Java并发编程核心方法与框架-Semaphore的使用

Semaphore中文含义是信号、信号系统，这个类的主要作用就是限制线程并发数量。如果不限制线程并发数量，CPU资源很快就会被耗尽，每个线程执行的任务会相当缓慢，因为CPU要把时间片分配给不同的线程对象，而且上下文切换也要耗时，最终造成系统运行效率大幅降低，所以限制并发线程的数量是很有必要的。

类Semaphore的同步性

类Semaphore的构造方法参数permits表示同一时间内，最多允许多少个线程同时执行acquire()和release()之间的代码。

public class Service {
	private Semaphore semaphore = new Semaphore(1);
	public void testMethod() {
		try {
			System.out.println(Thread.currentThread().getName() + "进入testMethod方法 time:" + System.currentTimeMillis());
			semaphore.acquire();
			System.out.println(Thread.currentThread().getName() + " begin time:" + System.currentTimeMillis());
			Thread.sleep(2000);
			System.out.println(Thread.currentThread().getName() + " end time:" + System.currentTimeMillis());
			semaphore.release();	
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}

public class ThreadA extends Thread {
	private Service service;
	public ThreadA(Service service) {
		super();
		this.service = service;
	}
	@Override
	public void run() {
		service.testMethod();
	}
}

public class ThreadB extends Thread {
	private Service service;
	public ThreadB(Service service) {
		super();
		this.service = service;
	}
	@Override
	public void run() {
		service.testMethod();
	}
}

public class ThreadC extends Thread {
	private Service service;
	public ThreadC(Service service) {
		super();
		this.service = service;
	}
	@Override
	public void run() {
		service.testMethod();
	}
}

public class Main {
	public static void main(String[] args) {
		Service service = new Service();
		ThreadA a = new ThreadA(service);
		a.setName("A");
		ThreadB b = new ThreadB(service);
		b.setName("B");
		ThreadC c = new ThreadC(service);
		c.setName("C");
		a.start();
		b.start();
		c.start();
	}
}

运行程序，控制台打印结果如下：

A进入testMethod方法 time:1468497756729
C进入testMethod方法 time:1468497756729
B进入testMethod方法 time:1468497756729
A begin time:1468497756729
A end time:1468497758733
C begin time:1468497758733
C end time:1468497760738
B begin time:1468497760738
B end time:1468497762742

从打印结果来看，A、B、C三个线程同时进入testMethod方法，三个线程排队执行acquire()和release()之间的代码。修改Semaphore的构造方法参数:

public class Service {
	private Semaphore semaphore = new Semaphore(2);
	public void testMethod() {
		try {
			System.out.println(Thread.currentThread().getName() + "进入testMethod方法 time:" + System.currentTimeMillis());
			semaphore.acquire();
			System.out.println(Thread.currentThread().getName() + " begin time:" + System.currentTimeMillis());
			Thread.sleep(2000);
			System.out.println(Thread.currentThread().getName() + " end time:" + System.currentTimeMillis());
			semaphore.release();	
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}

重新运行程序，控制台打印结果如下：

B进入testMethod方法 time:1468497974695
C进入testMethod方法 time:1468497974695
A进入testMethod方法 time:1468497974695
C begin time:1468497974695
B begin time:1468497974695
B end time:1468497976699
C end time:1468497976699
A begin time:1468497976699
A end time:1468497978703

可见，此A、B、C三个线程同时进入testMethod方法，时B、C线程同时开始执行acquire()和release()之间的代码。B、C线程执行完后，A线程开始执行acquire()和release()之间的代码。

方法acquire(int permits)参数作用及动态添加permits许可数量

有参方法acquire(int permits)的功能是每调用1次此方法，就使用permits个许可。

修改以上Service类：

//Semaphore的构造方法参数permits表示同一时间内
//最多允许多少个线程同时执行acquire()和release()之前的代码
public class Service {
	private Semaphore semaphore = new Semaphore(10);//一共有10个许可
	public void testMethod() {
		try {
			System.out.println(Thread.currentThread().getName() + "进入testMethod方法 time:" + System.currentTimeMillis());
			semaphore.acquire(2);//每次执行消耗掉2个许可
			System.out.println(Thread.currentThread().getName() + " begin time:" + System.currentTimeMillis());
			Thread.sleep(2000);
			System.out.println(Thread.currentThread().getName() + " end time:" + System.currentTimeMillis());
			semaphore.release(2);	
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}

public class Main {
	public static void main(String[] args) {
		Service service = new Service();
		ThreadA[] a = new ThreadA[10];//ThreadA类同上面的ThreadA类
		for (int i = 0; i < a.length; i++) {
			a[i] = new ThreadA(service);
			a[i].start();
		}
	}
}

