2

我正在尝试逐行读出我的文本文件

FILE *infile;
char line[1000];
infile = fopen("file.txt","r");
while(fgets(line,1000,infile) != NULL) 
{
    //....
}
fclose(infile);

然后我需要找到一个特定的单词，例如“the”，并且需要查看它出现了多少次以及它还出现在哪些行上。

我应该可以用这个数单词

int wordTimes = 0;
if((strcmp("the", currentWord) == 0)) 
{
    printf("'%s' appears in line %d  which is: \n%s\n\n", "the", line_num, line);
    wordTimes++;
}

whereline是字符串所在的文本行，是字符串line_num所在的行号。

然后显示单词的次数使用以下代码：

if(wordTimes > 0)
{
    printf("'%s' appears %d times\n", "the", wordTimes);
}
else
{
    printf("'%s' does not appear\n", "the");
}

问题是我不确定如何将行中的每个单词与“the”进行比较，并且仍然打印出它适用的行。

我必须为此使用非常基本的 C，这意味着我不能使用strtok()or strstr()。我只能使用strlen()和strcmp()。

4

1 回答 1

4

也许您需要编写这样的strword()函数。我假设您可以使用 中的分类函数（宏）<ctype.h>，但如果也不允许这样做，也有一些解决方法。

#include <assert.h>
#include <ctype.h>
#include <stdio.h>

char *strword(char *haystack, char *needle);

char *strword(char *haystack, char *needle)
{
    char *pos = haystack;
    char old_ch = ' ';
    while (*pos != '\0')
    {
        if (!isalpha(old_ch) && *pos == *needle)
        {
            char *txt = pos + 1;
            char *str = needle + 1;
            while (*txt == *str)
            {
                if (*str == '\0')
                    return pos;     // Exact match at end of haystack
                txt++, str++;
            }
            if (*str == '\0' && !isalpha(*txt))
                return pos;
        }
        old_ch = *pos++;
    }
    return 0;
}

int main(void)
{
    /*
    ** Note that 'the' appears in the haystack as a prefix to a word,
    ** wholly contained in a word, and at the end of a word - and is not
    ** counted in any of those places. And punctuation is OK.
    */
    char haystack[] =
        "the way to blithely count the occurrences (tithe)"
        " of 'the' in their line is the";
    char needle[] = "the";

    char *curpos = haystack;
    char *word;
    int count = 0;
    while ((word = strword(curpos, needle)) != 0)
    {
        count++;
        printf("Found <%s> at [%.20s]\n", needle, word);
        curpos = word + 1;
    }

    printf("Found %d occurrences of <%s> in [%s]\n", count, needle, haystack);

    assert(strword("the", "the") != 0);
    assert(strword("th", "the") == 0);
    assert(strword("t", "t") != 0);
    assert(strword("", "t") == 0);
    assert(strword("if t fi", "t") != 0);
    assert(strword("if t fi", "") == 0);
    return 0;
}

运行时，这会产生：

Found <the> at [the way to blithely ]
Found <the> at [the occurrences (tit]
Found <the> at [the' in their line i]
Found <the> at [the]
Found 4 occurrences of <the> in [the way to blithely count the occurrences (tithe) of 'the' in their line is the]

有没有办法在没有的情况下完成该strword功能<ctype.h>？

是的。我在开篇就说了这么多。由于唯一使用的函数/宏是isalpha()，因此您可以做出一些假设（您不在使用 EBCDIC 的系统上），以便拉丁字母是连续的，您可以使用它is_alpha()来代替- 并从包含的列表中isalpha()省略<ctype.h>标题：

static inline int is_alpha(int c)
{
    return (c >= 'A' && c <= 'Z') || (c >= 'a' && c <= 'z');
}
于 2015-05-23T05:19:10.400 回答