对包含n个元素的数组a进行洗牌

Question

我有洗牌数组的问题问题是，根据种子的随机再次是相同的结果，而不是2-交换和两个数组是相同的!我想要2个结果数组交换随机

代码为2-exchange

#include <stdio.h>
#include <cstdlib>
#include <ctime>
void swap(int *a, int *b)
{
    int temp = *a;
    *a = *b;
    *b = temp;
}
void printArray(int arr[], int n)
{
    for(int i = 0; i < n; i++)
        printf("%d ", arr[i]);
    printf("n");
}
// A function to generate a random exchange
int* randomized(int arr[], int n)
{
    // Use a different seed value so that we don't get same
    // result each time we run this program
    srand(time(NULL));
    // Start from the last element and swap one by one. We don't
    // need to run for the first element that's why i > 0
    for(int i = n - 1; i > 0; i--)
    {
        // Pick a random index from 1 to i-1
        int j = rand() % (i - 1 + 1) + 1;
        //int j = rand() % (i+1);
        // Swap arr[i] with the element at random index
        swap(&arr[i], &arr[j]);
    }
    return arr;
}
// Driver program to test above function.
int main()
{
    int *x1, *x2;
    int arr[] = {6, 1, 2, 3, 4, 5};
    int n = sizeof(arr) / sizeof(arr[0]);
    x1 = randomized(arr, n);
    x2 = randomized(x1, n);
    printArray(x1, n);
    printArray(x2, n);
    getchar();
}

Answer 1

显然，randomized在两种情况下都返回指向传入的相同数组的指针(x1 = x2)。"随机"听起来像是一个函数的名字，它应该返回数据的随机副本，而不是修改原始数据。

决定你想要哪个版本，并在此基础上，修复函数或测试程序。