In the getopt_long example here, why are the short options separated by a colon at abc:d:f: and why are abc are grouped together like that?
#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
/* Flag set by ‘--verbose’. */
static int verbose_flag;
int
main (int argc, char **argv)
{
int c;
while (1)
{
static struct option long_options[] =
{
/* These options set a flag. */
{"verbose", no_argument, &verbose_flag, 1},
{"brief", no_argument, &verbose_flag, 0},
/* These options don’t set a flag.
We distinguish them by their indices. */
{"add", no_argument, 0, 'a'},
{"append", no_argument, 0, 'b'},
{"delete", required_argument, 0, 'd'},
{"create", required_argument, 0, 'c'},
{"file", required_argument, 0, 'f'},
{0, 0, 0, 0}
};
/* getopt_long stores the option index here. */
int option_index = 0;
c = getopt_long (argc, argv, "abc:d:f:",
long_options, &option_index);
/* Detect the end of the options. */
if (c == -1)
break;
switch (c)
{
case 0:
/* If this option set a flag, do nothing else now. */
if (long_options[option_index].flag != 0)
break;
printf ("option %s", long_options[option_index].name);
if (optarg)
printf (" with arg %s", optarg);
printf ("\n");
break;
case 'a':
puts ("option -a\n");
break;
case 'b':
puts ("option -b\n");
break;
case 'c':
printf ("option -c with value `%s'\n", optarg);
break;
case 'd':
printf ("option -d with value `%s'\n", optarg);
break;
case 'f':
printf ("option -f with value `%s'\n", optarg);
break;
case '?':
/* getopt_long already printed an error message. */
break;
default:
abort ();
}
}
On GNU's page, it is said that
An option character in this string can be followed by a colon (:) to
indicate that it takes a required argument.
For example, xy:zq accepts the options x, y, z, q; y requires an additional argument.
Related
I would like to use getopt_long() correctly and am trying to understand why certain choices were made in the example of getopt_long. (I've copied the full code listing below).
I understand that case 0 occurs when the flag field of the struct is not null (find getopt_long returns 0), but why does the example show retesting long_options[option_index].flag for null in case 0? Isn't this guaranteed given that we are in case 0?
Asked another way: What input could I provide to cause the code to enter case 0 but fail the first if-statement within it? Is printf ("option %s", long_options[option_index].name); unreachable?
25.2.4 Example of Parsing Long Options with getopt_long
#include <stdio.h>
#include <stdlib.h>
#include <getopt.h>
/* Flag set by ‘--verbose’. */
static int verbose_flag;
int
main (int argc, char **argv)
{
int c;
while (1)
{
static struct option long_options[] =
{
/* These options set a flag. */
{"verbose", no_argument, &verbose_flag, 1},
{"brief", no_argument, &verbose_flag, 0},
/* These options don’t set a flag.
We distinguish them by their indices. */
{"add", no_argument, 0, 'a'},
{"append", no_argument, 0, 'b'},
{"delete", required_argument, 0, 'd'},
{"create", required_argument, 0, 'c'},
{"file", required_argument, 0, 'f'},
{0, 0, 0, 0}
};
/* getopt_long stores the option index here. */
int option_index = 0;
c = getopt_long (argc, argv, "abc:d:f:", long_options, &option_index);
/* Detect the end of the options. */
if (c == -1)
break;
switch (c)
{
case 0:
/* If this option set a flag, do nothing else now. */
if (long_options[option_index].flag != 0)
break;
printf ("option %s", long_options[option_index].name);
if (optarg)
printf (" with arg %s", optarg);
printf ("\n");
break;
case 'a':
puts ("option -a\n");
break;
case 'b':
puts ("option -b\n");
break;
case 'c':
printf ("option -c with value `%s'\n", optarg);
break;
case 'd':
printf ("option -d with value `%s'\n", optarg);
break;
case 'f':
printf ("option -f with value `%s'\n", optarg);
break;
case '?':
/* getopt_long already printed an error message. */
break;
default:
abort ();
}
}
/* Instead of reporting ‘--verbose’
and ‘--brief’ as they are encountered,
we report the final status resulting from them. */
if (verbose_flag)
puts ("verbose flag is set");
/* Print any remaining command line arguments (not options). */
if (optind < argc)
{
printf ("non-option ARGV-elements: ");
while (optind < argc)
printf ("%s ", argv[optind++]);
putchar ('\n');
}
exit (0);
}
I am new to c programming using command line arguments so despite finding few simmilar posts I am still stuck on this problem.
