java并发编程-Executor框架【转】

原文:http://mshijie.iteye.com/blog/366591

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:

《java并发编程-Executor框架【转】》

 

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。

ExecutorService executorService = (ExecutorService) executor;
while (!executorService.isShutdown()) {
    try {
        executorService.execute(task);
    } catch (RejectedExecutionException ignored) {
        
    }
}
executorService.shutdown();

 

 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。

Java代码

ExecutorService executorService = (ExecutorService) executor;
while (!executorService.isShutdown()) {
    try {
        executorService.execute(task);
    } catch (RejectedExecutionException ignored) {
        
    }
}
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。

Java代码

        Callable<Integer> func = new Callable<Integer>(){
            public Integer call() throws Exception {
                System.out.println("inside callable");
                Thread.sleep(1000);
                return new Integer(8);
            }        
        };        
        FutureTask<Integer> futureTask  = new FutureTask<Integer>(func);
        Thread newThread = new Thread(futureTask);
        newThread.start();
        
        try {
            System.out.println("blocking here");
            Integer result = futureTask.get();
            System.out.println(result);
        } catch (InterruptedException ignored) {
        } catch (ExecutionException ignored) {
        }

 

 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。

Java代码 

package executorservice;

import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;
import java.util.concurrent.FutureTask;

public class ConcurrentCalculator {

    private ExecutorService exec;
    private int cpuCoreNumber;
    private List<Future<Long>> tasks = new ArrayList<Future<Long>>();

    // 内部类
    class SumCalculator implements Callable<Long> {
        private int[] numbers;
        private int start;
        private int end;

        public SumCalculator(final int[] numbers, int start, int end) {
            this.numbers = numbers;
            this.start = start;
            this.end = end;
        }

        public Long call() throws Exception {
            Long sum = 0l;
            for (int i = start; i < end; i++) {
                sum += numbers[i];
            }
            return sum;
        }
    }

    public ConcurrentCalculator() {
        cpuCoreNumber = Runtime.getRuntime().availableProcessors();
        exec = Executors.newFixedThreadPool(cpuCoreNumber);
    }

    public Long sum(final int[] numbers) {
        // 根据CPU核心个数拆分任务,创建FutureTask并提交到Executor
        for (int i = 0; i < cpuCoreNumber; i++) {
            int increment = numbers.length / cpuCoreNumber + 1;
            int start = increment * i;
            int end = increment * i + increment;
            if (end > numbers.length)
                end = numbers.length;
            SumCalculator subCalc = new SumCalculator(numbers, start, end);
            FutureTask<Long> task = new FutureTask<Long>(subCalc);
            tasks.add(task);
            if (!exec.isShutdown()) {
                exec.submit(task);
            }
        }
        return getResult();
    }

    /**
     * 迭代每个只任务,获得部分和,相加返回
     * 
     * @return
     */
    public Long getResult() {
        Long result = 0l;
        for (Future<Long> task : tasks) {
            try {
                // 如果计算未完成则阻塞
                Long subSum = task.get();
                result += subSum;
            } catch (InterruptedException e) {
                e.printStackTrace();
            } catch (ExecutionException e) {
                e.printStackTrace();
            }
        }
        return result;
    }

    public void close() {
        exec.shutdown();
    }
}

Main

int[] numbers = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 10, 11 };
ConcurrentCalculator calc = new ConcurrentCalculator();
Long sum = calc.sum(numbers);
System.out.println(sum);
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:

Java代码

public class ConcurrentCalculator2 {

    private ExecutorService exec;
    private CompletionService<Long> completionService;


    private int cpuCoreNumber;

    // 内部类
    class SumCalculator implements Callable<Long> {
        ......
    }

    public ConcurrentCalculator2() {
        cpuCoreNumber = Runtime.getRuntime().availableProcessors();
        exec = Executors.newFixedThreadPool(cpuCoreNumber);
        completionService = new ExecutorCompletionService<Long>(exec);


    }

    public Long sum(final int[] numbers) {
        // 根据CPU核心个数拆分任务,创建FutureTask并提交到Executor
        for (int i = 0; i < cpuCoreNumber; i++) {
            int increment = numbers.length / cpuCoreNumber + 1;
            int start = increment * i;
            int end = increment * i + increment;
            if (end > numbers.length)
                end = numbers.length;
            SumCalculator subCalc = new SumCalculator(numbers, start, end);    
            if (!exec.isShutdown()) {
                completionService.submit(subCalc);


            }
            
        }
        return getResult();
    }

    /**
     * 迭代每个只任务,获得部分和,相加返回
     * 
     * @return
     */
    public Long getResult() {
        Long result = 0l;
        for (int i = 0; i < cpuCoreNumber; i++) {            
            try {
                Long subSum = completionService.take().get();
                result += subSum;            
            } catch (InterruptedException e) {
                e.printStackTrace();
            } catch (ExecutionException e) {
                e.printStackTrace();
            }
        }
        return result;
    }

    public void close() {
        exec.shutdown();
    }
}

 

    原文作者:Leo Forest
    原文地址: http://www.cnblogs.com/Leo-Forest/archive/2012/10/24/2737677.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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