程序运行结果如下：

Thread-0进入testMethod方法 time:1468587142883
Thread-3进入testMethod方法 time:1468587142883
Thread-2进入testMethod方法 time:1468587142883
Thread-1进入testMethod方法 time:1468587142883
Thread-5进入testMethod方法 time:1468587142883
Thread-2 begin time:1468587142883
Thread-3 begin time:1468587142883
Thread-4进入testMethod方法 time:1468587142883
Thread-0 begin time:1468587142883
Thread-8进入testMethod方法 time:1468587142883
Thread-7进入testMethod方法 time:1468587142883
Thread-6进入testMethod方法 time:1468587142883
Thread-5 begin time:1468587142883
Thread-1 begin time:1468587142883
Thread-9进入testMethod方法 time:1468587142884
Thread-0 end time:1468587144888
Thread-3 end time:1468587144888
Thread-2 end time:1468587144888
Thread-5 end time:1468587144888
Thread-1 end time:1468587144888
Thread-6 begin time:1468587144889
Thread-7 begin time:1468587144889
Thread-8 begin time:1468587144889
Thread-4 begin time:1468587144889
Thread-9 begin time:1468587144889
Thread-7 end time:1468587146892
Thread-4 end time:1468587146892
Thread-8 end time:1468587146892
Thread-6 end time:1468587146892
Thread-9 end time:1468587146892

由程序运行结果可见，10个线程同时进入testMethod()方法。由于一共有10个许可，每个线程acquire()时消耗2个许可，所以第一批有5个线程可以同时执行acquire()方法和release()方法之间的代码。第一批的5个线程执行完毕之后，每个线程释放掉2个许可，一共释放掉10个许可。剩下的5个线程一共获取10个许可同时开始执行。

方法acquireUninterruptibly()的使用

方法acquireUninterruptibly()的作用是使等待进入acquire()方法的线程不允许被中断。

先看一段能中断的代码

public class Service {
	private Semaphore semaphore = new Semaphore(1);
	public void testMethod() {
		try {
			semaphore.acquire();
			System.out.println(Thread.currentThread().getName() + " begin time=" + System.currentTimeMillis());
			for (int i = 0; i < Integer.MAX_VALUE/50; i++) {
				String newString = new String();
				Math.random();
			}
			System.out.println(Thread.currentThread().getName() + " end time=" + System.currentTimeMillis());
			semaphore.release();
		} catch (Exception e) {
			System.out.println(Thread.currentThread().getName() + " 进入了catch");
			e.printStackTrace();
		}
	}
}
//省略ThreadA和ThreadB的代码
public class Main {
	public static void main(String[] args) throws InterruptedException {
		Service service = new Service();
		ThreadA a = new ThreadA(service);
		a.setName("A");
		a.start();
		
		ThreadB b = new ThreadB(service);
		b.setName("B");
		b.start();
		
		Thread.sleep(1000);
		
		b.interrupt();
		System.out.println("main中断了a");
	}
}

程序运行结果如下：

A begin time=1468592693921
main中断了a
java.lang.InterruptedException
B 进入了catch
	at java.util.concurrent.locks.AbstractQueuedSynchronizer.doAcquireSharedInterruptibly(AbstractQueuedSynchronizer.java:996)
	at java.util.concurrent.locks.AbstractQueuedSynchronizer.acquireSharedInterruptibly(AbstractQueuedSynchronizer.java:1303)
	at java.util.concurrent.Semaphore.acquire(Semaphore.java:317)
	at com.concurrent.chapter1.concurrent02.Service.testMethod(Service.java:9)
	at com.concurrent.chapter1.concurrent02.ThreadB.run(ThreadB.java:11)
A end time=1468592695193

