【转】Java并发框架Executor学习笔记_JAVA_编程开发_程序员俱乐部

Java SE5的java.util.concurrent包中的执行器（Executor）将为你管理Thread对象，从而简化了并发编程。Executor在客户端和执行任务之间提供了一个间接层，Executor代替客户端执行任务。Executor允许你管理异步任务的执行，而无须显式地管理线程的生命周期。Executor在Java SE5/6中时启动任务的优选方法。Executor引入了一些功能类来管理和使用线程Thread，其中包括线程池，Executor，Executors，ExecutorService，CompletionService，Future，Callable等

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创建线程池

Executors类，提供了一系列工厂方法用于创先线程池，返回的线程池都实现了ExecutorService接口。

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public static ExecutorService newFixedThreadPool(int nThreads)

创建固定数目线程的线程池。

public static ExecutorService newCachedThreadPool()

创建一个可缓存的线程池，调用execute 将重用以前构造的线程（如果线程可用）。如果现有线程没有可用的，则创建一个新线程并添加到池中。终止并从缓存中移除那些已有 60 秒钟未被使用的线程。

public static ExecutorService newSingleThreadExecutor()

创建一个单线程化的Executor。

public static ScheduledExecutorService newScheduledThreadPool(int corePoolSize)

创建一个支持定时及周期性的任务执行的线程池，多数情况下可用来替代Timer类。

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见类图，接口Executor只有一个方法execute，接口ExecutorService扩展了Executor并添加了一些生命周期管理的方法，如shutdown、submit等。一个Executor的生命周期有三种状态，运行 ，关闭 ，终止。

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Callable，Future用于返回结果

Future<V>代表一个异步执行的操作，通过get()方法可以获得操作的结果，如果异步操作还没有完成，则，get()会使当前线程阻塞。FutureTask<V>实现了Future<V>和Runable<V>。Callable代表一个有返回值得操作。

实例：并行计算求和

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1. public?class?ConcurrentSum?{??
2. ????private?int?coreCpuNum;??
3. ????private?ExecutorService??executor;??
4. ????private?List<FutureTask<Long>>?tasks?=?new?ArrayList<FutureTask<Long>>();??
5. ????public?ConcurrentSum(){??
6. ????????coreCpuNum?=?Runtime.getRuntime().availableProcessors();??
7. ????????executor?=?Executors.newFixedThreadPool(coreCpuNum);??
8. ????}??
9. ????class?SumCalculator?implements?Callable<Long>{??
10. ????????int?nums[];??
11. ????????int?start;??
12. ????????int?end;??
13. ????????public?SumCalculator(final?int?nums[],int?start,int?end){??
14. ????????????this.nums?=?nums;??
15. ????????????this.start?=?start;??
16. ????????????this.end?=?end;??
17. ????????}??
18. ????????@Override??
19. ????????public?Long?call()?throws?Exception?{??
20. ????????????long?sum?=0;??
21. ????????????for(int?i=start;i<end;i++){??
22. ????????????????sum?+=?nums[i];??
23. ????????????}??
24. ????????????return?sum;??
25. ????????}??
26. ????}??
27. ????public?long?sum(int[]?nums){??
28. ????????int?start,end,increment;??
29. ????????//?根据CPU核心个数拆分任务，创建FutureTask并提交到Executor???
30. ????????for(int?i=0;i<coreCpuNum;i++){??
31. ????????????increment?=?nums.length?/?coreCpuNum+1;??
32. ????????????start?=?i*increment;??
33. ????????????end?=?start+increment;??
34. ????????????if(end?>?nums.length){??
35. ????????????????end?=?nums.length;???
36. ????????????}??
37. ????????????SumCalculator?calculator?=?new?SumCalculator(nums,?start,?end);??
38. ????????????FutureTask<Long>?task?=?new?FutureTask<Long>(calculator);??
39. ????????????tasks.add(task);??
40. ????????????if(!executor.isShutdown()){??
41. ????????????????executor.submit(task);??
42. ????????????}??
43. ????????}??
44. ????????return?getPartSum();??
45. ????}??
46. ????public?long?getPartSum(){??
47. ????????long?sum?=?0;??
48. ????????for(int?i=0;i<tasks.size();i++){??
49. ????????????try?{??
50. ????????????????sum?+=?tasks.get(i).get();??
51. ????????????}?catch?(InterruptedException?e)?{??
52. ????????????????e.printStackTrace();??
53. ????????????}?catch?(ExecutionException?e)?{??
54. ????????????????e.printStackTrace();??
55. ????????????}??
56. ????????}??
57. ????????return?sum;??
58. ????}??
59. ????public?void?close(){??
60. ????????executor.shutdown();??
61. ????}??
62. ??????
63. ????public?static?void?main(String[]?args)?{??
64. ????????int?arr[]?=?new?int[]{1,?22,?33,?4,?52,?61,?7,?48,?10,?11?};??
65. ????????long?sum?=?new?ConcurrentSum().sum(arr);??
66. ????????System.out.println("sum:?"?+?sum);??
67. ????}??
68. }??

