java并发编程-Executor框架+Future

Executor框架是指java 5中引入的一系列并发库中与executor相关的一些功能类，其中包括线程池，Executor，Executors，ExecutorService，CompletionService，Future，Callable等。他们的关系为：

并发编程的一种编程方式是把任务拆分为一些列的小任务，即Runnable，然后在提交给一个Executor执行，Executor.execute(Runnalbe) 。Executor在执行时使用内部的线程池完成操作。

一、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。

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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Executor executor = Executors.newFixedThreadPool(10); 

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Runnable task = new Runnable() { 

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    @Override

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    public void run() { 

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        System.out.println("task over"); 

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    } 

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}; 

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executor.execute(task); 

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executor = Executors.newScheduledThreadPool(10); 

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ScheduledExecutorService scheduler = (ScheduledExecutorService) executor; 

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scheduler.scheduleAtFixedRate(task, 10, 10, TimeUnit.SECONDS);

二、ExecutorService与生命周期

ExecutorService扩展了Executor并添加了一些生命周期管理的方法。一个Executor的生命周期有三种状态，运行，关闭，终止。Executor创建时处于运行状态。当调用ExecutorService.shutdown()后，处于关闭状态，isShutdown()方法返回true。这时，不应该再想Executor中添加任务，所有已添加的任务执行完毕后，Executor处于终止状态，isTerminated()返回true。

如果Executor处于关闭状态，往Executor提交任务会抛出unchecked exception RejectedExecutionException。

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ExecutorService executorService = (ExecutorService) executor; 

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while (!executorService.isShutdown()) { 

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    try { 

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        executorService.execute(task); 

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    } catch (RejectedExecutionException ignored) { 

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    } 

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} 

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executorService.shutdown();

三、使用Callable，Future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。

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Callable<Integer> func = new Callable<Integer>(){ 

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    public Integer call() throws Exception { 

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        System.out.println("inside callable"); 

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        Thread.sleep(1000); 

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        return new Integer(8); 

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    }        

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};       

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FutureTask<Integer> futureTask  = new FutureTask<Integer>(func); 

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Thread newThread = new Thread(futureTask); 

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newThread.start(); 

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try { 

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    System.out.println("blocking here"); 

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    Integer result = futureTask.get(); 

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    System.out.println(result); 

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} catch (InterruptedException ignored) { 

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} catch (ExecutionException ignored) { 

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}

ExecutoreService提供了submit()方法，传递一个Callable，或Runnable，返回Future。如果Executor后台线程池还没有完成Callable的计算，这调用返回Future对象的get()方法，会阻塞直到计算完成。

例子：并行计算数组的和。

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package executorservice; 

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import java.util.ArrayList; 

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import java.util.List; 

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import java.util.concurrent.Callable; 

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import java.util.concurrent.ExecutionException; 

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import java.util.concurrent.ExecutorService; 

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import java.util.concurrent.Executors; 

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import java.util.concurrent.Future; 

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import java.util.concurrent.FutureTask; 

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public class ConcurrentCalculator { 

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    private ExecutorService exec; 

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    private int cpuCoreNumber; 

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    private List<Future<Long>> tasks = new ArrayList<Future<Long>>(); 

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    // 内部类 

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    class SumCalculator implements Callable<Long> { 

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        private int[] numbers; 

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        private int start; 

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        private int end; 

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        public SumCalculator(final int[] numbers, int start, int end) { 

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            this.numbers = numbers; 

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            this.start = start; 

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            this.end = end; 

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        } 

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        public Long call() throws Exception { 

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            Long sum = 0l; 

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            for (int i = start; i < end; i++) { 

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                sum += numbers[i]; 

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            } 

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            return sum; 

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        } 

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    } 

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    public ConcurrentCalculator() { 

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        cpuCoreNumber = Runtime.getRuntime().availableProcessors(); 

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        exec = Executors.newFixedThreadPool(cpuCoreNumber); 

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    } 

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    public Long sum(final int[] numbers) { 

