我想从文件中读取单词，并知道何时开始新的一行。
我知道每行可以有三个，四个或零个单词，并且单词不能超过一定长度。但是带空格的行长度没有限制，因此不可能只读取一行到字符串，进行解析并继续。我想知道我在阅读每一行中是否有三个或四个单词。
目前，我使用fscanf和一些特定于问题的内部逻辑来确定我读取的第四个单词是换行还是上一行。但是这种方法很脆弱，很容易损坏。
我想我可以逐个读取char，忽略空格并查找'\ n'。有没有更优雅的方式？
谢谢
编辑：我仅限于使用C99和标准库。
答案 0 :(得分：2)
这是一些与您的要求紧密相关的代码。有两个主要区别：
在我发布之前，它已经通过了相当严格的测试。使用make UFLAGS=-DTEST进行编译可获得较短的行片段（默认为64字节vs 4096），这也为您提供了额外的诊断输出。我在MAX_LINE_LEN而不是6上用64做了很多测试-这对于调试单词在行的多个片段上连续的问题很有帮助。
#include <assert.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
enum { MAX_WORD_CNT = 8 };
#ifdef TEST
static int debug = 1;
enum { MAX_LINE_LEN = 64 };
#else
static int debug = 0;
enum { MAX_LINE_LEN = 4096 };
#endif /* TEST */
typedef struct Word
{
size_t length;
char *word;
} Word;
typedef struct WordList
{
size_t num_words;
size_t max_words;
Word *words;
} WordList;
typedef struct LineControl
{
size_t line_length;
bool part_word;
size_t part_len;
WordList list;
} LineControl;
static void init_wordlist(WordList *list)
{
list->num_words = 0;
list->max_words = 0;
list->words = 0;
}
static void free_wordlist(WordList *list)
{
assert(list != 0);
for (size_t i = 0; i < list->num_words; i++)
free(list->words[i].word);
free(list->words);
init_wordlist(list);
}
static void extend_word(const char *extn, size_t ext_len, Word *word)
{
if (debug)
printf("old (%zu) = [%s]; extra (%zu) = [%.*s]\n", word->length, word->word,
ext_len, (int)ext_len, extn);
size_t space = word->length + ext_len + 1;
char *new_space = realloc(word->word, space);
if (new_space == 0)
{
fprintf(stderr, "failed to reallocate %zu bytes of memory\n", space);
exit(EXIT_FAILURE);
}
word->word = new_space;
memmove(word->word + word->length, extn, ext_len);
word->length += ext_len;
word->word[word->length] = '\0';
if (debug)
printf("new (%zu) = [%s]\n", word->length, word->word);
}
static void addword_wordlist(const char *word, size_t word_len, WordList *list)
{
if (list->num_words >= list->max_words)
{
assert(list->num_words == list->max_words);
size_t new_max = list->max_words * 2 + 2;
Word *new_words = realloc(list->words, new_max * sizeof(*new_words));
if (new_words == 0)
{
fprintf(stderr, "failed to allocate %zu bytes of memory\n", new_max * sizeof(*new_words));
exit(EXIT_FAILURE);
}
list->max_words = new_max;
list->words = new_words;
}
list->words[list->num_words].word = malloc(word_len + 1);
if (list->words[list->num_words].word == 0)
{
fprintf(stderr, "failed to allocate %zu bytes of memory\n", word_len + 1);
exit(EXIT_FAILURE);
}
Word *wp = &list->words[list->num_words];
wp->length = word_len;
memmove(wp->word, word, word_len);
wp->word[word_len] = '\0';
list->num_words++;
}
static void init_linectrl(LineControl *ctrl)
{
ctrl->line_length = 0;
ctrl->part_word = false;
ctrl->part_len = 0;
init_wordlist(&ctrl->list);
}
static int parse_fragment(const char *line, LineControl *ctrl)
{
char whisp[] = " \t";
size_t offset = 0;
bool got_eol = false;
/* The only newline in the string is at the end, if it is there at all */
assert(strchr(line, '\n') == strrchr(line, '\n'));
assert(strchr(line, '\n') == 0 || *(strchr(line, '\n') + 1) == '\0');
if (debug && ctrl->part_word)
{
assert(ctrl->list.num_words > 0);
printf("Dealing with partial word on entry (%zu: [%s])\n",
ctrl->part_len, ctrl->list.words[ctrl->list.num_words - 1].word);
}
size_t o_nonsp = 0;
while (line[offset] != '\0')
{
size_t n_whisp = strspn(line + offset, whisp);
size_t n_nonsp = strcspn(line + offset + n_whisp, whisp);
if (debug)
printf("offset %zu, whisp %zu, nonsp %zu\n", offset, n_whisp, n_nonsp);
got_eol = false;
ctrl->line_length += n_whisp + n_nonsp;
if (line[offset + n_whisp + n_nonsp - 1] == '\n')
{
assert(n_nonsp > 0);
got_eol = true;
n_nonsp--;
}
if (n_whisp + n_nonsp == 0)
{
o_nonsp = 0;
break;
}
if (n_whisp != 0)
{
ctrl->part_word = false;
ctrl->part_len = 0;
}
/* Add words to list if the list is not already full */
if (n_nonsp > 0)
{
const char *word = line + offset + n_whisp;
if (ctrl->part_word)
{
assert(ctrl->list.num_words > 0);
extend_word(word, n_nonsp,
&ctrl->list.words[ctrl->list.num_words - 1]);
}
else
{
