捕获空格和引号内的单词？

huangapple
go评论91阅读模式
英文:

Capturing words within spaces and quotation marks?

问题

The code you provided aims to split a string into words separated by spaces. To achieve the desired output, you need to handle quotation marks correctly. Here's the modified code with the necessary changes:

  1. #include <stdlib.h>
  2. #include <stdio.h>
  3. #include <string.h>
  4. #include <ctype.h>
  6. int main () {
  7.    char command[BUFSIZ];
  8.    char *token;
  9.    fgets(command, BUFSIZ, stdin);
  10.    
  11.    // Initialize a variable to track whether we are inside quotation marks
  12.    int insideQuotes = 0;
  14.    // Split the string by space and quotation marks
  15.    token = strtok(command, " ");
  17.    while( token != NULL ) {
  18.       // Check if the token starts with a quotation mark
  19.       if (token[0] == '"') {
  20.          if (!insideQuotes) {
  21.             // If not inside quotes, remove the starting quotation mark
  22.             printf("%s\n", &token[1]);
  23.             insideQuotes = 1;
  24.          } else {
  25.             // If inside quotes, remove the ending quotation mark
  26.             printf("%s ", &token[0]);
  27.             insideQuotes = 0;
  28.          }
  29.       } else {
  30.          // If not within quotes, print the token
  31.          printf("%s\n", token);
  32.       }
  33.     
  34.       token = strtok(NULL, " ");
  35.    }
  36.    
  37.    return 0;
  38. }

With this code, you should achieve the desired output:

  1. The
  2. Brown
  3. Fox Jumps Over
  4. The Lazy
  5. Dog
英文:

The idea, explicit in the title, is to capture words within spaces and quotation marks here's an example of the input we are dealing with:

  1. Input:
  3. The Brown &quot;Fox Jumps Over&quot; &quot;The Lazy&quot; Dog

Currently my code can capture words within spaces, as many of you know, a basic strtok() is enough. Here's my code so far:

  1. #include &lt;stdlib.h&gt;
  2. #include &lt;stdio.h&gt;
  3. #include &lt;string.h&gt;
  4. #include &lt;ctype.h&gt;
  6. int main () {
  7.    char command[BUFSIZ];
  8.    char *token;
  9.    fgets(command,BUFSIZ,stdin);
  10.    
  11.    token = strtok(command, &quot; &quot;);
  13.    while( token != NULL ) {
  14.       printf( &quot; %s\n&quot;, token );
  15.     
  16.       token = strtok(NULL, &quot; &quot;);
  17.    }
  18.    
  19.    return 0;
  20. }

And as expected, my code prints the following:

  1. Current Output:
  3. The
  4. Brown
  5. &quot;Fox
  6. Jumps
  7. Over&quot;
  8. &quot;The
  9. Lazy&quot;
  10. Dog

But the whole idea and problem is to get the following output:

  1. The
  2. Brown
  3. Fox Jumps Over
  4. The Lazy
  5. Dog

All the help is welcome and I thank you in advance.
(PS: The included libraries are the only ones allowed.)

答案1

得分: 2

这个程序适用于您的输入，它使用一个小的状态机来防止在引号之间分割。对于比单个分割标记更复杂的情况，strtok 显然有一些限制。

  1. #include <stdio.h>
  2. #include <stdlib.h>
  4. void prn(char* str) {
  5.     printf("<< %s >>\n", str);
  6. }
  8. int main(){
  9.     char command[BUFSIZ];
  10.     char state = 0;
  11.     char *start = NULL;
  12.     char *cur = NULL;
  13.     
  14.     fgets(command, BUFSIZ, stdin);
  15.     start = cur = command;
  16.     
  17.     while (*cur) {
  18.         if (state == 0 && *cur == ' ') {
  19.             /* space outside quotes */
  20.             *cur = 0;
  21.             prn(start);
  22.             start = cur+1;
  23.             cur++;
  24.         } else if (*cur == '"') {
  25.             /* quote found */
  26.             *cur = 0;
  27.             if (state) {
  28.                 /* end quote -- print */
  29.                 prn(start);
  30.                 
  31.                 /* skip past spaces */
  32.                 cur++;
  33.                 while (*cur == ' ')
  34.                     cur++;
  35.             } else {
  36.                 /* in quote, move cursor forward */
  37.                 cur++;
  38.             }
  39.             /* flip state and reset start */
  40.             state ^= 1;
  41.             start = cur;
  42.         } else {
  43.             cur++;
  44.         }
  45.         if (cur - command >= BUFSIZ) {
  46.             fprintf(stderr, "Buffer overrun\n");
  47.             return -1;
  48.         }
  49.     }
  50.     /* print the last string */
  51.     prn(start);
  52.     
  53.     return 0;
  54. }