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        // 根据CPU核心个数拆分任务，创建FutureTask并提交到Executor 

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        for (int i = 0; i < cpuCoreNumber; i++) { 

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            int increment = numbers.length / cpuCoreNumber + 1; 

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            int start = increment * i; 

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            int end = increment * i + increment; 

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            if (end > numbers.length) 

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                end = numbers.length; 

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            SumCalculator subCalc = new SumCalculator(numbers, start, end); 

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            FutureTask<Long> task = new FutureTask<Long>(subCalc); 

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            tasks.add(task); 

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            if (!exec.isShutdown()) { 

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                exec.submit(task); 

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            } 

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        } 

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        return getResult(); 

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    } 

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    /** 

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     * 迭代每个只任务，获得部分和，相加返回 

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     *  

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     * @return 

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     */

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    public Long getResult() { 

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        Long result = 0l; 

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        for (Future<Long> task : tasks) { 

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            try { 

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                // 如果计算未完成则阻塞 

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                Long subSum = task.get(); 

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                result += subSum; 

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            } catch (InterruptedException e) { 

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                e.printStackTrace(); 

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            } catch (ExecutionException e) { 

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                e.printStackTrace(); 

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            } 

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        } 

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        return result; 

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    } 

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    public void close() { 

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        exec.shutdown(); 

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    } 

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}

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int[] numbers = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 10, 11 }; 

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ConcurrentCalculator calc = new ConcurrentCalculator(); 

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Long sum = calc.sum(numbers); 

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System.out.println(sum); 

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calc.close();

四、CompletionService

在刚在的例子中，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，如果我们希望任意字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CompletionService。生产者submit()执行的任务。使用者take()已完成的任务，并按照完成这些任务的顺序处理它们的结果 。也就是调用CompletionService的take方法是，会返回按完成顺序放回任务的结果，CompletionService内部维护了一个阻塞队列BlockingQueue，如果没有任务完成，take()方法也会阻塞。修改刚才的例子使用CompletionService：

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public class ConcurrentCalculator2 { 

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    private ExecutorService exec; 

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    private CompletionService<Long> completionService; 

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    private int cpuCoreNumber; 

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    // 内部类 

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    class SumCalculator implements Callable<Long> { 

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        ...... 

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    } 

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    public ConcurrentCalculator2() { 

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        cpuCoreNumber = Runtime.getRuntime().availableProcessors(); 

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        exec = Executors.newFixedThreadPool(cpuCoreNumber); 

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        completionService = new ExecutorCompletionService<Long>(exec); 

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    } 

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    public Long sum(final int[] numbers) { 

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        // 根据CPU核心个数拆分任务，创建FutureTask并提交到Executor 

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        for (int i = 0; i < cpuCoreNumber; i++) { 

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            int increment = numbers.length / cpuCoreNumber + 1; 

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            int start = increment * i; 

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            int end = increment * i + increment; 

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            if (end > numbers.length) 

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                end = numbers.length; 

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            SumCalculator subCalc = new SumCalculator(numbers, start, end);  

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            if (!exec.isShutdown()) { 

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                completionService.submit(subCalc); 

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            } 

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        } 

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        return getResult(); 

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    } 

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    /** 

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     * 迭代每个只任务，获得部分和，相加返回 

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     *  

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     * @return 

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     */

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    public Long getResult() { 

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        Long result = 0l; 

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        for (int i = 0; i < cpuCoreNumber; i++) {             

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            try { 

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                Long subSum = completionService.take().get(); 

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                result += subSum;            

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            } catch (InterruptedException e) { 

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                e.printStackTrace(); 

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            } catch (ExecutionException e) { 

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                e.printStackTrace(); 

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            } 

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        } 

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        return result; 

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    } 

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    public void close() { 

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        exec.shutdown(); 

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    } 

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}

posted @ 2011-06-01 16:42 跳刀的兔子阅读(10621) 评论(0) 编辑收藏举报