future_callable

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。

简单代码：

public class FutureTest { public static class Task implements Callable<String>{@Overridepublic String call() throws Exception {Thread.sleep(1800);System.out.println("ah");return "something";} } public static void main(String[] args) throws InterruptedException, ExecutionException {List<Future<String>> future = new ArrayList<Future<String>>();ExecutorService service = Executors.newCachedThreadPool();for (int i = 0; i < 100; i++) {future.add(service.submit(new Task()));}for (Future<String> future2 : future) {System.out.println(future2.get());} }}

稍微复杂： 集合里面的数和，查看cpu数量，分开开线程

public class FutureTest1 {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 FutureTest1() {        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();    }    public static void main(String[] args) {  int[] numbers = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 10, 11 };    FutureTest1 calc = new FutureTest1();    Long sum = calc.sum(numbers);    System.out.println(sum);    calc.close(); }}