Java线程的多线程问题，使用Runnable或Thread

Question

我正在尝试使用合并排序实现多线程。我让它在将一个数组切成两半的位置生成新线程
数组的排序取决于： [阵列大小]与[创建新线程的次数] 例如：如果我让它在一个大小为70的数组上只创建两个线程，那么数组将被排序，但是如果我让它创建6个线程，那么它将返回未排序。我认为可能的一点是，线程没有同步，但我使用了threadName。加入

下面是一些代码：合并。爪哇

import java.util.Random;

public class merge implements Runnable {
    int[] list;
    int length;
    int countdown;

    public merge(int size, int[] newList, int numberOfThreadReps, int firstMerge) {
        length = size;
        countdown = numberOfThreadReps;
        list = newList;
        if (firstMerge == 1)
            threadMerge(0, length - 1);
    }

    public void run() {
        threadMerge(0, length - 1);
    }

    public void printList(int[] list, int size) {
        for (int i = 0; i < size; i++) {
            System.out.println(list[i]);
        }
    }

    public void regMerge(int low, int high) {
        if (low < high) {
            int middle = (low + high) / 2;
            regMerge(low, middle);
            regMerge(middle + 1, high);
            mergeJoin(low, middle, high);
        }
    }

    public void mergeJoin(int low, int middle, int high) {
        int[] helper = new int[length];

        for (int i = low; i <= high; i++) {
            helper[i] = list[i];
        }

        int i = low;
        int j = middle + 1;
        int k = low;

        while (i <= middle && j <= high) {
            if (helper[i] <= helper[j]) {
                list[k] = helper[i];
                i++;
            } else {
                list[k] = helper[j];
                j++;
            }
            k++;
        }
        while (i <= middle) {
            list[k] = helper[i];
            k++;
            i++;
        }
        helper = null;
    }

    public void threadMerge(int low, int high) {
        if (countdown > 0) {
            if (low < high) {
                countdown--;
                int middle = (low + high) / 2;
                int[] first = new int[length / 2];
                int[] last = new int[length / 2 + ((length % 2 == 1) ? 1 : 0)];
                for (int i = 0; i < length / 2; i++)
                    first[i] = list[i];
                for (int i = 0; i < length / 2 + ((length % 2 == 1) ? 1 : 0); i++)
                    last[i] = list[i + length / 2];

                merge thread1 = new merge(length / 2, first, countdown, 0);// 0
                                                                            // is
                                                                            // so
                                                                            // that
                                                                            // it
                                                                            // doesn't
                                                                            // call
                                                                            // threadMerge
                                                                            // twice
                merge thread2 = new merge(length / 2
                        + ((length % 2 == 1) ? 1 : 0), last, countdown, 0);

                Thread merge1 = new Thread(thread1);
                Thread merge2 = new Thread(thread2);
                merge1.start();
                merge2.start();

                try {
                    merge1.join();
                    merge2.join();
                } catch (InterruptedException ex) {
                    System.out.println("ERROR");
                }

                for (int i = 0; i < length / 2; i++)
                    list[i] = thread1.list[i];
                for (int i = 0; i < length / 2 + ((length % 2 == 1) ? 1 : 0); i++)
                    list[i + length / 2] = thread2.list[i];

                mergeJoin(low, middle, high);
            } else {
                System.out.println("elsd)");
            }
        } else {
            regMerge(low, high);
        }
    }
}

项目4。爪哇

import java.util.Random;

public class proj4 {
    public static void main(String[] args) {
        int size = 70000;
        int threadRepeat = 6;
        int[] list = new int[size];
        list = fillList(list, size);
        list = perm(list, size);
        merge mergy = new merge(size, list, threadRepeat, 1);
        // mergy.printList(mergy.list,mergy.length);
        for (int i = 0; i < mergy.length; i++) {
            if (mergy.list[i] != i) {
                System.out.println("error)");
            }
        }
    }

    public static int[] fillList(int[] list, int size) {
        for (int i = 0; i < size; i++)
            list[i] = i;
        return list;
    }

    public static int[] perm(int[] list, int size) {
        Random generator = new Random();
        int rand = generator.nextInt(size);
        int temp;
        for (int i = 0; i < size; i++) {
            rand = generator.nextInt(size);
            temp = list[i];
            list[i] = list[rand];
            list[rand] = temp;
        }
        return list;
    }

}

所以TL；DR my array没有根据数组的大小和使用线程拆分数组的次数进行多线程合并排序。。。为什么

Answer 1

哇。这是一个关于受虐狂的有趣练习。我相信你已经离开了，但我想为了后代

代码中的错误在mergeJoin中，带有middle参数。这对regMerge来说没问题，但在threadMerge中，传递的中间是(low + high) / 2，而不是(length / 2) - 1。因为在threadMergelow总是0，high是length - 1，而first数组的大小是(length / 2)。这意味着，对于条目数为奇数的列表，它通常会因随机化而失败

还有一些风格问题使这个程序变得更加复杂和容易出错：

• 当Java有一个方便的list.length调用时，代码会传递一定大小的数组，这将更简单、更安全
• 代码在许多地方重复计算（参见length/2）
• 代码应该能够在数组内部排序，而无需创建子数组
• 类应该以大写字母（Merge而不是merge）开头
• firstMerge应该是布尔值
• 代码将Thread变量merge1和merge变量thread1命名。大口喝
• 调用threadMerge(0,length -1)的merge构造函数很奇怪。我会把那个电话放在new回proj4的电话后面。然后firstMerge可以被删除 我会考虑切换到{{CD9}}是最大值而不是最大值。我们更倾向于像for (int i = 0; i < 10; i++)而不是i <= 9那样思考。然后代码可以让j从low到< middle和k从middle到< high。更好的对称性

祝你好运