近日于LeetCode看题遇1114 按序打印，获悉一解法使用了Semaphore，顺势研究，记心得于此。
此解视Semaphore为锁，以保证同一时刻单线程的顺序执行。在此原题上，我作出如下更改。
package test; import java.util.concurrent.ExecutorService;import java.util.concurrent.Executors;import java.util.concurrent.Semaphore; public class SemaphoreDemo { static Semaphore A; static Semaphore B; static Semaphore C; public static void main(String[] args) throws InterruptedException { A = new Semaphore(1); B = new Semaphore(0); C = new Semaphore(0); ExecutorService ex=Executors.newFixedThreadPool(10); for (int i = 0; i <7; i++) { ex.execute(new R1()); ex.execute(new R2()); ex.execute(new R3()); } ex.shutdown(); } public static class R1 implements Runnable{ @Override public void run() { try {// A.acquire(); System.out.println("1"+Thread.currentThread().getName());// B.release(); } catch (Exception e) { e.printStackTrace(); } } } public static class R2 implements Runnable{ @Override public void run() { try {// B.acquire(); System.out.println("2"+Thread.currentThread().getName());// C.release(); } catch (Exception e) { e.printStackTrace(); } } } public static class R3 implements Runnable{ @Override public void run() { try {// C.acquire(); System.out.println("3"+Thread.currentThread().getName());// A.release(); } catch (Exception e) { e.printStackTrace(); } } } }
10个线程的常量池中，分别调用R1,R2,R3的方法多次，控制台输出对应各方法名拼接执行该方法的线程名。多次执行结果各不相同：
1pool-1-thread-12pool-1-thread-21pool-1-thread-43pool-1-thread-62pool-1-thread-53pool-1-thread-31pool-1-thread-72pool-1-thread-83pool-1-thread-93pool-1-thread-12pool-1-thread-81pool-1-thread-43pool-1-thread-11pool-1-thread-22pool-1-thread-91pool-1-thread-103pool-1-thread-12pool-1-thread-51pool-1-thread-63pool-1-thread-42pool-1-thread-8
1pool-1-thread-12pool-1-thread-23pool-1-thread-31pool-1-thread-42pool-1-thread-53pool-1-thread-61pool-1-thread-72pool-1-thread-83pool-1-thread-91pool-1-thread-103pool-1-thread-11pool-1-thread-42pool-1-thread-83pool-1-thread-32pool-1-thread-101pool-1-thread-22pool-1-thread-93pool-1-thread-41pool-1-thread-73pool-1-thread-62pool-1-thread-5
方法能调用，多线程下却无法保证方法的顺序执行。使用Semaphore后，代码为：
package test; import java.util.concurrent.ExecutorService;import java.util.concurrent.Executors;import java.util.concurrent.Semaphore; public class SemaphoreDemo { static Semaphore A; static Semaphore B; static Semaphore C; public static void main(String[] args) throws InterruptedException { A = new Semaphore(1); B = new Semaphore(0); C = new Semaphore(0); ExecutorService ex=Executors.newFixedThreadPool(10); for (int i = 0; i <7; i++) { ex.execute(new R1()); ex.execute(new R2()); ex.execute(new R3()); } ex.shutdown(); } public static class R1 implements Runnable{ @Override public void run() { try { A.acquire(); System.out.println("1"+Thread.currentThread().getName()); B.release(); } catch (Exception e) { e.printStackTrace(); } } } public static class R2 implements Runnable{ @Override public void run() { try { B.acquire(); System.out.println("2"+Thread.currentThread().getName()); C.release(); } catch (Exception e) { e.printStackTrace(); } } } public static class R3 implements Runnable{ @Override public void run() { try { C.acquire(); System.out.println("3"+Thread.currentThread().getName()); A.release(); } catch (Exception e) { e.printStackTrace(); } } } }
多次运行结果皆能保证1、2、3的顺序：
1pool-1-thread-12pool-1-thread-23pool-1-thread-31pool-1-thread-42pool-1-thread-53pool-1-thread-61pool-1-thread-72pool-1-thread-83pool-1-thread-91pool-1-thread-102pool-1-thread-13pool-1-thread-21pool-1-thread-32pool-1-thread-43pool-1-thread-51pool-1-thread-62pool-1-thread-93pool-1-thread-71pool-1-thread-102pool-1-thread-83pool-1-thread-1
附上api文档链接Semaphore
A = new Semaphore(1);B = new Semaphore(0);C = new Semaphore(0);
进入R2、R3方法的线程会执行acquire()方法，而B、C中的计数器为0获取不到许可，阻塞直到一个可用，或者线程被中断，不能继续执行。R1方法中A尚有1个许可可拿到，方法执行，并给B发布一个许可，若B先于A执行acquire()，此时B为阻塞状态，则获取到刚刚发布的许可，该线程被重新启用。
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本文标题: Java 信号量Semaphore的实现
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