号量，我们可以用它来实现互斥。

在下面的例子中，将使用一个简单的二进制信号量来构建一个计数器：

public class CounterUsingMutex { 
        private Semaphore mutex;
        private int count;
    
        public CounterUsingMutex() { 
            mutex = new Semaphore(1);
            count = 0;
        }
    
        public void increase() throws InterruptedException { 
            mutex.acquire();
            this.count = this.count + 1;
            Thread.sleep(1000);
            mutex.release();
    
        }
    
        public  int getCount() { 
            return this.count;
        }
    
        public boolean hasQueuedThreads() { 
            return mutex.hasQueuedThreads();
        }
    }

当许多线程尝试同时访问计数器时，它们只会被阻塞在队列中：

@Test
        public void test1() { 
            int count = 10;
            ExecutorService executorService
                    = Executors.newFixedThreadPool(count);
            CounterUsingMutex counter = new CounterUsingMutex();
            IntStream.range(0, count)
                    .forEach(user -> executorService.execute(() -> { 
                        try { 
                            counter.increase();
                        } catch (InterruptedException e) { 
                            e.printStackTrace();
                        }
                    }));
            executorService.shutdown();
            System.out.println(counter.getCount());
            System.out.println(counter.hasQueuedThreads());//true
        }

等待所有线程都访问计数器，队列中没有剩余线程：

@Test
        public void test2() throws InterruptedException { 
            final int count = 5;
            final ExecutorService executorService = Executors.newFixedThreadPool(count);
            final CounterUsingMutex counter = new CounterUsingMutex();
            CountDownLatch countDownLatch = new CountDownLatch(count);
    
            IntStream.range(0, count)
                    .forEach(user -> executorService.execute(() -> { 
                        try { 
                            counter.increase();
                            countDownLatch.countDown();
                        } catch (final InterruptedException e) { 
                            e.printStackTrace();
                        }
                    }));
            executorService.shutdown();
            countDownLatch.await();;
            System.out.println(counter.hasQueuedThreads());//false
            System.out.println(counter.getCount());
        }