线程B成功被中断。对以上代码做如下修改：

public class Service {
	private Semaphore semaphore = new Semaphore(1);
	public void testMethod() {
		try {
			semaphore.acquireUninterruptibly();//不允许被中断
			System.out.println(Thread.currentThread().getName() + " begin time=" + System.currentTimeMillis());
			for (int i = 0; i < Integer.MAX_VALUE/50; i++) {
				String newString = new String();
				Math.random();
			}
			System.out.println(Thread.currentThread().getName() + " end time=" + System.currentTimeMillis());
			semaphore.release();
		} catch (Exception e) {
			System.out.println(Thread.currentThread().getName() + " 进入了catch");
			e.printStackTrace();
		}
	}
}

重新运行程序，控制台的打印结果如下：

A begin time=1468592930516
main中断了a
A end time=1468592931792
B begin time=1468592931792
B end time=1468592932998

acquireUninterruptibly()方法还有重载的写法acquireUninterruptibly(int permits),作用是在等待许可的情况下不允许中断，如果成功获得锁，则取得指定permits个许可。

方法availablePermits()和drainPermits()

availablePermits()返回此Semaphore对象中当前可用的许可数。

public class Service {
	private Semaphore semaphore = new Semaphore(10);//一共有10个许可
	public void testMethod() {
		try {
			semaphore.acquire(2);//每次执行消耗掉2个许可
			System.out.println(semaphore.getQueueLength() + "个线程正在等待");
			System.out.println("是否有线程正在等待semaphore：" + semaphore.hasQueuedThreads());
			System.out.println("可用许可个数" + semaphore.availablePermits());
			Thread.sleep(2000);
			semaphore.release(2);
			System.out.println("可用许可个数" + semaphore.availablePermits());
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}

//省略ThreadA类代码

public class Main {
	public static void main(String[] args) throws InterruptedException {
		Service service = new Service();
		ThreadA[] a = new ThreadA[10];
		for (int i = 0; i < a.length; i++) {
			a[i] = new ThreadA(service);
			a[i].start();
			Thread.sleep(100);
		}
	}
}

运行程序，控制台打印结果如下：

0个线程正在等待
是否有线程正在等待semaphore：false
可用许可个数8
0个线程正在等待
是否有线程正在等待semaphore：false
可用许可个数6
0个线程正在等待
是否有线程正在等待semaphore：false
可用许可个数4
0个线程正在等待
是否有线程正在等待semaphore：false
可用许可个数2
0个线程正在等待
是否有线程正在等待semaphore：false
可用许可个数0
可用许可个数2
4个线程正在等待
是否有线程正在等待semaphore：true
可用许可个数0
可用许可个数0
3个线程正在等待
是否有线程正在等待semaphore：true
可用许可个数0
可用许可个数2
2个线程正在等待
是否有线程正在等待semaphore：true
可用许可个数0
可用许可个数2
1个线程正在等待
是否有线程正在等待semaphore：true
可用许可个数0
可用许可个数2
0个线程正在等待
是否有线程正在等待semaphore：false
可用许可个数0
可用许可个数2
可用许可个数4
可用许可个数6
可用许可个数8
可用许可个数10

availablePermits()通常用于调试，因为许可的数量有可能实时在改变。

drainPermits()可以获取并返回立即可用的许可数，并且将许可置为0.

getQueueLength()获取等待许可的线程的个数。

hasQueuedThreads()判断是否有线程在等待这个许可。

公平与非公平信号量的测试

公平信号量是获得锁的顺序与线程启动顺序有关，但不代表100%得获得信号量，仅仅是在概率上能得到保证。非公平信号量就是获得锁的顺序与线程启动顺序无关。

public class Service {
	private boolean isFair = false;
	private Semaphore semaphore = new Semaphore(1, isFair);
	public void testMethod() {
		try {
			semaphore.acquire();
			System.out.println(Thread.currentThread().getName());
		} catch (Exception e) {
			e.printStackTrace();
		} finally {
			semaphore.release();
		}
	}
}

public class MyThread extends Thread {
	private Service service;
	public MyThread(Service service) {
		super();
		this.service = service;
	}
	@Override
	public void run() {
		System.out.println(Thread.currentThread().getName() + "启动了");
		service.testMethod();
	}
}

public class Main {
	public static void main(String[] args) {
		Service service = new Service();
		MyThread thread = new MyThread(service);
		thread.start();
		MyThread[] threads = new MyThread[4];
		for (int i = 0; i < threads.length; i++) {
			threads[i] = new MyThread(service);
			threads[i].start();
		}
	}
}