输出：

  1. ➜ echo -n 'The Brown "Fox Jumps Over" "The Lazy" Dog' | ./a.out
  2. << The >>
  3. << Brown >>
  4. << Fox Jumps Over >>
  5. << The Lazy >>
  6. << Dog >>

[编辑：根据反馈进行了整理，只打印限定以捕获任何潜在的空格]

英文:

This program works for your input, it employs a tiny state machine that prevents splitting between quotes. strtok is pretty limited for cases more complicated than a single split token IMO:

  1. #include &lt;stdio.h&gt;
  2. #include &lt;stdlib.h&gt;
  4. void prn(char* str) {
  5.     printf(&quot;&lt;&lt; %s &gt;&gt;\n&quot;, str);
  6. }
  8. int main(){
  9.     char command[BUFSIZ];
  10.     char state = 0;
  11.     char *start = NULL;
  12.     char *cur = NULL;
  13.     
  14.     fgets(command, BUFSIZ, stdin);
  15.     start = cur = command;
  16.     
  17.     while (*cur) {
  18.         if (state == 0 &amp;&amp; *cur == &#39; &#39;) {
  19.             /* space outside quotes */
  20.             *cur = 0;
  21.             prn(start);
  22.             start = cur+1;
  23.             cur++;
  24.         } else if (*cur == &#39;&quot;&#39;) {
  25.             /* quote found */
  26.             *cur = 0;
  27.             if (state) {
  28.                 /* end quote -- print */
  29.                 prn(start);
  30.                 
  31.                 /* skip past spaces */
  32.                 cur++;
  33.                 while (*cur == &#39; &#39;)
  34.                     cur++;
  35.             } else {
  36.                 /* in quote, move cursor forward */
  37.                 cur++;
  38.             }
  39.             /* flip state and reset start */
  40.             state ^= 1;
  41.             start = cur;
  42.         } else {
  43.             cur++;
  44.         }
  45.         if (cur - command &gt;= BUFSIZ) {
  46.             fprintf(stderr, &quot;Buffer overrun\n&quot;);
  47.             return -1;
  48.         }
  49.     }
  50.     /* print the last string */
  51.     prn(start);
  52.     
  53.     return 0;
  54. }

The output:

  1. ➜ echo -n &#39;The Brown &quot;Fox Jumps Over&quot; &quot;The Lazy&quot; Dog&#39; |./a.out
  2. &lt;&lt; The &gt;&gt;
  3. &lt;&lt; Brown &gt;&gt;
  4. &lt;&lt; Fox Jumps Over &gt;&gt;
  5. &lt;&lt; The Lazy &gt;&gt;
  6. &lt;&lt; Dog &gt;&gt;

[edit: tidied following feedback, printing delimited to catch any sneaky spaces creeping through]

答案2

得分: 0

这也使用了状态机。与其他回答不同，它通过具有3个状态，即empty，word和quote，来忽略单词内部的引号。它还具有错误检测功能，处理制表符和多个空格，但更为复杂。

