C 编程 : Bag of words program, 需要帮助双指针

Question

我正在尝试完成一个单词包程序，但我被困在一个使用双指针的特定函数上。

char *get_word( char **string_ptr )
{
    char *word;
    int i;

    word = malloc(919); /* Assign memory space for storying gathered characters */
    i = 0;

    for(i = 0; i < strlen(*string_ptr); i++) /* sort through double pointer */
    {
        if(((**string_ptr) != '\0') && (isalpha(**string_ptr) == 0))
        {
            *string_ptr += 1;       
        }
        else if((**string_ptr) == '\0')
        {
            return NULL;
        }
    }

    *string_ptr += strlen(*string_ptr); 

    return word;
}

该函数被调用到main.c

int main()
{
  char *sentence = "#The quick brown fox jumped over 23&%^24 the lazy dogs."; /* test sentence */
  char *word;  /* pointer to a word */

  printf( "sentence = \"%s\"\n", sentence );  /* show the sentence */

  while (*sentence)  /* while sentence doesn't point to the '\0' character at the end of the string */
  {
    word = get_word( &sentence );  /* this will allocate memory for a word */
    printf( "word = \"%s\"; sentence = \"%s\"\n", word, sentence );  /* print out to see what's happening */

    free(word);  /* free the memory that was allocated in get_word */
  }

  return 0;
}

目前我能够打印出该函数:

sentence = "#The quick brown fox jumped over 23&%^24 the lazy dogs."
word = ""; sentence = ""

第一次通话后的句子是:“敏捷的棕色狐狸跳过了 23&%^24 只懒狗。”所以我相信我正确地删除了符号，但我也删除了需要保留的空格。

我正在寻找的最终产品是:

sentence = "#The quick brown fox jumped over 23&%^24 the lazy dogs."
word = "The"; sentence = " quick brown fox jumped over 23&%^24 the lazy dogs."
word = "quick"; sentence = " brown fox jumped over 23&%^24 the lazy dogs."
word = "brown"; sentence = " fox jumped over 23&%^24 the lazy dogs."
word = "fox"; sentence = " jumped over 23&%^24 the lazy dogs."
word = "jumped"; sentence = " over 23&%^24 the lazy dogs."
word = "over"; sentence = " 23&%^24 the lazy dogs."
word = "the"; sentence = " lazy dogs."
word = "lazy"; sentence = " dogs."
word = "dogs"; sentence = "."

word = "(null)"; sentence = ""

很抱歉这篇文章很长。我将不胜感激我能得到的任何帮助。

Answer 1

当前编写的函数存在一些问题。

这是循环字符串的糟糕方法。它在每次迭代结束时重新计算剩余字符串的长度。

for(i = 0; i < strlen(*string_ptr); i++) /* sort through double pointer */

更好的方法是将 strlen() 的结果存储在变量中，或者更好的是，利用字符串总是以 结尾的知识NUL 像这样

for(;**string_ptr!='\0';(*string_ptr)++)

在函数开头为word分配内存是一个坏主意。首先，你不知道你需要多少内存，虽然 919 字节可能就足够了，但你不需要猜测，因为一旦你有了一个单词，你应该知道它有多长。此外，在某些情况下，您的代码会返回 NULL，这意味着分配的内存将丢失。

word = malloc(919); /* Assign memory space for storying gathered characters */

这一行会跳过 string_ptr 到字符串的最末尾，因此您永远不会得到多个结果。

*string_ptr += strlen(*string_ptr); 

完成您想要做的事情的方法是，首先跟踪单词的开头位置，然后找到它的结尾。然后您将确切地知道需要分配多少内存(记住存储 NUL 的额外字节)。这是一个可能的示例，说明如何执行此操作

char *get_word(char **string_ptr)
   {
   char *start;
   char *word=NULL;
   unsigned int word_len;

   // Return NULL if there is no string to process
   if((string_ptr==NULL)||(*string_ptr==NULL)||(**string_ptr=='\0'))
      {
      return NULL;
      }

   // Skip over any non-alpha characters
   while((**string_ptr!='\0')&&(!isalpha(**string_ptr)))
      {
      (*string_ptr)++;
      }

   if(**string_ptr=='\0')
      {
      return NULL;
      }

   // Make note of where this word starts
   start=*string_ptr;

   while(isalpha(**string_ptr))
      {
      (*string_ptr)++;
      }


   word_len=(*string_ptr)-start;
   if(word_len)
      {
      word=malloc(word_len+1);
      strncpy(word,start,word_len);
      word[word_len]='\0'; // Always NUL terminate strings when using strncpy()
      }
   return word;
   }

由于它在某些情况下返回 NULL，因此您还应该确保调用它的代码检查结果是否为 NULL 并采取相应措施，而不是假设它始终为 NULL得到一根绳子。调用 free(NULL) 从来都不是一个好主意。