Pass multiple arguments in C command line

Question

How am i able to pass multiple arguments in a C program like this, using different switches

program -d <argument1> -p <argument2>

I'm using getopt to enable me to pass arguments.

int main(int argc, char **argv)
{
         while(1) 
            {
            unsigned int c = getopt(argc, argv, "-dD:hHgGp:");
            if( c == -1 ) break;

            switch( c ) 
            {
                    case 'D':
                    case 'd':
                    printf("\nd=");
                    strcpy(D,optarg);
                    printf(D);
                    break;

                    case 'g':
                    case 'G':
                    printf("g");
                    break;

                    case 'p':
                    printf("\nPath=");
                    strcpy(pathFile,optarg);
                    printf(pathFile);
                    break;

                    case 'H':
                    case 'h':
                    usage(); //For help
                    return 0;

                    default:
                    return 0;
            }
    }

}

EDIT: The code here is a dummy code which I use for testing. It returns the argument that is passed as a string.

What is your program supposed to do, exactly, and how is it behaving differently from how you expect? We can't answer your question if we don't know what the problem is. — murgatroid99
– murgatroid99, Commented Dec 7, 2012 at 7:27
Your code seems to handle multiple arguments. Do you have any problems with it? — codewarrior
– codewarrior, Commented Dec 7, 2012 at 7:28

kmkaplan · Accepted Answer · 2012-12-07 07:37:04Z

3

Is it just a case of you forgetting the “:” after the “d” in getopt arguments?

unsigned int c = getopt(argc, argv, "-d:D:hHgGp:");

answered Dec 7, 2012 at 7:37

kmkaplan

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Comments

Community · Accepted Answer · 2017-05-23 10:24:57Z

It seems rather odd to write this:

while (1) 
{
    unsigned int c = getopt(argc, argv, "-dD:hHgGp:");
    if( c == -1 ) break;

The return value of getopt() is an int; why would you save it in an unsigned int?

int opt;

while ((opt = getopt(argc, argv, "-dD:hHgGp:")) != -1)
{
    switch (opt)
    {
    case ...
    }
}

If you're going to make options case-insensitive (not a good idea, IMO), then be consistent about it and handle P: too. Also, as first noted by kmkaplan's answer, you have D: and d being handled by the same switch; they should both be followed by a colon for sanity's sake: "-d:D:hHgGp:P:" would at least be self-consistent.

Also, under most circumstances, you don't need to copy the argument string (optarg) anywhere; you simply save a pointer to its current value in a convenient variable. If you do copy the argument string, you must check the length of the argument to ensure you are not overflowing buffers.

The first character of the option string is not normally a dash; it isn't a standard behaviour. The Mac OS X documentation for getopt() does note that it is a GNU extension and advises against ever starting an option string with a dash (and the option string should only contain a dash for backwards compatibility, not in new code — again, on Mac OS X or BSD). Under GNU getopt(), the leading dash means that non-option arguments are reported as if they were options. As long as you're aware that you're using a GNU getopt() extension, there's no harm in doing so.

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