C编程: How to Access a Single Character from variable Type char**

Question

我这里有一个用于测试的主要内容，它通过一个简单的句子作为测试。

    int main()
{
   char *sentence = "Hello There."; /* test sentence */
   char *word;  /* pointer to a word */

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

  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;
}

这是我的 get_word 函数，它需要一个 char** 作为传递参数，并且必须输出一个指针。

char *get_word( char **string_ptr )
{
    char *word;
    word = (char*)malloc(919 * sizeof(char)); /* Assign memory space for storying gathered characters */
    strcpy(word,"")  /***********This is the line I need help with****/ 

  return word;
}

这是一个学校项目的作业问题，所以不能要求太多，我主要关心的是我想弄清楚如何将 main 中句子的第一个字符“H”传递给其中的变量单词我的 get_word 函数。由于此函数是作业的一部分，因此我无法更改签名。

这个问题还有更多内容，比如大写和小写之间的排序，但我想如果我理解这个概念，我应该能够完成剩下的事情。

感谢您的帮助。

Answer 1

在评论中，我问:

Do you want to pass the single character 'H', or a pointer to it? If you want to pass a single character, you need to review the signature of strcpy() — it would be a mistake to pass a single character as the second argument. What do you know about dereferencing pointers?

对此的回应是:

I know a little bit about dereferencing. It basically gathers a value from a pointer? If I use the code strcpy(word, *string_ptr);, I can copy the entire value of the string to the variable word.

I need to be able to pass a single character; the question requires me to loop over the entire string until a non-letter character is reached, convert any uppercase letters to lowercase and then return a single word. then I'll have to do this again for the next set of letter characters in the string.

我注意到:

Right — strcpy(word, *string_ptr) will copy the string "Hello world" into the array word. And if you only want to copy one character from a string, consider strncpy(), but don't forget to null-terminate the result string. It looks like you're going to need to update the value in *string_ptr to allow for the characters copied to the return function — that's why it needs to be a char ** and not just a char *.

但是，考虑到要求，最好不要使用 strcpy()或strncpy() .

您可以访问H使用(*string_ptr)[0]甚至 **string_ptr ，但是带有指针的数组表示法经常输入(不止一次？)是相当笨拙的。如果 (*string_ptr) 中的指针，它们都容易崩溃。是一个空指针——或者无效；但是，下面的代码也是如此。您可以通过assert(*string_ptr != NULL);查看如果您愿意，您可以添加 <assert.h> ，无论您选择什么其他错误处理。

有很多选项，但我可能会使用以下变体:

char *src = *string_ptr;   /* Starting position */
char *dst = word;
int c;

while (*src != '\0')
{
    int c = (unsigned char)*src++;
    …break on space or other characters as required…
    …map c as required…
    *dst++ = c;
}
*dst = '\0';  /* Null terminate the string */
*string_ptr = src;
return word;

这可以避免各种问题。使用本地指针src (源字符串 — 匹配 dst ，目标字符串)避免编写 (*string_ptr)每时每刻。使用c准备将 <ctype.h> 中的值传递给函数(宏) 。类型转换(unsigned char)处理 当纯 char 时，单字节代码集中的重音字符类型是有符号的(当普通 char 类型无符号时是无害的)。循环中的增量避免超出字符串末尾。