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CompletionService

在上述例子中，getResult()方法的实现过程中，迭代了FutureTask的数组，如果任务还没有完成则当前线程会阻塞，如果我们希望任意任务完成后就把其结果加到result中，而不用依次等待每个任务完成，可以使用CompletionService。

它与ExecutorService最主要的区别在于submit的task不一定是按照加入时的顺序完成的。CompletionService对ExecutorService进行了包装，内部维护一个保存Future对象的BlockingQueue。只有当这个Future对象状态是结束的时候，才会加入到这个Queue中，take()方法其实就是Producer-Consumer中的Consumer。它会从Queue中取出Future对象，如果Queue是空的，就会阻塞在那里，直到有完成的Future对象加入到Queue中。所以，先完成的必定先被取出。这样就减少了不必要的等待时间。

实例：并行计算求和

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1. public?class?ConcurrentSum2?{??
2. ????private?int?coreCpuNum;??
3. ????private?ExecutorService??executor;??
4. ????private?CompletionService<Long>?completionService;??
5. ??????
6. ????public?ConcurrentSum2(){??
7. ????????//.....??
8. ????}??
9. ????class?SumCalculator?implements?Callable<Long>{??
10. ????????//.....??
11. ????}??
12. ????public?long?sum(int[]?nums){??
13. ????????int?start,end,increment;??
14. ????????//?根据CPU核心个数拆分任务，创建FutureTask并提交到Executor???
15. ????????for(int?i=0;i<coreCpuNum;i++){??
16. ????????????increment?=?nums.length?/?coreCpuNum+1;??
17. ????????????start?=?i*increment;??
18. ????????????end?=?start+increment;??
19. ????????????if(end?>?nums.length){??
20. ????????????????end?=?nums.length;???
21. ????????????}??
22. ????????????SumCalculator?task?=?new?SumCalculator(nums,?start,?end);??
23. ????????????if(!executor.isShutdown()){??
24. ????????????????completionService.submit(task);??
25. ????????????}??
26. ????????}??
27. ????????return?getPartSum();??
28. ????}??
29. ????public?long?getPartSum(){??
30. ????????long?sum?=?0;??
31. ????????for(int?i=0;i<coreCpuNum;i++){??
32. ????????????try?{??
33. ????????????????sum?+=?completionService.take().get();??
34. ????????????}?catch?(InterruptedException?e)?{??
35. ????????????????e.printStackTrace();??
36. ????????????}?catch?(ExecutionException?e)?{??
37. ????????????????e.printStackTrace();??
38. ????????????}??
39. ????????}??
40. ????????return?sum;??
41. ????}??
42. ????public?void?close(){??
43. ????????executor.shutdown();??
44. ????}??
45. } ?

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转自：http://willsunforjava.iteye.com/blog/1631353