UNIX for Advanced & Expert Users

Is there a way to get the command line arguments.

I am using getopt(3) but if the arguments are more than one for a particular option than it just ignores the second argument. For eg

./a.out -x abc def

now abd will be got with -x using getopt "( x : )" and string abc\0def will get stored in optarg.

Now while parsing I get the \0 in between so how can I get this sorted out.

Basically I want to enter the command like this

./a.out -k abc def -a junk -b junk1

now I want to store "abc" and "def" in some string. And parse further the options with -a and -b

I am using solaris 5.8

Regards
JK

Use $1 to $9 to get up to 9 parameters... in your case, "-k" would be stored in $1, "abc" would be stored in $2, etc...

So you want 'abc' and 'def' in the same string, why don't you simply put a quote around it, i.e.

./a.out -k 'abc def' -a junk -b junk2

then getopt(argc, argv, "k:a") will work ....

I wan' t a system call and need not want to tokenize either, I just want to execute the way as normal program does.

./a.out -k abc def -x adf ghi

Regards
JK

My earlier post wasn't what you wanted.. I don't think I understood your question at first. A "normal" program executes the way cbkihong suggested. For example, if you use the find command to find a file called "some file" with a space in it, then you have to surround the name of that file with quotes or the command will fail. From the timestamp on the posts, I can see you most likely didn't see cbkihong's post before you posted...

Yeah that is a workaround that will do,

But I am really wondering how to do the following

suppose I am using getopt function, in a C program and try to caputre the arguments with the -x option then how can I do that.

Like I have a command ls abc def, this is just an example now I want to wrap it up in some other function. So basically i write this

Code:

main()
{
      char ch;
      char buf[100];
      while ( (ch = getopt (x:a:b:)) != NULL) {
               switch (ch) {
                          case 'x' :
                                   strcpy (buf, optarg);

now here comes the problem how will I copy optarg to buf, because \0 will be there at the end of the first string so only the first string will be stored in buf

I tried to do this

Code:

                  for ( i = 0; i < 50; i++) {
                          dest[i ] = optarg[i ];
                  }

and then print dest[i ] its showing that optarg has all the arguments whatever we pass with -x, now how to determine how many arguments are passed with -x,

Here I am giving the i value randomly, how will I determine the exact i value.

Now this arguments needs to be passed to ls which i will call in some other part like system ("ls arguments) got from above which will stored in some variable

Hoping that you understood the real problem

Regards
JK

[i]added code tags and spaces to so rest of post wasn't italicized --oombera

Yes, I understand your problem. I don't know if any Unix variants have special extensions. Otherwise, from my interpretation of the getopt manpage I believe getopt doesn't expect multiple parameters to a single option, at all.

So, for example, if you invoke this as
./a.out -x abc def -a

Then argv[] is an array consisting of

-x
abc
def
-a

(not showing argv[0] above)

What getopt does is that it monotonically scans forward. It identifies '-x', associates 'abc' as optarg. It doesn't know what 'def' is, and throw it to the end. Then it identifies '-a', and so on. 'def' is left at the end and regarded as a non-option (the index of which you get with optind). Therefore, as getopt proceeds it permutes the order of arguments that appear in argv[].

I made this test program to illustrate the idea:

Code:

#include <stdio.h>
#include <getopt.h>

extern char *optarg;
extern int optind;

int main(int argc, char** argv) {
	int optchr; int i;
	while ((optchr = getopt(argc, argv, "x:a")) != -1) {
		printf("Option '%c' detected.\n", optchr);
		if (optarg) {
			printf("\tArgument: '%s'\n", optarg);
		}
	}
	printf("Other arguments:\n");
	for(i=optind; i<argc; i++) {
		printf("\t%s\n", argv[i]);
	}
	return 0;
}

./a.out -x abc def -a

gives

Code:

Option 'x' detected.
        Argument: 'abc'
Option 'a' detected.
Other arguments:
        def

I think conventional Unix tradition is that there is at most one argument for each option. I think this is because getopt allows options to be given in any order. So if you would like to break this convention you can always write your own routine to parse argv[]. Of course that will not be easy if your program has to accommodate a variety of options. Formally speaking, that requires parser construction knowledge (lexical analysis) that are beyond the scope of this forum.