运行程序，控制台打印结果如下：

Thread-0启动了
Thread-3启动了
Thread-2启动了
Thread-1启动了
Thread-4启动了
Thread-0
Thread-3
Thread-2
Thread-1
Thread-4

此时的信号量为非公平信号量，线程的启动顺序与其调用semaphore.acquire()无关。先启动的线程不一定先获得许可。

对以上程序做如下修改：

public class Service {
	private boolean isFair = true;//公平锁
	private Semaphore semaphore = new Semaphore(1, isFair);
	public void testMethod() {
		try {
			semaphore.acquire()
			System.out.println(Thread.currentThread().getName());
		} catch (Exception e) {
			e.printStackTrace();
		} finally {
			semaphore.release();
		}
	}
}

重新运行程序，控制台打印结果如下：

Thread-0启动了
Thread-3启动了
Thread-2启动了
Thread-1启动了
Thread-0
Thread-4启动了
Thread-3
Thread-2
Thread-1
Thread-4

此时线程启动的顺序与线程执行semaphore.acquire()的顺序一致。先启动的线程先获得许可（非100%）。

方法tryAcquire()的使用

无参方法tryAcquire()的作用是尝试获得1一个许可，如果获取不到则返回false，此方法通常与if语句结合使用，具有无阻塞的特点。

public class Service {
	private Semaphore semaphore = new Semaphore(1);
	public void testMethod() {
		if (semaphore.tryAcquire()) {
			System.out.println(Thread.currentThread().getName() + "首选进入");
			for (int i = 0; i < Integer.MAX_VALUE; i++) {
				String newString = new String();
				Math.random();
			}
			semaphore.release();
		} else {
			System.out.println(Thread.currentThread().getName() + "未成功进入");
		}
	}
}
//省略ThreadA、ThreadB代码
public class Main {
	public static void main(String[] args) throws InterruptedException {
		Service service = new Service();
		ThreadA a = new ThreadA(service);
		a.setName("A");
		ThreadB b = new ThreadB(service);
		b.setName("B");
		a.start();
		b.start();
	}
}

程序运行结果如下：

A首选进入
B未成功进入

方法tryAcquire(long timeout, TimeUnit unit)的使用

有参方法tryAcquire(long timeout, TimeUnit unit)的作用是在指定的时间内尝试获得1个许可，如果获取不到就返回false。

public class Service {
	private Semaphore semaphore = new Semaphore(1);
	public void testMethod() {
		try {
			if (semaphore.tryAcquire(3, TimeUnit.SECONDS)) {
				System.out.println(Thread.currentThread().getName() + "首选进入");
				for (int i = 0; i < Integer.MAX_VALUE; i++) {
					String newString = new String();
					Math.random();
				}
				semaphore.release();
			} else {
				System.out.println(Thread.currentThread().getName() + "未成功进入");
			}
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}
}
//省略ThreadA、ThreadB
//省略面函数

运行程序，控制台打印结果如下：

A首选进入
B未成功进入

对以上代码做如下修改：

public class Service {
	private Semaphore semaphore = new Semaphore(1);
	public void testMethod() {
		try {
			if (semaphore.tryAcquire(3, TimeUnit.SECONDS)) {
				System.out.println(Thread.currentThread().getName() + "首选进入");
				for (int i = 0; i < Integer.MAX_VALUE; i++) {
					//String newString = new String();
					//Math.random();
				}
				semaphore.release();
			} else {
				System.out.println(Thread.currentThread().getName() + "未成功进入");
			}
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}
}

重新运行程序，控制台打印结果如下：

A首选进入
B首选进入

方法tryAcquire(int permits, long timeout, TimeUnit unit)的使用

有参方法tryAcquire(int permits, long timeout, TimeUnit unit)的作用是在指定的时间内尝试去的permits个许可，如果获取不到则返回false。

public class Service {
	private Semaphore semaphore = new Semaphore(3);
	public void testMethod() {
		try {
			if (semaphore.tryAcquire(3, 3, TimeUnit.SECONDS)) {
				System.out.println(Thread.currentThread().getName() + "首选进入");
				for (int i = 0; i < Integer.MAX_VALUE; i++) {
					String newString = new String();
					Math.random();
				}
				semaphore.release(3);
			} else {
				System.out.println(Thread.currentThread().getName() + "未成功进入");
			}
		} catch (InterruptedException e) {
			e.printStackTrace();
		}
	}
}
//省略ThreadA、ThreadB
//省略面函数

