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。
packagetest;
importjava.util.concurrent.ExecutorService;
importjava.util.concurrent.Executors;
importjava.util.concurrent.Semaphore;
publicclassSemaphoreDemo{
staticSemaphoreA;
staticSemaphoreB;
staticSemaphoreC;
publicstaticvoidmain(String[]args)throwsInterruptedException{
A=newSemaphore(1);
B=newSemaphore(0);
C=newSemaphore(0);
ExecutorServiceex=Executors.newFixedThreadPool(10);
for(inti=0;i<7;i++){
ex.execute(newR1());
ex.execute(newR2());
ex.execute(newR3());
}
ex.shutdown();
}
publicstaticclassR1implementsRunnable{
@Override
publicvoidrun(){
try{
//A.acquire();
System.out.println("1"+Thread.currentThread().getName());
//B.release();
}catch(Exceptione){
e.printStackTrace();
}
}
}
publicstaticclassR2implementsRunnable{
@Override
publicvoidrun(){
try{
//B.acquire();
System.out.println("2"+Thread.currentThread().getName());
//C.release();
}catch(Exceptione){
e.printStackTrace();
}
}
}
publicstaticclassR3implementsRunnable{
@Override
publicvoidrun(){
try{
//C.acquire();
System.out.println("3"+Thread.currentThread().getName());
//A.release();
}catch(Exceptione){
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：
packagetest;
importjava.util.concurrent.ExecutorService;
importjava.util.concurrent.Executors;
importjava.util.concurrent.Semaphore;
publicclassSemaphoreDemo{
staticSemaphoreA;
staticSemaphoreB;
staticSemaphoreC;
publicstaticvoidmain(String[]args)throwsInterruptedException{
A=newSemaphore(1);
B=newSemaphore(0);
C=newSemaphore(0);
ExecutorServiceex=Executors.newFixedThreadPool(10);
for(inti=0;i<7;i++){
ex.execute(newR1());
ex.execute(newR2());
ex.execute(newR3());
}
ex.shutdown();
}
publicstaticclassR1implementsRunnable{
@Override
publicvoidrun(){
try{
A.acquire();
System.out.println("1"+Thread.currentThread().getName());
B.release();
}catch(Exceptione){
e.printStackTrace();
}
}
}
publicstaticclassR2implementsRunnable{
@Override
publicvoidrun(){
try{
B.acquire();
System.out.println("2"+Thread.currentThread().getName());
C.release();
}catch(Exceptione){
e.printStackTrace();
}
}
}
publicstaticclassR3implementsRunnable{
@Override
publicvoidrun(){
try{
C.acquire();
System.out.println("3"+Thread.currentThread().getName());
A.release();
}catch(Exceptione){
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=newSemaphore(1);
B=newSemaphore(0);
C=newSemaphore(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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