Java 信号量Semaphore的实现

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-threhttp://ad-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.TASueDunGexecute(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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