运行程序，控制台打印结果如下：

A首选进入
B未成功进入

注释掉for循环中的两行代码，A、B就都可以获得许可。

多进路-多处理-多出路实验

public class Service {
	private Semaphore semaphore = new Semaphore(3);
	public void sayHello() {
		try {
			semaphore.acquire();
			System.out.println(Thread.currentThread().getName() + "准备：" + System.currentTimeMillis());
			for (int i = 0; i < 5; i++) {
				System.out.println(Thread.currentThread().getName() + "打印：" + i);
			}
			System.out.println(Thread.currentThread().getName() + "结束：" + System.currentTimeMillis());
			semaphore.release();
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}
//省略MyThread代码
public class Main {
	public static void main(String[] args) {
		Service service = new Service();
		MyThread[] threads = new MyThread[10];
		for (int i = 0; i < threads.length; i++) {
			threads[i] = new MyThread(service);
			threads[i].start();
		}
	}
}

运行程序，控制台打印结果如下：

Thread-0准备：1469151758503
Thread-2准备：1469151758503
Thread-1准备：1469151758503
Thread-2打印：0
Thread-2打印：1
Thread-0打印：0
Thread-2打印：2
Thread-1打印：0
Thread-2打印：3
Thread-0打印：1
Thread-2打印：4
Thread-1打印：1
Thread-1打印：2
Thread-2结束：1469151758504
Thread-0打印：2
Thread-3准备：1469151758504
Thread-1打印：3
Thread-3打印：0
Thread-0打印：3
Thread-0打印：4
Thread-3打印：1
Thread-1打印：4
Thread-3打印：2
Thread-0结束：1469151758505
Thread-3打印：3
Thread-4准备：1469151758505
Thread-1结束：1469151758505
Thread-4打印：0
Thread-4打印：1
Thread-3打印：4
Thread-4打印：2
Thread-5准备：1469151758505
Thread-4打印：3
Thread-3结束：1469151758505
Thread-4打印：4
Thread-5打印：0
Thread-4结束：1469151758505
Thread-6准备：1469151758505
Thread-7准备：1469151758505
Thread-6打印：0
Thread-5打印：1
Thread-5打印：2
Thread-6打印：1
Thread-7打印：0
Thread-6打印：2
Thread-5打印：3
Thread-6打印：3
Thread-7打印：1
Thread-6打印：4
Thread-5打印：4
Thread-6结束：1469151758506
Thread-7打印：2
Thread-8准备：1469151758506
Thread-5结束：1469151758506
Thread-8打印：0
Thread-9准备：1469151758506
Thread-7打印：3
Thread-9打印：0
Thread-8打印：1
Thread-8打印：2
Thread-9打印：1
Thread-7打印：4
Thread-9打印：2
Thread-9打印：3
Thread-9打印：4
Thread-8打印：3
Thread-8打印：4
Thread-9结束：1469151758507
Thread-7结束：1469151758506
Thread-8结束：1469151758507

可见，在某一时刻最多有三个线程同时在执行。

多进路-单处理-多出路实验

对以上代码做如下修改：

public class Service {
	private Semaphore semaphore = new Semaphore(3);
	private ReentrantLock lock = new ReentrantLock();
	public void sayHello() {
		try {
			semaphore.acquire();
			System.out.println(Thread.currentThread().getName() + "准备：" + System.currentTimeMillis());
			lock.lock();//加锁
			for (int i = 0; i < 5; i++) {
				System.out.println(Thread.currentThread().getName() + "打印：" + i);
			}
			System.out.println(Thread.currentThread().getName() + "结束：" + System.currentTimeMillis());
			lock.unlock();
			semaphore.release();
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}

