java - 如何让这个程序在Java中最后显示正确的值？

Question

我在函数末尾获取正确的值时遇到问题。我不明白如何让主线程等待执行程序完成，这样我就可以显示正确的值。

每个任务计算数组中索引 i 处的值的乘积。我将产品变量设置为 volatile ，并同步计算产品的函数，以确保只有一个线程可以访问它并修改它。但我无法安排线程等待执行程序完成然后显示答案。

 public class NewBetterDot {
static double[] vectorOne;
static double[] vectorTwo;
private static volatile double product;;

public NewBetterDot(double[] vectorOne, double[] vectorTwo, double product) {
    super();
    this.vectorOne = vectorOne;
    this.vectorTwo = vectorTwo;
    this.product = product;

}

public static synchronized double smallDot(int i) {

    double multi = vectorOne[i] * vectorTwo[i];
    return multi;
}

static class Task implements Runnable {
    int i;

    @Override
    public void run() {

        // System.out.println("Before operation: " + product);

        product = product + NewBetterDot.smallDot(i);

        System.out.println(Thread.currentThread().getName() + " Product is equal to: " + product);

    }

    public Task(int i) {
        super();
        this.i = i;
    }

}

static class PrintTask implements Runnable {
    @Override
    public void run() {
        System.out.println("Product equals to: " + product);
    }
}

public static void main(String[] args) throws NullPointerException {

    double sum = 0.0;
    NewBetterDot objectOne = new NewBetterDot(new double[] { 5.0, 4.0, 1.0 }, new double[] { 1.0, 2.0, 3.0 }, 0.0);

    int threshhold = vectorOne.length;

    ExecutorService executor = Executors.newFixedThreadPool(2);
    ;

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

        NewBetterDot.Task task = new NewBetterDot.Task(i);

        executor.submit(task);

    }

    executor.shutdown();

    NewBetterDot.PrintTask pt = new NewBetterDot.PrintTask();
    pt.run();
}

}

Blockquote

Answer 1

shutdown() 的文档说:

This method does not wait for previously submitted tasks to complete execution. Use awaitTermination to do that.

此代码最多等待 1 分钟才能停止计算:

executor.shutdown();
executor.awaitTermination(1, TimeUnit.MINUTES);