Andrew Andrew - 2 months ago 8
C Question

how to take integers as command line arguments?

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...

Answer

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

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