Java多线程加法计算

题意：要求开6条线程计算累加1 -> 10000000

难点：如何获取子线程的执行结果并聚合

思路一

生产者-消费者 经典模型：

1. 多个生产者负责生产(累加)作业
2. 生产者将生产结果存入共享仓库中
3. 消费者（主线程）从共享仓库中取出结果

/**
 * 多线程计算累加数
 */
public class Accumulate {
    public static void main(String[] args) {
        Storage storage = new Storage();
        // 为多个计算器对象创建线程
        Thread calThread1 = new Thread(new Calculate(1, storage), "Thread-1");
        Thread calThread2 = new Thread(new Calculate(2, storage), "Thread-2");
        Thread calThread3 = new Thread(new Calculate(3, storage), "Thread-3");
        Thread calThread4 = new Thread(new Calculate(4, storage), "Thread-4");
        Thread calThread5 = new Thread(new Calculate(5, storage), "Thread-5");
        Thread calThread6 = new Thread(new Calculate(6, storage), "Thread-6");

        calThread1.start();
        calThread2.start();
        calThread3.start();
        calThread4.start();
        calThread5.start();
        calThread6.start();

        // 打印最终结果
        storage.printTotal();
    }
}

/**
 * 计算器对象，负责计算start -> end
 */
class Calculate implements Runnable {
    private Storage storage;
    private long start;

    public Calculate(long start, Storage storage) {
        this.start = start;
        this.storage = storage;
    }

    @Override
    public void run() {
        long num = start;
        long sum = 0;
        while (num <= 10000000) {
            System.out.println(Thread.currentThread().getName() + " add num " + num);
            sum += num;
            num += 6;
        }
        // 线程计算完毕， 调用累加器进行累加
        storage.push(sum);
    }
}

/**
 * 仓库对象，负责累加
 */
class Storage {
    private long total = 0;
    private int count = 0;

    public synchronized void push(long sum) {
        total += sum;
        count++;
        notifyAll();
    }

    public synchronized void printTotal() {
        while (count < 6) {
            try {
                System.out.println(Thread.currentThread().getName() + " is wait");
                wait();
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
        System.out.println("storage result = " + total);
    }
}

思路二

线程异步返回：

1. 利用线程池并发处理多个任务
2. 使用Future+Callable获取异步执行结果
3. 待线程池中所有任务结束，计算累加结果

import java.util.ArrayList;
import java.util.List;
import java.util.Random;
import java.util.concurrent.*;

/**
 * 线程池计算累加数
 */
public class Accumulate {
    public static void main(String[] args) {
        // 建立线程池 与 动态结果数组
        ThreadPoolExecutor executor = (ThreadPoolExecutor) Executors.newFixedThreadPool(6);
        List<Future<Long>> resultList = new ArrayList<>();

        // 定义10个任务分别负责一定范围内的元素累计
        for (int i = 0; i < 10; i++) {
            CalTask calTask = new CalTask(i*100000000+1, (i+1)*100000000);
            Future<Long> result = executor.submit(calTask);
            resultList.add(result);
        }

        // 每隔50毫秒遍历一遍所有动态结果，直到所有任务执行完毕
        do {
            System.out.printf("Main: 已经完成多少个任务: %d\n",executor.getCompletedTaskCount());
            for (int i = 0; i < resultList.size(); i++) {
                Future<Long> result = resultList.get(i);
                System.out.printf("Main: Task %d is %s\n",i,result.isDone());
            }
            try {
                Thread.sleep(50);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }

        } while (executor.getCompletedTaskCount() < resultList.size());

        // 若所有任务执行完毕，则对执行结果进行累计
        long total = 0;
        for (int i = 0; i < resultList.size(); i++) {
            Future<Long> result = resultList.get(i);
            long sum = 0;
            try {
                sum = result.get();
                total += sum;
            } catch (InterruptedException e) {
                e.printStackTrace();
            } catch (ExecutionException e) {
                e.printStackTrace();
            }
        }
        System.out.printf("total is: %d\n", total);
        executor.shutdown();
    }
}

class CalTask implements Callable<Long> {
    private int startNum;
    private int endNum;

    public CalTask(int startNum, int endNum) {
        this.startNum = startNum;
        this.endNum = endNum;
    }

    @Override
    public Long call() throws Exception {
        long sum = 0;
        for (int i = startNum; i <= endNum; i++) {
            sum += i;
        }
        Thread.sleep(new Random().nextInt(100));
        System.out.printf("%s: %d\n", Thread.currentThread().getName(), sum);
        return sum;
    }
}