  1. #include <stdlib.h>
  2. #include <stdio.h>
  3. #include <string.h>
  4. #include <ctype.h>
  5. #include <errno.h>
  6. #include <assert.h>
  8. /*!conditions:re2c*/
  10. static int scan(char **text, const char **token) {
  11.     char *YYCURSOR = *text, *yyt1, *open = 0, *close = 0;
  12.     enum YYCONDTYPE condition = yycempty;
  13.     assert(text && token);
  14.     /*!re2c /**/
  15.     re2c:define:YYCTYPE = char;
  16.     re2c:yyfill:enable = 0;
  17.     re2c:define:YYGETCONDITION = "condition";
  18.     re2c:define:YYSETCONDITION = "condition = @@;";
  19.     re2c:define:YYGETCONDITION:naked = 1;
  20.     re2c:define:YYSETCONDITION:naked = 1;
  22.     space = [ \t\v\n];
  23.     nul = "\x00";
  24.     quote = "\"";
  25.     */
  26.     for( ; ; ) { /*!re2c /**/
  27.         <empty> nul { return *token = 0, 0; }
  28.         <empty> space+ { continue; } /* Leading space. */
  29.         <empty> quote @open => quote
  30.         <empty> @open * => word
  32.         <word> nul { return *token = open, 1; }
  33.         <word> @close space
  34.             { *close = '\0'; *text = close + 1; *token = open; return 1; }
  35.         <word> * { continue; }
  37.         <quote> @close quote
  38.             { *close = '\0'; *text = close + 1; *token = open; return 1; }
  39.         <quote> nul { return errno = EILSEQ, 0; }
  40.         <quote> * { continue; }
  41.         */
  42.     }
  43. }
  45. int main(void) {
  46.     int success = EXIT_SUCCESS;
  47.     char command[BUFSIZ], *input;
  48.     const char *token;
  49.     errno = 0;
  50.     if(!(input = fgets(command,BUFSIZ,stdin))) goto catch;
  51.     while(scan(&input, &token)) printf( "%s\n", token );
  52.     if(errno) goto catch;
  53.     goto finally;
  54. catch:
  55.     success = EXIT_FAILURE;
  56.     if(errno) perror("capture");
  57. finally:
  58.     return success;
  59. }

使用re2c作为工具，运行以下命令来生成代码：re2c -W -T -c -o main.re.c main.c。

英文:

This also uses a state machine. Unlike the other answer, it ignores quotes inside words by having 3 states, empty, word, and quote. It also has error detection, and handles tabs and multiple spaces, but is more complex.

  1. #include &lt;stdlib.h&gt;
  2. #include &lt;stdio.h&gt;
  3. #include &lt;string.h&gt;
  4. #include &lt;ctype.h&gt;
  5. #include &lt;errno.h&gt;
  6. #include &lt;assert.h&gt;
  7. /*!conditions:re2c*/
  8. static int scan(char **text, const char **token) {
  9. char *YYCURSOR = *text, *yyt1, *open = 0, *close = 0;
  10. enum YYCONDTYPE condition = yycempty;
  11. assert(text &amp;&amp; token);
  12. /*!re2c /**/
  13. re2c:define:YYCTYPE = char;
  14. re2c:yyfill:enable = 0;
  15. re2c:define:YYGETCONDITION = &quot;condition&quot;;
  16. re2c:define:YYSETCONDITION = &quot;condition = @@;&quot;;
  17. re2c:define:YYGETCONDITION:naked = 1;
  18. re2c:define:YYSETCONDITION:naked = 1;
  19. space = [ \t\v\n];
  20. nul = &quot;\x00&quot;;
  21. quote = &quot;\&quot;&quot;;
  22. */
  23. for( ; ; ) { /*!re2c /**/
  24. &lt;empty&gt; nul { return *token = 0, 0; }
  25. &lt;empty&gt; space+ { continue; } /* Leading space. */
  26. &lt;empty&gt; quote @open :=&gt; quote
  27. &lt;empty&gt; @open * :=&gt; word
  28. &lt;word&gt; nul { return *token = open, 1; }
  29. &lt;word&gt; @close space
  30. { *close = &#39;\0&#39;; *text = close + 1; *token = open; return 1; }
  31. &lt;word&gt; * { continue; }
  32. &lt;quote&gt; @close quote
  33. { *close = &#39;\0&#39;; *text = close + 1; *token = open; return 1; }
  34. &lt;quote&gt; nul { return errno = EILSEQ, 0; }
  35. &lt;quote&gt; * { continue; }
  36. */
  37. }
  38. }
  39. int main(void) {
  40. int success = EXIT_SUCCESS;
  41. char command[BUFSIZ], *input;
  42. const char *token;
  43. errno = 0;
  44. if(!(input = fgets(command,BUFSIZ,stdin))) goto catch;
  45. while(scan(&amp;input, &amp;token)) printf( &quot;%s\n&quot;, token );
  46. if(errno) goto catch;
  47. goto finally;
  48. catch:
  49. success = EXIT_FAILURE;
  50. if(errno) perror(&quot;capture&quot;);
  51. finally:
  52. return success;
  53. }

Uses re2c as re2c -W -T -c -o main.re.c main.c to generate the code.

huangapple
  • 本文由 发表于 2023年3月21日 01:53:41
  • 转载请务必保留本文链接：https://go.coder-hub.com/75793687-2.html
  • c
  • string
  • token
  • tokenize
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