删除字符串后的空格（指针）

发布于 2024-11-03 11:18:33 字数 599 浏览 5 评论 0原文

我有一个包含姓名、姓氏、地址的结构列表...所有内容都以单独的 CHAR 字符串输入。问题是，每当我粘贴这些字符串之一时，我也会得到一大堆空白。例如，对于名称和姓氏字符串，我得到：布兰妮……………………约翰逊…… ...................................... （点代表每个字符串末尾的空格字符）

这是我的代码：

void table_names (void)
{
ADDRESS *tmp_ptr;
printf ("All names in address book:\n");
tmp_ptr= hol; 
printf(" ______________________________\n");
printf("|  FIRST NAMES   | LAST NAMES  |\n");
printf("|________________|_____________|\n");
while(tmp_ptr!=NULL)
   {   
    printf ("%s%s",tmp_ptr->name,tmp_ptr->lname);
    tmp_ptr= tmp_ptr->next;
   }
}

有关如何消除空格的任何帮助吗？

原文

I have a struct list of names, last names, addresses... All inputted in separate CHAR strings.
The problem is whenever I paste one of these strings I also get a whole lot of whitespace afterwards.
For example, for Name and Last name strings, I get:
Britney............................................Johnson...................................
(dots represent space characters at the end of each string)

Here is my code:

void table_names (void)
{
ADDRESS *tmp_ptr;
printf ("All names in address book:\n");
tmp_ptr= hol; 
printf(" ______________________________\n");
printf("|  FIRST NAMES   | LAST NAMES  |\n");
printf("|________________|_____________|\n");
while(tmp_ptr!=NULL)
   {   
    printf ("%s%s",tmp_ptr->name,tmp_ptr->lname);
    tmp_ptr= tmp_ptr->next;
   }
}

Any help on how to get rid of the whitespace?

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苍风燃霜 2024-11-10 11:18:33

我认为粘贴是指显示。

有了这个假设，我还假设每当 tmp_ptr->name 和 tmp_ptr->lname 形成时，它们的整个缓冲区都充满空格，并且只有在末尾才有 NUL 终止符（'\0'）。

无论在何处创建它们，您都需要通过在所有空白处放置 '\0' 来消除所有多余的空白。也许一个聪明的方法是向后工作以允许名称中存在空格。

int i;

for (i = BUFFER_LENGTH - 1; i > -1; --i)
{
    if (value[i] != ' ')
    {
        if (i + 1 != BUFFER_LENGTH)
        {
            value[i + 1] = '\0';
        }
        break;
    }
}

这也可以使用原始指针来完成，并且假设这是通过类似于以下内容的函数传入的：

void rtrim(char *value, const int BUFFER_LENGTH);

I assume by paste that you mean display.

With that assumption, I also assume that whenever tmp_ptr->name and tmp_ptr->lname are formed, their entire buffer is filled with spaces and only at the end is there a NUL terminator ('\0').

Wherever those are created, you need to chop off all of the extra whitespace by putting a '\0' at the first sight of all blanks. Probably a smart approach would be to work backwards to allow for spaces in names.

int i;

for (i = BUFFER_LENGTH - 1; i > -1; --i)
{
    if (value[i] != ' ')
    {
        if (i + 1 != BUFFER_LENGTH)
        {
            value[i + 1] = '\0';
        }
        break;
    }
}

This could be done with the raw pointer as well, and it assumes that this is passed in through a function similar to:

void rtrim(char *value, const int BUFFER_LENGTH);

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只有一腔孤勇 2024-11-10 11:18:33

要打印时不带尾随空格，请使用格式 .*s 和 strcspn()。

.*s 采用 int 值作为要打印的 char 的最大数量。
strcspn(s," ") 计算不包含 ' ' 的初始前缀的长度。

不需要修改s。如果姓氏中出现空格，则此方法不起作用。

int main(void) {
  const char *s = "Johnson        ";
  printf("'%.*s'\n", (int) strcspn(s, " "), s);
  return 0;
}

“约翰逊”

否则的话，工作量就有点大了。搜索最后一个 ' '（如果有）。

int main(void) {
  const char *s = "Johnson Smith  ";
  size_t len = strlen(s);
  while (len > 0 && s[len-1] == ' ') {
    len--;
  }
  printf("'%.*s'\n", (int) len, s);
  return 0;
}

“约翰逊·史密斯”

To print without trailing spaces, use format .*s and strcspn().

.*s takes an int value as the maximum number of char to print.
strcspn(s," ") computes the length of the initial prefix not containing ' '.

Modification of s is not needed. This method does not work should a space occur within the last name.

int main(void) {
  const char *s = "Johnson        ";
  printf("'%.*s'\n", (int) strcspn(s, " "), s);
  return 0;
}

'Johnson'

Otherwise it is a bit more work. Search for the last ' ', if any.

int main(void) {
  const char *s = "Johnson Smith  ";
  size_t len = strlen(s);
  while (len > 0 && s[len-1] == ' ') {
    len--;
  }
  printf("'%.*s'\n", (int) len, s);
  return 0;
}