1、编码案例
基于1+2..+100的计算案例演示Fork/Join框架基础用法。
import java.util.concurrent.ForkJoinPool;
import java.util.concurrent.ForkJoinTask;
import java.util.concurrent.RecursiveTask;
public class ForkJoin01 {
public static void main (String[] args) {
int[] numArr = new int[100];
for (int i = 0; i < 100; i++) {
numArr[i] = i + 1;
}
ForkJoinPool pool = new ForkJoinPool();
ForkJoinTask
pool.submit(new SumTask(numArr, 0, numArr.length));
System.out.println("合并计算结果: " + forkJoinTask.invoke());
pool.shutdown();
}
}
/**
* 线程任务
*/
class SumTask extends RecursiveTask
/*
* 切分任务块的阈值
* 如果THRESHOLD=100
* 输出：main【求和：(0...100)=5050】 合并计算结果: 5050
*/
private static final int THRESHOLD = 100;
private int arr[];
private int start;
private int over;
public SumTask(int[] arr, int start, int over) {
this.arr = arr;
this.start = start;
this.over = over;
}
// 求和计算
private Integer sumCalculate () {
Integer sum = 0;
for (int i = start; i < over; i++) {
sum += arr[i];
}
String task = "【求和：(" + start + "..." + over + ")=" + sum +"】";
System.out.println(Thread.currentThread().getName() + task);
return sum ;
}
@Override
protected Integer compute() {
if ((over - start) <= THRESHOLD) {
return sumCalculate();
}else {
int middle = (start + over) / 2;
SumTask left = new SumTask(arr, start, middle);
SumTask right = new SumTask(arr, middle, over);
left.fork();
right.fork();
return left.join() + right.join();
}
}
}
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