c - 输入字符串并以随机顺序打印它们

Question

我在以随机顺序打印出我的文字时遇到了一些麻烦。我插入的 8 个单词需要以随机顺序打印出来。现在我只能以随机顺序生成其中的一些，因为如果生成两次相同的数字，它会覆盖以前的空间。我怎样才能消除这个问题？

    #include <stdio.h>
    #include <stdlib.h>
    #include <time.h>
    #include <string.h>
    #include <ctype.h>
    #define MAX 50

    int readLine(char string[]);

    int main(void)
    {
        int i;
        char str[8][MAX], temp[8][MAX];
        srand((unsigned)time(NULL));
        int r_num;

        printf("Enter 8 words:\n");

        for(i=0; i<8; ++i)
        {
            readLine(str[i]);
        }


        for (int i = 0; i < 8; i++)
        {
            r_num = rand()%8+1;
            strcpy(temp[r_num], str[i]);
        }

    printf("\nRandom order of words: \n");
    for(i=0; i<8; ++i)
    {
        printf("%s\n", temp[i]);
    }
    printf("\n");

    return 0;
}

int readLine(char string[])
{
    int ch;
    int i=0;
    while (isspace(ch = getchar()))
        ;
    while (ch != '\n' && ch != EOF)
    {
        if (i < MAX)
        {
            string[i++] = ch;
            ch = getchar();
        }
    }
    string[i] = '\0';
    return i;
}

Answer 1

打乱数据的一种方法是从尚未选择的内容中挑选。

int main(void)
{
    int i;
    char str[8][MAX];
    int order[8]; /* order list */
    srand((unsigned)time(NULL));
    int r_num;

    printf("Enter 8 words:\n");

    for(i=0; i<8; ++i)
    {
        readLine(str[i]);
    }

    /* initialize order list */
    for (i = 0; i < 8; i++)
    {
        order[i] = i;
    }
    for (i = 7; i > 0; i--) /* select from back to front */
    {
        int pos = rand() % (i + 1); /* pick one position randomly */
        /* pick selected element by swapping */
        int temp = order[pos];
        order[pos] = order[i];
        order[i] = temp;
    }

    printf("\nRandom order of words: \n");
    for(i=0; i<8; ++i)
    {
        printf("%s\n", str[order[i]]); /* print strings in the shuffled order */
    }
    printf("\n");

    return 0;
}