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