Novice question.

[Bryan Morse]

In article <336 at brat.UUCP> donn at brat.UUCP (Donn Pedro) writes:
>A question for you that know, because I dont.
>
>
>I understand that the following is a pointer.
>
>> 	char *s;
>
>But what is this doing?
>Why the two "*". 
>
>> 	STRING **s_array;
>
>Please e-mail.

I'm posting my reply since the mail I sent you bounced.  Besides, maybe
someone else wants to know (and if someone doesn't, my apologies).

First, the real way to read C declarations.

Any statement of the form

        type <list of expressions>

says that all expressions in the list are of the specified type.

Thus, the statement

        char *s;

does not (repeat, does *not*) say that s is a pointer to a char.  What
it says is that *s is a char.  It may seem the same (and in reality is)
but it's important to make the distinction.  That's why you can say

        char c, *s;

and declare both a char and a pointer to one.  A very common mistake is to
say "hmm, I want to declare a few char pointers" and type

        char *  s,t;
        ------  ---
        type    list of variables

and this is WRONG!  Be very careful to think in terms of this C style
instead of the Pascal style where you specify the type (including pointers)
and a list of variables.  Does this make sense?

Having said this, let's get to your question:

        STRING **s;

says that **s is of type STRING.  Since *s is what s points to and **s is
what *s points to (you can chain dereferences in C like what--similar to
p^^ in Pascal if you've every seen it) that means that s is a pointer to
a pointer to a STRING.  Make sense?

Seriously, once you learn this simple rule of how to read C declarations,
you can understand even the most convoluted declarations.

Hope this helps..

Bryan Morse                University of North Carolina at Chapel Hill
morse at cs.unc.edu           Department of Computer Science