Java 信号量Semaphore的实现

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-1
2pool-1-thread-2
1pool-1-thread-4
3pool-1-thread-6
2pool-1-thread-5
3pool-1-thread-3
1pool-1-thread-7
2pool-1-thread-8
3pool-1-thread-9
3pool-1-thread-1
2pool-1-thread-8
1pool-1-thread-4
3pool-1-thread-1
1pool-1-thread-2
2pool-1-thread-9
1pool-1-thread-10
3pool-1-thread-1
2pool-1-thread-5
1pool-1-thread-6
3pool-1-thread-4
2pool-1-thread-8

1pool-1-thread-1
2pool-1-thread-2
3pool-1-thread-3
1pool-1-thread-4
2pool-1-thread-5
3pool-1-thread-6
1pool-1-thread-7
2pool-1-thread-8
3pool-1-thread-9
1pool-1-thread-10
3pool-1-thread-1
1pool-1-thread-4
2pool-1-thread-8
3pool-1-thread-3
2pool-1-thread-10
1pool-1-thread-2
2pool-1-thread-9
3pool-1-thread-4
1pool-1-thread-7
3pool-1-thread-6
2pool-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-1
2pool-1-thread-2
3pool-1-thread-3
1pool-1-thread-4
2pool-1-thread-5
3pool-1-thread-6
1pool-1-thread-7
2pool-1-thread-8
3pool-1-thread-9
1pool-1-thread-10
2pool-1-thread-1
3pool-1-thread-2
1pool-1-thread-3
2pool-1-thread-4
3pool-1-thread-5
1pool-1-thread-6
2pool-1-thread-9
3pool-1-thread-7
1pool-1-thread-10
2pool-1-thread-8
3pool-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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