如何把整数作为命令行参数？

问题描述

我读过但它并没有说明如何接受整数作为参数的选项，如 cvalue 将在例子中的code：

I've read a getopt() example but it doesn't show how to accept integers as argument options, like cvalue would be in the code from the example:

 #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;
 }

如果我跑了上面 testop -c富， cvalue 将富，但如果我想要的东西 testop -c 42 ？由于 cvalue 的类型的char * 的，可我只投 OPTARG 是（INT）？我试着这样做，而无需使用 getopt的（）和访问的argv [无论] 直接，铸造它作为一个整数，但我总是用大负数与％d个打印时结束。我假设我不取消引用的argv [] 正确或东西，不知道...

If I ran the above as testop -c foo, cvalue would be foo, but what if I wanted testop -c 42? Since cvalue is of type char *, could I just cast optarg to be (int)? I've tried doing this without using getopt() and accessing argv[whatever] directly, and casting it as an integer, but I always end up with a large negative number when printing with %d. I'm assuming I'm not dereferencing argv[] correctly or something, not sure...

推荐答案

您需要使用的atoi（）来从字符串转换为整数。

You need to use atoi() to convert from string to integer.

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