Optarg and Command Line Arguments

Question

I am getting command line arguments from the user.

I then do switch cases for the commands, for example:

    case 'f':
         *file_name = optarg;
         break;

I am unsure if I need to malloc for the pointer since I don't fully understand optarg.

This is how file_name is declared:

char **file_name;

Should I do

int length = strlen(optarg); // This gives a warning about types when compiling.

Then malloc for the string length + 1?

How should the malloc be done for this kind of problem? Remember the user is typing in the file_name into **argv.

EDIT: This is how I am calling this function and still getting segmentation fault.

int main(int argc, char **argv)
{
   char **file_name;
   parser(argc, argvm file_name);
}

void parser(int argc, char **argv, char **file_name)
{
  // Switch cases.
}

Answer 1

'optarg' is just a pointer to an element in argv[]. So, it's safe to not allocate memory and copy the value which 'optarg' points to.

Suppose your programm is called with the following arguments:

myapp -a "hello" -b "world"

and your code is:

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

void parse_options(int argc, char* argv[], char ** first_arg, char ** second_arg)
{
  const char* opt_string = "a:b:";
  int opt = -1;
  opt = getopt(argc, argv, opt_string);
  while (opt != -1) {
    switch(opt) {
      case 'a':
        *first_arg = optarg; /* points to argv[2]="hello" */
        break;
      case 'b':
        *second_arg = optarg; /* points to argv[4]="world" */
        break;
      default:
        break;
      }
    opt = getopt(argc, argv, opt_string);
  }
}

int main(int argc, char* argv[])
{
  char* first = 0;
  char* second = 0;
  parse_options(argc, argv, &first, &second);
  printf("first=%s, second=%s\n", first, second);
  return 0;
}

My output:

freebsd% gcc -Wall main.c
freebsd% ./a.out -a hello -b world
first=hello, second=world