Question

我正在尝试比较2个字符串之间的相似字符数，并遇到了strpbrk（）函数。但是我找不到任何将搜索字符串拆分为字符数组的方法。

char search[] = "chzi";
char arr[2][20] = {"cheang", "wai"};
float lengthSearch = strlen(search);
float count = 0;

for(int i = 0; i < 2; i++){
    int lengthArr = strlen(arr[i]);

    for(int j = 0; j < lengthSearch; j++){
      if(strpbrk(&search[j], arr[i])){
        printf("%c is the similarity between %s and %s\n", *strpbrk(&search[j], arr[i]), &search[j], arr[i]);

        count++;
        printf("count is now %.1f\n", count);
      }
    }

    float probability = ((count/lengthSearch) * (count/lengthArr)) * 100;

    printf("%s has a probability of %.2f\n\n", arr[i], probability);
    count = 0;
  }

问题在这里

i is the similarity between chzi and wai
count is now 1.0
i is the similarity between hzi and wai
count is now 2.0
i is the similarity between zi and wai
count is now 3.0
i is the similarity between i and wai
count is now 4.0

我只想比较c和wai，而不是chzi

Answer 1

我想检查search和 arr[i]

然后您可以使用strpbrk，但是与您尝试的方式有些相反。 man 3 strpbrk，声明为

char *strpbrk(const char *s, const char *accept);

locates the first occurrence in the string s
of any of the bytes in the string accept.

因此，您要做的就是简单地与strpbrk循环并找出search共有多少个arr[i]字符。使用指针可以简化操作，例如

#include <stdio.h>
#include <string.h>

int main (void) {

    char search[] = "chzi",
        arr[][20] = {"cheang", "wai"};
    size_t n = sizeof arr / sizeof *arr;

    for (size_t i = 0; i < n; i++) {
        size_t count = 0;
        char *p = arr[i];
        while ((p = strpbrk (p, search)))
            count++, p++;
        printf ("%s contains %zu occurrence of characters in %s.\n",
                arr[i], count, search);
    }
}

请注意，您只需使用指向arr[i]的指针，然后使用strpbrk来定位search中arr[i]中出现的第一个字符。然后，您将指针增加到arr[i]中的下一个字符（使用p++;），并再次进行操作，直到strpbrk返回NULL为止，指示{{1} }与arr[i]中的任何字符匹配，上面的代码就是这样做的，例如

search

如果您想避免使用逗号运算符，它将是：

        char *p = arr[i];
        while ((p = strpbrk (p, search)))
            count++, p++;

使用/输出示例

运行字符串会导致：

        char *p = arr[i];
        while ((p = strpbrk (p, search))) {
            count++;
            p++;
        }

仔细研究一下，让我知道您是否打算这样做。您将需要添加概率代码，但这留给您。

如何分割字符串并将每个字符存储到数组中

1 个答案: