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From: decot@hpisod2.UUCP
Newsgroups: comp.lang.c
Subject: Re: static char (*b)[6]; /* an unusual declaration */
Message-ID: <2550018@hpisod2.HP.COM>
Date: Fri, 12-Jun-87 20:05:17 EDT
Article-I.D.: hpisod2.2550018
Posted: Fri Jun 12 20:05:17 1987
Date-Received: Sun, 14-Jun-87 21:59:01 EDT
References: <761@bsu-cs.UUCP>
Organization: Hewlett Packard, Cupertino
Lines: 44

> What is the meaning of the following declaration?
> 
>      static char (*b)[6];

b has static storage and is a pointer to a (pseudo-named) array of
6 characters (not to the first character of an array, mind you, but
to something which may be used in any context where a name of an
array is allowed).  Initially, b is a null-pointer.  sizeof(*b) is
6*sizeof(char).  For instance, ++b increases the pointer address by
six characters.  *b is equivalent to the name of a particular array,
and since it has no initial value, it never will.

> The compiler and lint checker used by the magazine columnist seem to
> treat the declaration of b as equivalent to char ** but the following
> strange things happen.

Not really, since sizeof(*b) is not sizeof(char *), although (*b)[0] is
a character.

> main()
> {
>      static char (*b)[6];	/* an unusual declaration */
>      static char a[] = "Hello";
>      char *c = a;
>      char **d = &c;
>      b = d;			/* So b == d ... Watch. */

This statement should have encountered an error, since b and d are not
of compatible types.

>      printf ("%s\n", *d);	/* prints "Hello" */
>      printf ("%s\n", *b);	/* prints 0 */
> }

> Other code fragments omitted.  The columnist continues, "...This means
> that *b is a constant and while (b) can be modified, *b never will
> change its original value.  Thus, setting b = d does not mean that 
> *b == *d.  Also, while *b is treated as a constant (unchangeable), the
> value of b can vary."

This is accurate.

Dave Decot
hpda!decot