How do I use getopt_long to parse multiple arguments?
-
15-04-2021 - |
문제
#include <iostream>
#include <getopt.h>
#define no_argument 0
#define required_argument 1
#define optional_argument 2
int main(int argc, char * argv[])
{
std::cout << "Hello" << std::endl;
const struct option longopts[] =
{
{"version", no_argument, 0, 'v'},
{"help", no_argument, 0, 'h'},
{"stuff", required_argument, 0, 's'},
{0,0,0,0},
};
int index;
int iarg=0;
//turn off getopt error message
opterr=1;
while(iarg != -1)
{
iarg = getopt_long(argc, argv, "s:vh", longopts, &index);
switch (iarg)
{
case 'h':
std::cout << "You hit help" << std::endl;
break;
case 'v':
std::cout << "You hit version" << std::endl;
break;
case 's':
std::cout << "You hit stuff" << std::endl;
if(optarg)
std::cout << "Your argument(s): " << optarg << std::endl;
break;
}
}
std::cout << "GoodBye!" << std::endl;
return 0;
}
Desired output:
./a.out --stuff someArg1 someArg2
Hello
You hit stuff
Your agument(s): someArg1 someArg2
GoodBye!
해결책
getopt returns -1 when all option args have been processed. The --stuff
is recognized as an option that takes an argument, in this case someArg1
. The someArg2
arg does not start with -
or --
, so it is not an option. By default, this will be permuted to the end of argv
. After getopt returns -1, all non-option args will be in argv
from optind
to argc-1
:
while (iarg != -1) {
iarg = getopt_long(argc, argv, "s:vh", longopts, &index);
// ...
}
for (int i = optind; i < argc; i++) {
cout << "non-option arg: " << argv[i] << std::endl;
}
If you add a single -
to the start of optstring
, getopt
will return 1 (not '1') and point optarg
to the non-option parameter:
while (iarg != -1) {
iarg = getopt_long(argc, argv, "-s:vh", longopts, &index);
switch (iarg)
{
// ...
case 1:
std::cout << "You hit a non-option arg:" << optarg << std::endl;
break;
}
}
다른 팁
In the line ./a.out --stuff someArg1 someArg2
the shell interprets three arguments to a.out. You want the shell to interpret "someArg1 someArg2" as one argument - so put the words in quotes:
./a.out --stuff "someArg1 someArg2"
I'm working on windows, so I had to compile getopt and getopt_long from this excellent source
I modified getopt_long.c (below) to accommodate two input arguments. I didn't bother with the more general case of multiple arguments since that would require more (and cleaner) rework than I had time/need for. The second argument is placed in another global, "optarg2".
If you don't need to compile getopt from source, Frank's answer above is more elegant.
extern char * optarg2
.
.
.
int getopt_long(nargc, nargv, options, long_options, index)
{
.
.
.
if (long_options[match].has_arg == required_argument ||
long_options[match].has_arg == optional_argument ||
long_options[match].has_arg == two_req_arguments) {
if (has_equal)
optarg = has_equal;
else
optarg = nargv[optind++];
if (long_options[match].has_arg == two_req_arguments) {
optarg2 = nargv[optind++];
}
}
if ((long_options[match].has_arg == required_argument ||
long_options[match].has_arg == two_req_arguments)
&& (optarg == NULL)) {
/*
* Missing argument, leading :
* indicates no error should be generated
*/
if ((opterr) && (*options != ':'))
(void)fprintf(stderr,
"%s: option requires an argument -- %s\n",
__progname(nargv[0]), current_argv);
return (BADARG);
}
if ((long_options[match].has_arg == two_req_arguments)
&& (optarg2 == NULL)) {
/*
* Missing 2nd argument, leading :
* indicates no error should be generated
*/
if ((opterr) && (*options != ':'))
(void)fprintf(stderr,
"%s: option requires 2nd argument -- %s\n",
__progname(nargv[0]), current_argv);
return (BADARG);
}
You'll also need to add a define in getopt.h for "two_required_args" or "multiple_args" as you see fit.
edit: I'm bad at markdown
optarg points to "someArg1" and argv[optind] is "someArg2" if it exists and is not an option. You can simply use it and then consume it by incrementing optind.
case 's':
std::cout << "You hit stuff" << std::endl;
if (optind < argc && argv[optind][0] != '-') {
std::cout << "Your argument(s): " << optarg << argv[optind] << std::endl;
optind++;
} else {
printusage();
}
break;
Note this can work for an arbitrary number of arguments:
case 's':
std::cout << "You hit stuff." << std::endl;
std::cout << "Your arguments:" std::endl << optarg << std::endl;
while (optind < argc && argv[optind][0] != '-') {
std::cout << argv[optind] << std::endl;
optind++;
}
break;