I need to pass a command line option that have more than 1 argument, for example
./testProgram --command arg1 arg2 arg3
here is my code
static struct option long_options[] = {
{"rdonly", required_argument, 0, 'a'},
{"wronly", required_argument, 0, 'b'},
{"command", required_argument, 0, 'c'},
{"verbose", no_argument, 0, 'd'},
{0,0,0,0}
};
c=getopt_long(argc,argv, "", long_options, &option_index);
if(c == -1)
break;
switch(c) {
case 'a':
rdonly(fd,optarg);
break;
case 'b':
wronly(fd,optarg);
break;
case 'c':
command(optind,optarg,argc,argv);
break;
case 'd':
printf("its verbose\n");
break;
default:
printf("WRONG OPTION\n");
}
}
//command function
int command(int optind, char *optarg, int argc, char* argv)
{
printf("Its option command with argument %s\n",optarg);
return 0;
}
I have a program which takes in multiple command line arguments so I am using getopt. One of my arguments takes in a string as a parameter. Is there anyway to obtain that string through the getopt function or would I have to obtain it through the argv[] array? Also can getopt read args like -file ? All the arguments I have seen till now have only one character such as -a
EDIT
From the below answers I have written a program to use getopt_long(), but the switch statement only recognizes the argument when I use the character argument and not the long argument. I'm not sure why this happening. On passing the arguments -mf -file sample I do not see the print statements.
EDIT
I tried entering the command arguments as --file and then it worked. Is it not possible to do this with just -file ?
static struct option long_options[] =
{
{"mf", required_argument, NULL, 'a'},
{"md", required_argument, NULL, 'b'},
{"mn", required_argument, NULL, 'c'},
{"mw", required_argument, NULL, 'd'},
{"lf", required_argument, NULL, 'e'},
{"ld", required_argument, NULL, 'f'},
{"ln", required_argument, NULL, 'g'},
{"lw", required_argument, NULL, 'h'},
{"rf", required_argument, NULL, 'i'},
{"rd", required_argument, NULL, 'j'},
{"rn", required_argument, NULL, 'k'},
{"rw", required_argument, NULL, 'l'},
{"df", required_argument, NULL, 'm'},
{"dd", required_argument, NULL, 'n'},
{"dn", required_argument, NULL, 'o'},
{"dw", required_argument, NULL, 'p'},
{"file", required_argument, NULL, 'q'},
{NULL, 0, NULL, 0}
};
int ch=0;
while ((ch = getopt_long(argc, argv, "abcdefghijklmnopq:", long_options, NULL)) != -1)
{
// check to see if a single character or long option came through
switch (ch){
case 'a':
cout<<"title";
break;
case 'b':
break;
case 'c':
break;
case 'd':
break;
case 'e':
break;
case 'f':
break;
case 'g':
break;
case 'h':
break;
case 'i':
break;
case 'j':
break;
case 'k':
break;
case 'l':
break;
case 'm':
break;
case 'n':
break;
case 'o':
break;
case 'p':
break;
case 'q':
cout<<"file";
break;
case '?':
cout<<"wrong message"
break;
}
}
Read man getopt http://linux.die.net/man/3/getopt
optstring is a string containing the legitimate option characters. If
such a character is followed by a colon, the option requires an
argument, so getopt() places a pointer to the following text in the
same argv-element, or the text of the following argv-element, in
optarg. Two colons mean an option takes an optional arg; if there is
text in the current argv-element (i.e., in the same word as the option
name itself, for example, "-oarg"), then it is returned in optarg,
otherwise optarg is set to zero.
A sample code:
#include <stdio.h>
#include <unistd.h>
int main (int argc, char *argv[])
{
int opt;
while ((opt = getopt (argc, argv, "i:o:")) != -1)
{
switch (opt)
{
case 'i':
printf("Input file: \"%s\"\n", optarg);
break;
case 'o':
printf("Output file: \"%s\"\n", optarg);
break;
}
}
return 0;
}
Here in the optstring is "i:o:" the colon ':' character after each character in the string tells that those options will require an argument. You can find argument as a string in the optarg global var. See manual for detail and more examples.
