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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。
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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。
以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持创新互联。
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