运行程序，控制台打印结果如下：

Thread-1准备：1469151895747
Thread-0准备：1469151895747
Thread-2准备：1469151895747
Thread-1打印：0
Thread-1打印：1
Thread-1打印：2
Thread-1打印：3
Thread-1打印：4
Thread-1结束：1469151895748
Thread-0打印：0
Thread-3准备：1469151895748
Thread-0打印：1
Thread-0打印：2
Thread-0打印：3
Thread-0打印：4
Thread-0结束：1469151895748
Thread-2打印：0
Thread-4准备：1469151895748
Thread-2打印：1
Thread-2打印：2
Thread-2打印：3
Thread-2打印：4
Thread-2结束：1469151895748
Thread-3打印：0
Thread-5准备：1469151895748
Thread-3打印：1
Thread-3打印：2
Thread-3打印：3
Thread-3打印：4
Thread-3结束：1469151895748
Thread-6准备：1469151895748
Thread-4打印：0
Thread-4打印：1
Thread-4打印：2
Thread-4打印：3
Thread-4打印：4
Thread-4结束：1469151895749
Thread-5打印：0
Thread-7准备：1469151895749
Thread-5打印：1
Thread-5打印：2
Thread-5打印：3
Thread-5打印：4
Thread-5结束：1469151895749
Thread-6打印：0
Thread-8准备：1469151895749
Thread-6打印：1
Thread-6打印：2
Thread-6打印：3
Thread-6打印：4
Thread-6结束：1469151895749
Thread-7打印：0
Thread-9准备：1469151895749
Thread-7打印：1
Thread-7打印：2
Thread-7打印：3
Thread-7打印：4
Thread-7结束：1469151895749
Thread-8打印：0
Thread-8打印：1
Thread-8打印：2
Thread-8打印：3
Thread-8打印：4
Thread-8结束：1469151895750
Thread-9打印：0
Thread-9打印：1
Thread-9打印：2
Thread-9打印：3
Thread-9打印：4
Thread-9结束：1469151895750

此时，某一时刻最多只有一个线程在运行，执行任务的顺序是同步的。

使用Semaphore创建字符串池

Semaphore类可以有效地对并发执行任务的线程数量进行限制，这个功能可以应用在pool池技术中，可以设置同时访问pool池中数据的线程数量。

public class ListPool {
	private int poolMaxSize = 3;
	private int semaphorePermits = 5;
	private List<String> list = new ArrayList<>();
	private Semaphore concurrentSemaphore = new Semaphore(semaphorePermits);
	private ReentrantLock lock = new ReentrantLock();
	private Condition condition = lock.newCondition();
	
	public ListPool() {
		super();
		for (int i = 0; i < poolMaxSize; i++) {
			list.add("test-" + (i + 1));
		}
	}
	
	public String get() {
		String string = null;
		try {
			concurrentSemaphore.acquire();
			lock.lock();
			while (list.size() == 0) {
				condition.await();
			}
			string = list.remove(0);
			lock.unlock();
		} catch (Exception e) {
			e.printStackTrace();
		}
		return string;
	}
	
	public void put(String string) {
		lock.lock();
		list.add(string);
		condition.signalAll();
		lock.unlock();
		concurrentSemaphore.release();
	}
}

public class MyThread extends Thread {
	private ListPool listPool;
	public MyThread(ListPool listPool) {
		super();
		this.listPool = listPool;
	}
	@Override
	public void run() {
		for (int i = 0; i < Integer.MAX_VALUE; i++) {
			String string = listPool.get();
			System.out.println(Thread.currentThread().getName() + "取值：" + string);
			listPool.put(string);
		}
	}
}

public class Main {
	public static void main(String[] args) {
		ListPool pool = new ListPool();
		MyThread[] threads = new MyThread[12];
		for (int i = 0; i < threads.length; i++) {
			threads[i] = new MyThread(pool);
		}
		for (int i = 0; i < threads.length; i++) {
			threads[i].start();;
		}
	}
}

程序运行结果如下：

......
Thread-10取值：test-3
Thread-10取值：test-3
Thread-10取值：test-3
Thread-10取值：test-3
Thread-2取值：test-2
Thread-10取值：test-3
Thread-2取值：test-2
Thread-8取值：test-1
Thread-2取值：test-2
Thread-10取值：test-3
Thread-10取值：test-3
Thread-10取值：test-3
Thread-10取值：test-3
Thread-10取值：test-3
......

