Question

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

int count (const char *str, const char *search);

int count (const char *str, const char *search)
{
    int counter = 0,flag = 0;
    const char *holdstr, *holdsearch;
    do
    {
        for (holdstr = str, holdsearch = search ;
             *holdsearch != '\0' && *holdstr == *holdsearch ;
             holdsearch++,holdstr++) {}

            if ( *holdsearch == '\0' )
            {
                counter++;
                flag = 1;
            }

    }while (*(str++));

    if (flag) return counter;
    return -1;
}

int main()
{
    const char *find = "the";
    const char *arr1 = "the cat sat on the mat";
    const char *arr2 = "theqqqqq cat sat on the mat";
    const char *arr3 = "-qthe-- cat sat on theok mat";

    int val = count(arr1, find);
    printf("%d\n", val);
    val = count(arr2, find);
    printf("%d\n", val);
    val = count(arr3, find);
    printf("%d\n", val);

    return 0;
}

我试图找到一对一的对应词。但是，我还得到了一些包含指定单词的其他单词。该函数类似于strstr()。我也意识到它适用于strstr()如何修复问题？

Output                  Expected Output
-------                 ---------------
2                       2
2                       1
2                       0

Answer 1

修复想法的样本

int count (const char *str, const char *search)
{
    int counter = 0;
    const char *holdstr, *holdsearch, *wordtop, *s=str;
    do
    {
        for (wordtop = holdstr = s, holdsearch = search ;
             *holdsearch != '\0' && *holdstr == *holdsearch ;
             holdsearch++,holdstr++)
             ;
        //                   top               space                     end          space
        if (!*holdsearch && (wordtop == str || isspace(wordtop[-1])) && (!*holdstr || isspace(*holdstr)))
        {
            counter++;
        }

    }while (*s++);

    return counter;//no needs flag if return 0;
}

使用指针在字符串中出现单词

1 个答案: