CODEWITHSUNDEEP
cpointerscommand-line-argumentsargv
CS Student
CS Student

Reputation: 1633

Understanding argv and *++argv[0]

From chapter 5.10 of K&R's C book the idea of argv has been introduced to allow for command line arguments.

argv is a pointer to an array of character pointers. Taking this, how does the following code check that the argument supplied starts with a hyphen?

(*++argv)[0] == '-'

From my understanding the [0] is the same as *(argv + 0), so if I take the value of ( *++argv) then combine it with *(argv + 0) what do I actually get?

I know it returns the first characters of the argument string, but how? From my understanding:

  1. (*++argv) - The value argv points to is retrieved (which gives another pointer)
  2. [0] or *(argv + 0) - Then the next pointer to the next arguments pointer is returned from this statement.
  3. I don't see how the first char of the argument is obtained from this statement.

Upvotes: 3

Views: 945

Answers (1)

M.M
M.M

Reputation: 141618

Let's separate out the ++ for clarity:

argv = argv + 1;

Then we have:

(*argv)[0] == '-'

(I'm presuming this appears in an if statement).

Bear in mind that the definition of p[0] is *(p+0), i.e. *p . So this is the same as argv[0][0] == '-' .

argv[0] is a char * which points to the first character of a string. So argv[0][0] is the first character of that string.

The effect of argv = argv + 1; is to make argv point to the next char * which follows directly in memory after the previous one.

argv is a thing which points to char * (it is not a char * itself). There are some char * in adjacent memory, each of which points to a string located somewhere else in memory. Using argv you can iterate over that list of char *.

Upvotes: 3

Related Questions