Command line options???

**waiteg** · 05-08-2006

How do I define the options I want to get from the command line? Any ideas?

**Rash** · 05-08-2006

You can specify the names of the array which stores command line arguments in your definition of the main procedure. E.g.

Code:

int main(int argc, char ** argv) { ...

Then argc tells you how many arguments you have, and argv is an array of memory addresses, which point to individual command line argument strings. So if you run a program with the command

Code:

./foo bar a b -c

then argc will be 5, and argv[0], argv[1], ..., argv[4] will point to the charactar arrays "./foo", "bar", "a", "b", and "-c", respectively. Yes, the program name gets passed as the first command line argument, and it's usually ignored.

You can define the options you want by writing code that does things, like complaining about missing options and exiting, based on the command line arguments.

**OnionKnight** · 05-08-2006

Note: argv[0] on Windows will point to the absolute path of the exectuable whereas in unix argv[0] will be how the program was called.

**tuxinator** · 05-08-2006

I got this method from the supertux source code

Code:

#include <stdio.h>
#include <string.h>

int main(int argc, char *argv[])
{
  /* Parse command line arguments */
  int i;
  for (i = 1;i < argc, ++i)
  {
    if (strcmp(argv[i], "--test") == 0)
    {
      printf("Test Succesful\n");
      printf("The string you entered is %s\n", argv[++i]);
    }
    else if (strcmp(argv[i], "--help") == 0 || strcmp(argv[i], "-h") == 0)
    {
      printf("Usage %s [options]\n", argv[0]);
      printf("Options:\n");
      printf("  --test str     Display the the value str\n");
      printf("  -h, --help     Display this help message\n");
    }
    else
    {
      printf("Unknown option %s\n", argv[i]);
      printf("Use %s --help for a list of command line arguments\n", argv[0]);
      return -1;
    }
  }
  /* End of Command Line Parser */
  return 0;
}

Just add whatever commands you want the user to enter.

**quzah** · 05-08-2006

Originally Posted by OnionKnight

Note: argv[0] on Windows will point to the absolute path of the exectuable whereas in unix argv[0] will be how the program was called.

Actually it may point to it. In fact, it probably will point to it. However, it is not guarinteed. If the name is not available, argv[0][0] will be '\0'.

Quzah.

**dwks** · 05-10-2006

Can argv[0] be NULL? (Instead of pointing to ""?)

**Dave_Sinkula** · 05-10-2006

If argc is 0, given that argv[argc] is a null pointer, then argv[0] would be a null pointer.

http://web.archive.org/web/200502070...t.html#2.1.2.2

If they are defined, the parameters to the main function shall obey the following constraints:

* The value of argc shall be nonnegative.

* argv[argc] shall be a null pointer.

**dwks** · 05-13-2006

Okay, thanks. I always use

Code:

print_usage(argc ? argv[0] : "*unknown*");

but I see I don't have to.

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