How to use command line options in a c command line tool? - c

Question

I am trying to understand how to use command line options with a command line c tool and I came accross this example.Can some one explain how the code flow works,I am not able to understand it,also I understand that it uses a getopt() function which is inbuilt.
The exe called is rocket_to and it has two command line options, e and a. e option takes 4 as an argument and a option takes Brasalia,Tokyo,London as argument.
Can some one explain how the code works?
This is the actual code:
command line:
rocket_to -e 4 -a Brasalia Tokyo London
code:
#include<unistd.h>
..
while((ch=getopt(argc,argv,"ae:"))!=EOF)
switch(ch){
..
case 'e':
engine_count=optarg;
..
}
argc -=optind;
argv +=optind;

Answer 1

There are many manual pages for getopt() including the POSIX specification. They describe what the getopt() function does. You can also read the POSIX Utility Conventions which describes how arguments are handled by most programs (but there are plenty of exceptions to the rules, usually because of historical, pre-POSIX precedent).
In the example outline code, the -e option takes an argument, and that is the 4 in the example command line. You can tell because of the e: in the third argument to getopt() (the colon following the letter indicates that the option takes an argument). The -a option takes no argument; you can tell because it is not followed by a colon in the third argument to getopt(). The names Brasilia, Tokyo, London are non-option arguments after the option processing is complete. They're the values in argv[0] .. argv[argc-1] after the two compound assignments outside the loop.
The use of EOF is incorrect; getopt() returns -1 when there are no more options for it to process. You don't have to include <stdio.h> to be able to use getopt().
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main(int argc, char **argv)
{
int ch;
int aflag = 0;
char *engine_count = "0";
while ((ch = getopt(argc, argv, "ae:")) != -1)
{
switch (ch)
{
case 'a':
aflag = 1;
break;
case 'e':
engine_count = optarg;
break;
default:
fprintf(stderr, "Usage: %s [-a][-e engine] [name ...]\n", argv[0]);
exit(EXIT_FAILURE);
}
}
argc -= optind;
argv += optind;
printf("A flag = %d\n", aflag);
printf("Engine = %s\n", engine_count);
for (int i = 0; i < argc; i++)
printf("argv[%d] = %s\n", i, argv[i]);
return 0;
}
That is working code which, if compiled to create a program rocket_to, produces:
$ ./rocket_to -e 4 -a Brasilia Tokyo London
A flag = 1
Engine = 4
argv[0] = Brasilia
argv[1] = Tokyo
argv[2] = London
$ ./rocket_to -a -e 4 Brasilia Tokyo London
A flag = 1
Engine = 4
argv[0] = Brasilia
argv[1] = Tokyo
argv[2] = London
$ ./rocket_to -e -a 4 Brasilia Tokyo London
A flag = 0
Engine = -a
argv[0] = 4
argv[1] = Brasilia
argv[2] = Tokyo
argv[3] = London
$

Answer 2

From the getopt man page:
The getopt() function parses the command-line arguments. Its arguments argc and argv are the argument count and array as passed to
the main() function on program invocation. An element of argv that starts with '-' (and is not exactly "-" or "--") is an option element. The characters of this element (aside from the initial '-') are option characters. If getopt() is called repeatedly, it
returns successively each of the option characters from each of the option elements.
The 3rd argument to getopt() are the valid options. If the option is followed by a colon it requires an argument. The argument can be accessed through the optarg variable. So in your example you have two options: 'a' which takes no argument and 'e' which takes an argument.
If getopt() finds an options it returns the character. If all options are parsed it returns -1 and if an unknown option is found it returns -1.
So your code loops through all options and processes them in a switch statement.
Next time when you have trouble understanding something like this try to run man <unknown function> first.