使用Semaphore实现多生产者/多消费者模式

使用Semaphore可以限制生产者与消费者的数量。Semaphore提供了限制并发线程数的功能，synchronized不提供这个功能。

public class Service {
	volatile private Semaphore setSemaphore = new Semaphore(10);//厨师 生产者
	volatile private Semaphore getSemaphore = new Semaphore(20);//就餐者 消费者
	volatile private ReentrantLock lock = new ReentrantLock();
	volatile private Condition setCondition = lock.newCondition();
	volatile private Condition getCondition = lock.newCondition();
	volatile private Object[] producePosition = new Object[4];//4个盒子存放菜品
	
	private boolean isEmpty() {
		boolean isEmpty = true;
		for (int i = 0; i < producePosition.length; i++) {
			if (producePosition[i] != null) {
				isEmpty = false;
				break;
			}
		}
		return isEmpty;
	}
	
	private boolean isFull() {
		boolean isFull = true;
		for (int i = 0; i < producePosition.length; i++) {
			if (producePosition[i] == null) {
				isFull = false;
				break;
			}
		}
		return isFull;
	}
	
	public void set() {//生产
		try {
			setSemaphore.acquire();//最多允许10个厨师同时生产
			lock.lock();
			while (isFull()) {
				setCondition.await();
			}
			for (int i = 0; i < producePosition.length; i++) {
				if (producePosition[i] == null) {
					producePosition[i] = "数据";
					System.out.println(Thread.currentThread().getName() + "生产了：" + producePosition[i]);
					break;
				}
			}
			getCondition.signalAll();
			lock.unlock();
		} catch (Exception e) {
			e.printStackTrace();
		} finally {
			setSemaphore.release();
		}
	}
	
	public void get() {//消费
		try {
			getSemaphore.acquire();
			lock.lock();
			while (isEmpty()) {
				getCondition.await();
			}
			for (int i = 0; i < producePosition.length; i++) {
				if (producePosition[i] != null) {
					System.out.println(Thread.currentThread().getName() + "消费了：" + producePosition[i]);
					producePosition[i] = null;
					break;
				}
			}
			setCondition.signalAll();
			lock.unlock();
		} catch (Exception e) {
			e.printStackTrace();
		} finally {
			getSemaphore.release();
		}
	}
}

public class ThreadP extends Thread {
	private Service service;
	public ThreadP(Service service) {
		super();
		this.service = service;
	}
	@Override
	public void run() {
		service.set();
	}
}

public class ThreadC extends Thread {
	private Service service;
	public ThreadC(Service service) {
		super();
		this.service = service;
	}
	@Override
	public void run() {
		service.get();
	}
}

public class Main {
	public static void main(String[] args) throws InterruptedException {
		Service service = new Service();
		ThreadP[] threadPs = new ThreadP[60];
		ThreadC[] threadCs = new ThreadC[60];
		for (int i = 0; i < threadCs.length; i++) {
			threadPs[i] = new ThreadP(service);
			threadCs[i] = new ThreadC(service);
		}
		Thread.sleep(2000);
		for (int i = 0; i < threadCs.length; i++) {
			threadPs[i].start();;
			threadCs[i].start();;
		}
	}
}

运行程序，控制台打印结果如下：

......
Thread-19消费了：数据
Thread-4生产了：数据
Thread-20生产了：数据
Thread-5消费了：数据
Thread-23消费了：数据
Thread-24生产了：数据
Thread-6生产了：数据
Thread-7消费了：数据
Thread-28生产了：数据
......

posted @ 2016-07-14 20:19 商商-77 阅读(465) 评论(0) 编辑收藏举报

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Java并发编程核心方法与框架-Semaphore的使用

类Semaphore的同步性

方法acquire(int permits)参数作用及动态添加permits许可数量

方法acquireUninterruptibly()的使用

方法availablePermits()和drainPermits()

公平与非公平信号量的测试

方法tryAcquire()的使用

方法tryAcquire(long timeout, TimeUnit unit)的使用

方法tryAcquire(int permits, long timeout, TimeUnit unit)的使用

多进路-多处理-多出路实验

多进路-单处理-多出路实验

使用Semaphore创建字符串池

使用Semaphore实现多生产者/多消费者模式

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