Linux 从命令行将参数传递给 C 程序

Pass arguments into C program from command line

提问by BlackCow

So I'm in Linux and I want to have a program accept arguments when you execute it from the command line.

所以我在 Linux 中，我想让程序在从命令行执行时接受参数。

For example,

例如，

./myprogram 42 -b -s

./myprogram 42 -b -s

So then the program would store that number 42 as an int and execute certain parts of code depending on what arguments it gets like -b or -s.

因此，程序会将数字 42 存储为 int 并根据它获得的参数（如 -b 或 -s）执行某些代码部分。

采纳答案by CMS

You could use getopt.

您可以使用getopt。

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

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

   opterr = 0;

   while ((c = getopt (argc, argv, "bs")) != -1)
     switch (c)
       {
       case 'b':
         bflag = 1;
         break;
       case 's':
         sflag = 1;
         break;
       case '?':
         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 ("bflag = %d, sflag = %d\n", bflag, sflag);

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

回答by Greg Hewgill

In C, this is done using arguments passed to your main()function:

在 C 中，这是使用传递给main()函数的参数完成的：

int main(int argc, char *argv[])
{
    int i = 0;
    for (i = 0; i < argc; i++) {
        printf("argv[%d] = %s\n", i, argv[i]);
    }
    return 0;
}

More information can be found online such as this Arguments to mainarticle.

更多信息可以在网上找到，比如这篇主要文章的论据。

回答by womble

Take a look at the getopt library; it's pretty much the gold standard for this sort of thing.

看看 getopt 库；这几乎是这类事情的黄金标准。

回答by jfs

Consider using getopt_long(). It allows both short and long options in any combination.

考虑使用getopt_long(). 它允许任意组合的短期和长期期权。

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

/* Flag set by `--verbose'. */
static int verbose_flag;

int
main (int argc, char *argv[])
{
  while (1)
    {
      static struct option long_options[] =
    {
      /* This option set a flag. */
      {"verbose", no_argument,       &verbose_flag, 1},
      /* These options don't set a flag.
         We distinguish them by their indices. */
      {"blip",    no_argument,       0, 'b'},
      {"slip",    no_argument,       0, 's'},
      {0,         0,                 0,  0}
    };
      /* getopt_long stores the option index here. */
      int option_index = 0;

      int c = getopt_long (argc, argv, "bs",
               long_options, &option_index);

      /* Detect the end of the options. */
      if (c == -1)
    break;

      switch (c)
    {
    case 0:
      /* If this option set a flag, do nothing else now. */
      if (long_options[option_index].flag != 0)
        break;
      printf ("option %s", long_options[option_index].name);
      if (optarg)
        printf (" with arg %s", optarg);
      printf ("\n");
      break;
    case 'b':
      puts ("option -b\n");
      break;
    case 's':
      puts ("option -s\n");
      break;
    case '?':
      /* getopt_long already printed an error message. */
      break;

    default:
      abort ();
    }
    }

  if (verbose_flag)
    puts ("verbose flag is set");

  /* Print any remaining command line arguments (not options). */
  if (optind < argc)
    {
      printf ("non-option ARGV-elements: ");
      while (optind < argc)
    printf ("%s ", argv[optind++]);
      putchar ('\n');
    }

  return 0;
}

Related:

有关的：

• Which command line commands style do you prefer?
• What is the general syntax of a Unix shell command?

• 你更喜欢哪种命令行命令风格？
• Unix shell 命令的一般语法是什么？

回答by sastanin

Instead of getopt(), you may also consider using argp_parse()(an alternative interface to the same library).

除了getopt()，您还可以考虑使用argp_parse()（同一库的替代接口）。

From libc manual:

从libc 手册：

getoptis more standard (the short-option only version of it is a part of the POSIX standard), but using argp_parseis often easier, both for very simple and very complex option structures, because it does more of the dirty work for you.

getopt更标准（它的仅短选项版本是 POSIX 标准的一部分），但是argp_parse对于非常简单和非常复杂的选项结构，使用 通常更容易，因为它为您做了更多的脏活。

But I was always happy with the standard getopt.

但我总是对标准感到满意getopt。

N.B. GNU getoptwith getopt_longis GNU LGPL.

NB GNUgetopt与getopt_long是GNU LGPL。

回答by dmckee --- ex-moderator kitten

Other have hit this one on the head:

其他人打了这个头：

• the standard arguments to main(int argc, char **argv)give you direct access to the command line (after it has been mangled and tokenized by the shell)
• there are very standard facility to parse the command line: getopt()and getopt_long()

• main(int argc, char **argv)使您可以直接访问命令行的标准参数（在它被 shell 修改和标记后）
• 有非常标准的设施，以解析命令行：getopt()和getopt_long()

but as you've seen the code to use them is a bit wordy, and quite idomatic. I generally push it out of view with something like:

但是正如您所看到的，使用它们的代码有点冗长，而且非常惯用。我通常会使用以下内容将其从视图中移除：

typedef
struct options_struct {
   int some_flag;
   int other_flage;
   char *use_file;
} opt_t;
/* Parses the command line and fills the options structure, 
 * returns non-zero on error */
int parse_options(opt_t *opts, int argc, char **argv);

Then first thing in main:

然后主要的第一件事：

int main(int argc, char **argv){
   opt_t opts;
   if (parse_options(&opts,argc,argv)){
      ...
   } 
   ...
}

Or you could use one of the solutions suggested in Argument-parsing helpers for C/UNIX.

或者您可以使用Argument-parsing helpers for C/UNIX 中建议的解决方案之一。