使用 getchar 从字符串中间删除换行符

Removing a newline from the middle of a string using getchar

现在我有一个看起来像这样的字符串:

A sentence        
with a newline.

我正在通过 控制台输入读取 中的字符串,如下所示:

ch = getchar();

while (ch != '.') {
  msg[i] = ch;
  i++;
  ch = getchar();
}

并且,读入后,我通过在函数中执行此操作并应用于 msg 字符数组来删除存在的空白:

char *remove_white_spaces(char *str) {
    int i = 0, j = 0;
    while (str[i]) {
        if (str[i] != ' ')
      str[j++] = str[i];
        i++;
    }
    str[j] = '[=12=]';
    return str;
}

我试过遍历它并在 \n 处停止,但这留下了“Asentence”的输出,因为字符串在 \n 设置为 0 时终止。

整个主线:

int main(void) {
  char msg[MAX_MSG_LEN+1];
  char ch;
  int i = 0;
  ch = getchar();

  while (ch != '.') {
    msg[i] = ch;
    i++;
    ch = getchar();
  }

  msg[i] = '.';
  msg[i + 1] = '[=13=]';
  remove_white_spaces(msg);

  printf("%s\n", msg);
  return 0;
}

您可以使用 isspace function 来测试和跳过 any/all 白色 space 字符,包括正常的 space 和换行符:

#include <ctype.h> // For definition of "isspace"

char *remove_white_spaces(char *str) {
    int i = 0, j = 0;
    while (str[i]) {
        if (!isspace((unsigned char)(str[i])))
            str[j++] = str[i];
        i++;
    }
    str[j] = '[=10=]';
    return str;
}

关于将 isspace 的参数转换为 unsigned char 的原因,请参阅 this discussion

函数删除和替换字符串中的任何字符。

  • toRemove - 要删除的字符
  • addSpace - 替换为 space
  • allowMultiple - 替换更多相邻
    时允许多个space 字符
  • allowEdges - 允许在开头和结尾添加 spaces
char *removeChars(char *str, const char *toRemove, const int addSpace, const int allowMultiple, int const allowEdges)
{
    char *rd = str, *wr = str;
    int replaced = 0;
    if(rd)
    {
        while(*rd)
        {
            if(strchr(toRemove, *rd))
            {
                if(addSpace)
                {
                    if(!replaced || allowMultiple)
                    {
                        if(wr != str || (wr == str && allowEdges))
                        {
                           *wr++ = ' ';
                           replaced = 1;
                        }
                    }
                }
            }
            else 
            {
                *wr++ = *rd;
                replaced = 0;
            }
            rd++;
        }
        if(allowEdges) *wr = 0;
        else 
        while((wr - 1) > str) 
        {
            if(*(wr - 1) == ' ') {*(wr - 1) = 0; wr--;}
            else break;
        }
    }
    return str;
}

int main(void)
{
    char str[] = "%%%%%A sentence\n\n\nwith!@#$%^a newline.%%%%%%%";

    printf("`%s`\n", removeChars(str,"\n!@#$%^", 1, 0, 0));
}

按照@MarkBenningfield 的建议,我执行了以下操作并检查了 '\n' 并将其替换为 space.

while (ch != '.') {
  msg[i] = ch;
  i++;
  ch = getchar();

  if (ch == '\n') {
    ch = ' ';
  }
}