问题
I am working on trying to take in command line arguments. If I want to have multiple optional command line arguments how would I go about doing that? For example you can run the program in the following ways: (a is required every instance but -b -c -d can be given optionally and in any order)
./myprogram -a
./myprogram -a -c -d
./myprogram -a -d -b
I know that getopt()'s third argument is options. I can set these options to be "abc" but will the way I have my switch case set up causes the loop to break at each option.
回答1:
The order doesn't matter so far as getopt()
is concerned. All that matters is your third argument to getopt()
(ie: it's format string) is correct:
The follow format strings are all equivalent:
"c:ba"
"c:ab"
"ac:b"
"abc:"
In your particular case, the format string just needs to be something like "abcd"
, and the switch()
statement is properly populated.
The following minimal example¹ will help.
#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;
}
¹Example taken from the GNU manual
来源:https://stackoverflow.com/questions/28129098/using-getopt-in-c-for-command-line-arguments