For more than one character option switches, see the long options getopt_long. Check the manual for examples.
EDIT in response to the single '-' long options:
From the man pages
getopt_long_only() is like getopt_long(), but '-' as well as "--" can indicate a long option. If an option that starts with '-'
(not "--") doesn't match a long option, but does match a short option,
it is parsed as a short option instead.
Check the manual and try it.
To specify that a flag requires an argument, add a ':' right after the flag in the short_opt variable. To use long arguments, use getopt_long().
Here is a quick example program:
#include <getopt.h>
#include <stdio.h>
int main(int argc, char * argv[]);
int main(int argc, char * argv[])
{
int c;
const char * short_opt = "hf:";
struct option long_opt[] =
{
{"help", no_argument, NULL, 'h'},
{"file", required_argument, NULL, 'f'},
{NULL, 0, NULL, 0 }
};
while((c = getopt_long(argc, argv, short_opt, long_opt, NULL)) != -1)
{
switch(c)
{
case -1: /* no more arguments */
case 0: /* long options toggles */
break;
case 'f':
printf("you entered \"%s\"\n", optarg);
break;
case 'h':
printf("Usage: %s [OPTIONS]\n", argv[0]);
printf(" -f file file\n");
printf(" -h, --help print this help and exit\n");
printf("\n");
return(0);
case ':':
case '?':
fprintf(stderr, "Try `%s --help' for more information.\n", argv[0]);
return(-2);
default:
fprintf(stderr, "%s: invalid option -- %c\n", argv[0], c);
fprintf(stderr, "Try `%s --help' for more information.\n", argv[0]);
return(-2);
};
};
return(0);
}
I'm having problems getting my command-line arguments to work properly. The arguments appear to fail when the argc is higher than 3 and after that you can only get them to work if you write -fwhatever and not -f whatever as it should be. -t also copies the value of -o.
int main(int argc,char **argv)
{
int c;
/* Are we in root? */
if(geteuid() !=0)
{
printf("Root access is required to run this program. \n");
exit(0);
}
/* How do we use the program */
if ((argc < 6) || (argc > 8))
{
usage(argv[0]);
exit (0);
}
while (1)
{
static struct option long_options[] =
{
/* Options */
{"send", no_argument, 0, 's'},
{"recieve", no_argument, 0, 'r'},
{"file", required_argument, 0, 'f'},
{"data", required_argument, 0, 'd'},
{"destip", required_argument, 0, 'i'},
{"destport", required_argument, 0, 'p'},
{"sourceip", required_argument, 0, 'o'},
{"sourceport", required_argument, 0, 't'},
{0, 0, 0, 0}
};
int option_index = 0;
c = getopt_long (argc, argv, "srf:d:i:p:o:t:",
long_options, &option_index);
/* Detect the end of the options. */
if (c == -1)
break;
switch (c)
{
case 0:
/* If this option set a flag, do nothing else now. */
if (long_options[option_index].flag != 0)
break;
printf ("option %s", long_options[option_index].name);
if (optarg)
printf (" with arg %s", optarg);
printf ("\n");
break;
case 's':
puts ("option -s\n");
break;
case 'r':
puts ("option -r\n");
break;
case 'f':
printf ("option -f with value `%s'\n", optarg);
break;
case 'd':
printf ("option -d with value `%s'\n", optarg);
break;
case 'i':
printf ("option -i with value `%s'\n", optarg);
break;
case 'p':
printf ("option -p with value `%s'\n", optarg);
break;
case 'o':
printf ("option -o with value `%s'\n", optarg);
case 't':
printf ("option -t with value `%s'\n", optarg);
case '?':
/* Error message printed */
break;
default:
abort ();
}
}
/* Print any remaining command line arguments (not options). */
if (optind < argc)
{
printf ("non-option ARGV-elements: ");
while (optind < argc)
printf ("%s ", argv[optind++]);
putchar ('\n');
}
getchar ();
exit (0);
}
There's obviously something horribly wrong here, but it can't seem to find it.
Regarding the -o copied to -t, you forgot to put a break; at the end of the case.
Also, I would have removed the argc checking. Let getopt do its magic. You can check at the end of parsing if all mandatory options were set. You can also check for unknown arguments.
