java - 使用 10 个线程处理一个数组

Question

我正在努力提高我的 Java 技能，但有点不确定如何处理这个多线程应用程序。基本上，该程序读取一个文本文件并找到最大的数字。我在我的搜索算法中添加了一个 for 循环来创建 10 个线程，但我不确定它是否真的创建了 10 个线程。这个想法是为了缩短执行时间，或者至少这是我认为应该发生的事情。有没有办法检查我是否做对了以及执行时间是否确实有所改善？

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;

public class ProcessDataFile {

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

        int max = Integer.MIN_VALUE;
        int i = 0;
        int[] numbers = new int[100000];
        String datafile = "dataset529.txt"; //string which contains datafile
        String line; //current line of text file

        try (BufferedReader br = new BufferedReader(new FileReader(datafile))) { //reads in the datafile
            while ((line = br.readLine()) != null) { //reads through each line
                numbers[i++] = Integer.parseInt(line); //pulls out the number of each line and puts it in numbers[]
            }
        }

        for (i = 0; i < 10000; i++){ //loop to go through each number in the file and compare it to find the largest int.
            for(int j = 0; j < 10; j++) { //creates 10 threads
                new Thread();
            }
            if (max < numbers[i]) //As max gets bigger it checks the array and keeps increasing it as it finds a larger int.
                max = numbers[i]; //Sets max equal to the final highest value found.
        }


        System.out.println("The largest number in DataSet529 is: " + max);
    }
}

Answer 1

这是一个非常基本的示例，它演示了创建和运行线程的基本概念，这些线程处理特定数组中给定范围的值。该示例做了一些假设(例如只有偶数个元素)。这个例子也有点冗长，是故意这样做的，目的是为了演示所需的基本步骤

先看看 the Concurrency Trail了解更多详情

import java.util.Random;

public class ThreadExample {

    public static void main(String[] args) {
        int[] numbers = new int[100000];
        Random rnd = new Random();
        for (int index = 0; index < numbers.length; index++) {
            numbers[index] = rnd.nextInt();
        }

        Thread[] threads = new Thread[10];
        Worker[] workers = new Worker[10];

        int range = numbers.length / 10;
        for (int index = 0; index < 10; index++) {
            int startAt = index * range;
            int endAt = startAt + range;
            workers[index] = new Worker(startAt, endAt, numbers);
        }

        for (int index = 0; index < 10; index++) {
            threads[index] = new Thread(workers[index]);
            threads[index].start();
        }

        boolean isProcessing = false;
        do {
            isProcessing = false;
            for (Thread t : threads) {
                if (t.isAlive()) {
                    isProcessing = true;
                    break;
                }
            }
        } while (isProcessing);

        for (Worker worker : workers) {
            System.out.println("Max = " + worker.getMax());
        }

    }

    public static class Worker implements Runnable {

        private int startAt;
        private int endAt;
        private int numbers[];

        private int max = Integer.MIN_VALUE;

        public Worker(int startAt, int endAt, int[] numbers) {
            this.startAt = startAt;
            this.endAt = endAt;
            this.numbers = numbers;
        }

        @Override
        public void run() {
            for (int index = startAt; index < endAt; index++) {
                max = Math.max(numbers[index], max);
            }
        }

        public int getMax() {
            return max;
        }

    }

}

一个稍微简单的解决方案将涉及 ExecutorService API，它允许您向服务提供一系列 Callable，然后返回一个 List future 的。这里的好处是，在所有 Callable 完成(或失败)之前服务不会返回，因此您不需要经常检查线程的状态

import java.util.Arrays;
import java.util.List;
import java.util.Random;
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;

public class ThreadExample {

    public static void main(String[] args) {
        int[] numbers = new int[100000];
        Random rnd = new Random();
        for (int index = 0; index < numbers.length; index++) {
            numbers[index] = rnd.nextInt();
        }

        ExecutorService executor = Executors.newFixedThreadPool(10);

        Worker[] workers = new Worker[10];

        int range = numbers.length / 10;
        for (int index = 0; index < 10; index++) {
            int startAt = index * range;
            int endAt = startAt + range;
            workers[index] = new Worker(startAt, endAt, numbers);
        }

        try {
            List<Future<Integer>> results = executor.invokeAll(Arrays.asList(workers));
            for (Future<Integer> future : results) {
                System.out.println(future.get());
            }
        } catch (InterruptedException | ExecutionException ex) {
            ex.printStackTrace();
        }

    }

    public static class Worker implements Callable<Integer> {

        private int startAt;
        private int endAt;
        private int numbers[];


        public Worker(int startAt, int endAt, int[] numbers) {
            this.startAt = startAt;
            this.endAt = endAt;
            this.numbers = numbers;
        }

        @Override
        public Integer call() throws Exception {
            int max = Integer.MIN_VALUE;
            for (int index = startAt; index < endAt; index++) {
                max = Math.max(numbers[index], max);
            }
            return max;
        }

    }

}