addword_wordlist(word, n_nonsp, &ctrl->list);
}
}
offset += n_whisp + n_nonsp;
if (line[offset] != '\0')
{
ctrl->part_word = false;
ctrl->part_len = 0;
}
o_nonsp = n_nonsp;
if (got_eol)
break;
}
/* Partial word detection */
if (o_nonsp > 0 && !got_eol)
{
ctrl->part_word = true;
ctrl->part_len += o_nonsp;
}
else
{
ctrl->part_word = false;
ctrl->part_len = 0;
}
/* If seen newline; line complete */
/* If No newline; line incomplete */
return !got_eol;
}
int main(void)
{
char line[MAX_LINE_LEN];
size_t lineno = 0;
while (fgets(line, sizeof(line), stdin) != 0)
{
LineControl ctrl;
init_linectrl(&ctrl);
lineno++;
if (debug)
printf("Line %zu: (%zu) [[%s]]\n", lineno, strlen(line), line);
int extra = 0;
while (parse_fragment(line, &ctrl) != 0 &&
fgets(line, sizeof(line), stdin) != 0)
{
if (debug)
printf("Extra %d for line %zu: (%zu) [[%s]]\n",
++extra, lineno, strlen(line), line);
}
WordList *list = &ctrl.list;
printf("Line %zu: length %zu, words = %zu\n",
lineno, ctrl.line_length, list->num_words);
size_t num_words = list->num_words;
if (num_words > MAX_WORD_CNT)
num_words = MAX_WORD_CNT;
for (size_t i = 0; i < num_words; i++)
{
printf(" %zu: (%zu) %s\n",
i + 1, list->words[i].length, list->words[i].word);
}
putchar('\n');
free_wordlist(&ctrl.list);
}
return 0;
}
我有一个更简单的版本，没有动态内存分配，但是当一个单词被分成一行的两个片段时，它无法正常工作（因此，如果行片段的大小为6（5个字符加空字节），并且例如，一个单词的最大长度为16，那么代码在组装片段时就遇到了困难，因此，我采用了一种更简单的方法-存储所有单词，从这个问题尚不清楚最大单词的大小是多少。如果代码应反对0、3或4个单词以外的任何东西，则可以使用该数据来提出这些抱怨；如果代码应反对长度超过某个长度的单词（例如32个），则可以使用这些数据来提出这些抱怨。
一个更简单的测试文件是test-data.1：
a b
a b c d
1123xxsdfdsfsfdsfdssa 1234ddfxxyff frrrdds
1123dfdffdfdxxxxxxxxxas 1234ydfyyyzm knsaaass 1234asdafxxfrrrfrrrsaa
1123werwetrretttrretertre aaaa bbbbbb ccccc
k
apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper
其中包含各种标签，如同一数据的该版本所示，其中标签显示为\t：
a b
a b c d
\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t
1123xxsdfdsfsfdsfdssa 1234ddfxxyff frrrdds
1123dfdffdfdxxxxxxxxxas 1234ydfyyyzm knsaaass 1234asdafxxfrrrfrrrsaa
1123werwetrretttrretertre aaaa bbbbbb ccccc
k
\t\t \t \t\t\t \t \t \t\t\t\tapoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper\t\t\t \t\t\t\tapoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper \t \t \t \t\t\t\t \t \tapoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper\t\t \t\t\t\t \t \t \t \t\tapoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper\t\t\t\t\t\t \t \t \t \t \t \t \t
运行此awk脚本可分析数据：
$ awk '{ printf "%3d %d [%s]\n", length($0) + 1, NF, $0 }' test-data.1
1 0 []
5 0 [ ]
11 2 [ a b ]
81 4 [ a b c d ]
20 0 [ ]
63 3 [1123xxsdfdsfsfdsfdssa 1234ddfxxyff frrrdds]
103 4 [1123dfdffdfdxxxxxxxxxas 1234ydfyyyzm knsaaass 1234asdafxxfrrrfrrrsaa ]
82 4 [ 1123werwetrretttrretertre aaaa bbbbbb ccccc ]
2 1 [k]
494 4 [ apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper ]
$
该数据文件上程序的输出为：
Line 1: length 1, words = 0
Line 2: length 5, words = 0
Line 3: length 11, words = 2
1: (1) a
2: (1) b
Line 4: length 81, words = 4
1: (1) a
2: (1) b
3: (1) c
4: (1) d
Line 5: length 20, words = 0
Line 6: length 63, words = 3
1: (21) 1123xxsdfdsfsfdsfdssa
2: (12) 1234ddfxxyff
3: (7) frrrdds
Line 7: length 103, words = 4
1: (23) 1123dfdffdfdxxxxxxxxxas
2: (12) 1234ydfyyyzm
3: (8) knsaaass
4: (22) 1234asdafxxfrrrfrrrsaa
Line 8: length 82, words = 4
1: (25) 1123werwetrretttrretertre
2: (4) aaaa
3: (6) bbbbbb
4: (5) ccccc
Line 9: length 2, words = 1
1: (1) k
Line 10: length 494, words = 4
1: (98) apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper
2: (98) apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper
3: (98) apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper
4: (98) apoplectic-catastrophe-mongers-of-the-world-unite-for-you-have-nothing-to-lose-but-your-bad-temper
您可以看到awk脚本中的数据出现在输出中。
此代码可在我在GitHub上的SOQ（堆栈溢出问题）存储库中以scan59.c，test-data.1，test-data.2和test-data.3中的文件test-data.3 {3}}子目录。 makefile文件特别包含一行包含9955个字符和693个单词的行，以及其他不严格测试的行。
使用GCC 8.2.0和Valgrind 3.14.0.GIT，代码可以在运行macOS 10.13.6 High Sierra的Mac上正常编译和运行。 （尽管make SFLAGS='-std=c99 -pedantic'规定了C11，但是此代码中没有C11特有的东西；它与C99完全兼容。它也可以与xhr.addEventListener("error", transferFailed);
一起干净地编译。）