c - getopt 中的 opterr 声明

Question

以下是来自 http://www.gnu.org 的示例代码.正如你们大多数人肯定会看到的那样，它是 getopt，我对变量声明有疑问。为什么

前面没有写什么type什么的

opterr = 0;

我以前从未见过。

#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>

int
main (int argc, char **argv)
{
  int aflag = 0;
  int bflag = 0;
  char *cvalue = NULL;
  int index;
  int c;

  opterr = 0;


  while ((c = getopt (argc, argv, "abc:")) != -1)
    switch (c)
      {
      case 'a':
        aflag = 1;
        break;
      case 'b':
        bflag = 1;
        break;
      case 'c':
        cvalue = optarg;
        break;
      case '?':
        if (optopt == 'c')
          fprintf (stderr, "Option -%c requires an argument.\n", optopt);
        else if (isprint (optopt))
          fprintf (stderr, "Unknown option `-%c'.\n", optopt);
        else
          fprintf (stderr,
                   "Unknown option character `\\x%x'.\n",
                   optopt);
        return 1;
      default:
        abort ();
      }
  printf ("aflag = %d, bflag = %d, cvalue = %s\n",
          aflag, bflag, cvalue);

  for (index = optind; index < argc; index++)
    printf ("Non-option argument %s\n", argv[index]);
  return 0;
}

Answer 1

opterr(3)在 unistd.h 中声明为外部变量:

extern int optind, opterr, optopt;

所以它是在不同的翻译单元中定义的全局变量，在本例中是您的标准 C 库。

设置为0的原因在manpage中也有说明:

If getopt() does not recognize an option character, it prints an error message to stderr, stores the character in optopt, and returns '?'. The calling program may prevent the error message by setting opterr to 0.