I'm trying to use the getopt_long() function for the first time only I'm having problems with arguments that aren't flags. For instance, in my code when a unknown argument is given I want to use it as a input file. When I run this with only a file name it is not printed, if I first use a flag, any flag, then I can print it.
How can I fix this?
#include <stdio.h>
#include <getopt.h>
static struct option long_options[] = {
{"help", no_argument, 0, 'h'},
{"input", required_argument, 0, 'i'},
{"output", required_argument, 0, 'o'},
{"algorithm", required_argument, 0, 'a'},
{0, 0, 0, 0}
};
int main(int argc, char *argv[]) {
int c;
int option_index = 0;
while(42) {
c = getopt_long(argc, argv, "hi:o:a:", long_options,
&option_index);
if(c == -1)
break;
switch(c) {
case 'h': /* --help */
printf("--help flag\n");
break;
case 'i': /* --input */
printf("--input flag\n");
break;
case 'o': /* --output */
printf("--output flag\n");
break;
case 'a': /* --algorithm */
printf("--algorithm flag \n");
break;
default: /* ??? */
fprintf(stderr, "Invalid option");
return 1;
}
if(optind < argc) {
printf("other arguments: ");
while(optind < argc) {
printf ("%s ", argv[optind]);
optind++;
}
printf("\n");
}
}
return 0;
}
The loop should only contain the switch. Process the residual arguments in a separate (not nested) loop:
#include <stdio.h>
#include <getopt.h>
static struct option long_options[] =
{
{"help", no_argument, 0, 'h'},
{"input", required_argument, 0, 'i'},
{"output", required_argument, 0, 'o'},
{"algorithm", required_argument, 0, 'a'},
{0, 0, 0, 0}
};
int main(int argc, char *argv[])
{
int opt;
int option_index = 0;
int i;
while ((opt = getopt_long(argc, argv, "hi:o:a:", long_options, &option_index)) != -1)
{
switch(opt)
{
case 'h': /* --help */
printf("--help flag\n");
break;
case 'i': /* --input */
printf("--input flag (%s)\n", optarg);
break;
case 'o': /* --output */
printf("--output flag (%s)\n", optarg);
break;
case 'a': /* --algorithm */
printf("--algorithm flag (%s)\n", optarg);
break;
default: /* ??? */
fprintf(stderr, "Invalid option %c\n", opt);
return 1;
}
}
for (i = optind; i < argc; i++)
printf("Process: %s\n", argv[i]);
return 0;
}
There is a way to have GNU getopt() and getopt_long() return file name arguments as if they were options with the 'letter' ^A '\1'; use '-' as the first character of the short options string, and trap '\1' in the switch; the value of optarg is the name of the file.
#include <stdio.h>
#include <getopt.h>
static struct option long_options[] =
{
{"help", no_argument, 0, 'h'},
{"input", required_argument, 0, 'i'},
{"output", required_argument, 0, 'o'},
{"algorithm", required_argument, 0, 'a'},
{0, 0, 0, 0}
};
int main(int argc, char *argv[])
{
int opt;
int option_index = 0;
int i;
while ((opt = getopt_long(argc, argv, "-hi:o:a:", long_options, &option_index)) != -1)
{
switch(opt)
{
case 'h': /* --help */
printf("--help flag\n");
break;
case 'i': /* --input */
printf("--input flag (%s)\n", optarg);
break;
case 'o': /* --output */
printf("--output flag (%s)\n", optarg);
break;
case 'a': /* --algorithm */
printf("--algorithm flag (%s)\n", optarg);
break;
case '\1':
printf("File: %s\n", optarg);
break;
default: /* ??? */
fprintf(stderr, "Invalid option %c\n", opt);
return 1;
}
}
for (i = optind; i < argc; i++)
printf("Process: %s\n", argv[i]);
return 0;
}
However, you do need the loop after the argument processing loop in case your cantankerous user types:
program -- abc def
The '--' terminates the while() loop without processing the file name arguments.
It doesn't work because you break out of the outer while loop when getopt_long returns -1 (which indicates that there are no more options).
You need to move the if (optind < argc) block out of the outer while loop; it doesn't belong there anyway. This might be clearer if you wrote that outer while loop as:
while ((c = getopt_long(...)) != -1)
{
switch (c)
{
/* Deal with flags. */
}
}