c - 在没有最大缓冲区长度的情况下从 C 中的标准输入读取

Question

以下代码设置从标准输入读取的最大行大小。我宁愿不对特定的行长度进行硬编码，而是可以灵活地处理任何缓冲区长度。允许处理任何规模的好策略是什么？

如果这些策略复杂得多，有没有办法至少保证 getline 不会溢出？谢谢。

 #include<stdlib.h>
 #include<stdio.h>
 #include<string.h>

 #define P 20

 int main()
 {
   size_t size = 1920;
   char *line;
   // record row; /* structure to store fields */
   char tokens[P][41];
   int p;
   char delims[] = ",";     /* ", |" */
   char *result = NULL;

   line = ( char * ) malloc( size + 1 );

   while( getline(&line, &size, stdin) != -1 )
   {
      /* chomp */
      line[strlen(line)-1] = '\0';

      /* load char array */
      result = strtok( line , delims );
      p = 0;
      while( result != NULL && ( p < P ) ) 
      {
         strcpy( tokens[p++] , result );
         result = strtok( NULL, delims );
      }

      if (p != P)
      {
         fprintf(stderr,"Wrong number of input fields.\nFormat: ID,x1, ... ,x%d\n",P);
     exit(-1);
      }

      /* load record ( atol, atof etc... , skipped for brevity ) and work with record */

      return 0;
 }

Answer 1

您可以让 getline 为您分配内存(这是使用 non-standard getline function over the standard fgets function 的全部要点)。来自 getline 手册页:

If *lineptr is NULL, then getline() will allocate a buffer for storing the line, which should be freed by the user program. (The value in *n is ignored.)

Alternatively, before calling getline(), *lineptr can contain a pointer to a malloc-allocated buffer *n bytes in size. If the buffer is not large enough to hold the line, getline() resizes it with realloc, updating *lineptr and *n as necessary.

所以你可以这样做:

line = NULL;
while (getline(&line, &size, stdin))
{
    // ... Do stuff with `line`...
}
free(line);

(或者让您的代码保持原样，因为 getline 将为您调整分配的缓冲区大小。)