c - 尝试用空格和引号标记字符串

Question

所以我已经为此做了一些工作，但遇到了一些奇怪的问题。最终目标是通过空格和引号拆分输入字符串(即这个"is"非常“非常复杂”的示例转到{this，是一个非常非常复杂的示例)。现在，除了第一个字符串外，它似乎已将所有内容正确拆分。

它在这里(buff 正在使用来自 getline 的值传入):

char **tokens = (char **)malloc(sizeof(char));
char *temp;
int count = 0;
int prev = 0;
// Get tokens
for (int i = 0; i <= strlen(command) && running; i++) {
    if (i > prev && strncmp((buff + i), " ", 1) == 0) {
        temp = (char **)realloc(tokens, (sizeof(char)) * WORD_SIZE * (++count));
        if (temp == NULL) {
            fprintf(stderr, "Error in parsing: ran out of memory\n");
            running = false;
            free(tokens);
        }
        else {
            tokens = temp;
            *(temp) = (buff + i);
            strncpy(*(temp), "\0", 1);
            temp = tokens + WORD_SIZE * (count - 1);
            *(temp) = buff+prev;
            prev = i+1;
        }
    }
    else if (strncmp((buff + i), "\"", 1) == 0) {
        *(temp) = (buff + i);
            strncpy(*(temp), "\0", 1);
        i++;
        prev = i;
        for (; strncmp((buff + i), "\"", 1) != 0; i++) { }
        temp = (char **)realloc(tokens, (sizeof(char)) * WORD_SIZE * (++count));
        if (temp == NULL) {
            fprintf(stderr, "Error in parsing: ran out of memory\n");
            running = false;
            free(tokens);
        }
        else {
            tokens = temp;
            *(temp) = (buff + i);
            strncpy(*(temp), "\0", 1);
            temp = tokens + WORD_SIZE * (count - 1);
            *(temp) = buff+prev;
            prev = i+1;
        }
    }
    else if (strncmp((buff + i), "\0", 1) == 0) {
        temp = (char **)realloc(tokens, (sizeof(char)) * WORD_SIZE * (++count));
        if (temp == NULL) {
            fprintf(stderr, "Error in parsing: ran out of memory\n");
            running = false;
            free(tokens);
        }
        else {
            tokens = temp;
            temp = tokens + WORD_SIZE * (count - 1);
            *(temp) = buff+prev;
            prev = i+1;
        }
    }
}
for (int i = 0; i < count; i++)
     printf("\t%i: %s\n", i, *tokens + sizeof(char) * WORD_SIZE * i);

现在，如果我输入“这是一个测试”(没有引号)，我会得到:
0:
1: 是
2: 一个
3: 测试

引用有点乱，因为“这是一个”非常“非常复杂”的测试”我得到:
0:
1: 是一个
2:
3:非常复杂
4: 测试

Answer 1

您说替代代码可以。如果您使用确定性有限自动机模型来考虑简单的字符串解析算法，它们几乎总是更容易生成更易于维护的代码。网络上有许多 DFA 的免费引用资料。

这是一个可以解决您的问题的 DFA。

[any] 的意思是“所有其他”。换句话说，如果没有其他转换匹配，则采用这个。它成为 C switch 中的 default 情况。 [eos] 的含义是“字符串结尾”或空字符。

请注意，DFA 可让您系统地处理所有情况，例如在单词中间出现的引号。在这里，我将其视为当前单词的结尾和新引用单词的开头。如果规范发生变化，DFA 也很容易更改，并且无需深思熟虑即可将更改转化为代码。

剩下的就是添加“操作代码”以捕获 token 开始并覆盖明显位置的空终止符。在 C 中，我们有:

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

char **tokenize(char *str, int *n_tokens_rtn)
{
  // State of the DFA.
  enum { Error = -1, Start, InQuoted, InWord } state = Start;

  // String pointer and current character
  int cp = 0;

#define CURRENT_CHAR (str[cp])
#define ADVANCE_TO_NEXT_CHAR do { ++cp; } while (0)
#define MARK_END_OF_TOKEN do { str[cp] = '\0'; } while (0)

  // Token pointer and buffer. Allocate biggest possible and shrink at end.
  int tp = 0;
  char **tokens = safe_malloc((1 + strlen(str) / 2) * sizeof *tokens);

#define SAVE_TOKEN do { tokens[tp++] = &str[cp]; } while (0)

  // Each iteration is one DFA transition.
  for (;;) {
    switch (state) {
    case Start:
      switch (CURRENT_CHAR) {
      case '\0':
        goto done_scanning;

      case ' ': case '\t': case '\n':
        ADVANCE_TO_NEXT_CHAR;
        break;

      case '"':
        state = InQuoted;
        ADVANCE_TO_NEXT_CHAR;
        SAVE_TOKEN;
        break;

      default:
        state = InWord;
        SAVE_TOKEN;
        ADVANCE_TO_NEXT_CHAR;
        break;
      }
      break;

    case InQuoted:
      switch (CURRENT_CHAR) {
      case '\0':
        state = Error; // Missing close quote.
        break;

      case '"':
        state = Start;
        MARK_END_OF_TOKEN;
        ADVANCE_TO_NEXT_CHAR;
        break;

      default:
        ADVANCE_TO_NEXT_CHAR;
        break;
      }
      break;

    case InWord:
      switch (CURRENT_CHAR) {

      case '\0':
        goto done_scanning;

      case ' ': case '\t': case '\n':
        state = Start;
        MARK_END_OF_TOKEN;
        ADVANCE_TO_NEXT_CHAR;
        break;

      case '"': // Word ended in quote, not space.
        state = InQuoted;
        MARK_END_OF_TOKEN;
        ADVANCE_TO_NEXT_CHAR;
        SAVE_TOKEN;
        break;

      default:
        ADVANCE_TO_NEXT_CHAR;
        break;
      }
      break;

    case Error:
      fprintf(stderr, "Syntax error.\n");
      goto done_scanning;
    }
  }

 done_scanning:
  // Return number of tokens if caller is interested.
  if (n_tokens_rtn) *n_tokens_rtn = tp;

  // Append a null terminator for good measure.
  tokens[tp++] = NULL;

  // Trim the returned value to the right size.
  return realloc(tokens, tp * sizeof *tokens);
}

int main(void)
{
  char str[] = "this \"is a\" very \"very complex\" example";
  char **tokens = tokenize(str, NULL);
  for (int i = 0; tokens[i]; i++)
    printf("%s\n", tokens[i]);
  return 0;
}