Path: utzoo!attcan!uunet!lll-winken!lll-tis!ames!nrl-cmf!cmcl2!adm!smoke!gwyn
From: gwyn@smoke.ARPA (Doug Gwyn )
Newsgroups: comp.lang.c
Subject: Re: Double inderection question
Message-ID: <8302@smoke.ARPA>
Date: 3 Aug 88 04:40:59 GMT
References: <2001@tulum.cs.swarthmore.edu>
Reply-To: gwyn@brl.arpa (Doug Gwyn (VLD/VMB) <gwyn>)
Organization: Ballistic Research Lab (BRL), APG, MD.
Lines: 21

In article <2001@tulum.cs.swarthmore.edu> pomeranz@cs.swarthmore.edu (Hal Pomeranz) writes:
>char array[10], **ptr;
>   ptr = &array;
>From my limited understanding 'array' is a pointer to the first element
>(i.e. a pointer to array[0]), so &array should be a pointer to the pointer.

NO.  "array" is the name of the array.  In MOST expression contexts (but
not all) it is converted to a pointer to the first member of the array
before it is used.

"&array" should give you a pointer to the array (not to its first element),
but many older compilers treat it the same as an unadorned "array".

>char array[10], **ptr, *bogus_ptr;
>   bogus_ptr = array;
>   ptr = &bogus_ptr;

"bogus_ptr" IS a pointer to char; in the context of the first assignment,
"array" is converted to a pointer to array[0] before being copied into
"bogus_ptr".  "ptr" is a pointer to "bogus_ptr" and only has anything to
do with "array" by virtue of the fact that you made "bogus_ptr" have
something to do with "array".