# 08 ForkJoin

> 本文为个人学习摘要笔记。\
> 原文地址：[廖雪峰 Java 教程 - 使用 ForkJoin](https://www.liaoxuefeng.com/wiki/1252599548343744/1306581226487842)

Java 7 开始引入了一种新的 `Fork/Join` 线程池，它可以执行一种特殊的任务：把一个大任务拆成多个小任务并行执行。

`Fork/Join` 任务的原理：判断一个任务是否足够小，如果是，直接计算，否则，就分拆成几个小任务分别计算。这个过程可以反复“裂变”成一系列小任务。

下面以大数组求和为例：

```java
public class T {
    private static Random random = new Random(0);

    private static long random() {
        return random.nextInt(10000);
    }

    public static void main(String[] args) throws Exception {
        // 创建2000个随机数组成的数组:
        long[] array = new long[2000];
        long expectedSum = 0;
        for (int i = 0; i < array.length; i++) {
            array[i] = random();
            expectedSum += array[i];
        }
        System.out.println("Expected sum: " + expectedSum);
        // fork/join:
        ForkJoinTask<Long> task = new SumTask(array, 0, array.length);
        long startTime = System.currentTimeMillis();
        Long result = ForkJoinPool.commonPool().invoke(task);
        long endTime = System.currentTimeMillis();
        System.out.println("Fork/join sum: " + result + " in " + (endTime - startTime) + " ms.");
    }
}

class SumTask extends RecursiveTask<Long> {
    static final int THRESHOLD = 500;
    long[] array;
    int start;
    int end;

    SumTask(long[] array, int start, int end) {
        this.array = array;
        this.start = start;
        this.end = end;
    }

    @Override
    protected Long compute() {
        if (end - start <= THRESHOLD) {
            // 如果任务足够小,直接计算:
            long sum = 0;
            for (int i = start; i < end; i++) {
                sum += this.array[i];
            }
            return sum;
        }
       // 任务太大，一分为二:
        int middle = (end + start) / 2;
        System.out.println(String.format("split %d~%d ==> %d~%d, %d~%d", start, end, start, middle, middle, end));
        // “分裂”子任务:
        SumTask subtask1 = new SumTask(this.array, start, middle);
        SumTask subtask2 = new SumTask(this.array, middle, end);
        // invokeAll会并行运行两个子任务:
        invokeAll(subtask1, subtask2);
        // 获得子任务的结果并汇总:
        Long subresult1 = subtask1.join();
        Long subresult2 = subtask2.join();
        Long result = subresult1 + subresult2;
        System.out.println("result = " + subresult1 + " + " + subresult2 + " ==> " + result);
        return result;
    }
}
```


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