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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