Newsgroups: comp.os.msdos.djgpp
From: design AT netcom DOT com (Chris Waters)
Subject: Re: Pointers
Message-ID: <designDz6oys.FLt@netcom.com>
Organization: Design and Delivery
References: <53oa2v$kr4 AT miracle DOT cybernex DOT net> <53oql4$qg7 AT news DOT ox DOT ac DOT uk>
Date: Sat, 12 Oct 1996 22:25:40 GMT
Lines: 51
Sender: design AT netcom20 DOT netcom DOT com
To: djgpp AT delorie DOT com
DJ-Gateway: from newsgroup comp.os.msdos.djgpp

In article <53oql4$qg7 AT news DOT ox DOT ac DOT uk>,
George Foot <mert0407 AT sable DOT ox DOT ac DOT uk> wrote:
>mrichman AT cybernex DOT net wrote:
>: Is it safe to assume that all pointer variables take up the same
>: number of bytes, or is that compiler dependent.

>If you're writing really portable code, it's not safe to assume anything. 

You can assume what the standard guarantees:  that any pointer can be
cast to void *, and then back _to the same type of pointer it was
originally_.

>If you make a #define at the start of your source, e.g.

>#define POINTER_SIZE sizeof(*void)

You mean sizeof(void *).

>then use this, e.g.

>int *my_array[20];

>for (int a=0;a<20;a++)
> printf("%d\n",*(my_array+a*POINTER_SIZE));

This is so far off base that it's truly frightening!

First of all, pointer addition doesn't work that way.  You don't need to
multiply the offset by anything:

   &my_array[a] == (my_array + a)

Thus, the above is the equivalent to:

   for (a = 0; a < 20; a++)
     printf ("%d\n", my_array[a * POINTER_SIZE]);

Which is not (one hopes) what is desired.  Not to mention the fact that
you're using "%d" to print a pointer.

Secondly, the standard does not guarantee that an int * is the same size
as a void *!  All it guarantees is that an int * can be cast to a void *
and back to an int * safely.  Thus, the code above wouldn't work even if
it did (if you get what I mean).  

Finally, the fact that this fellow got two incorrect answers before I
came along shows why one should ask general questions about C or C++ in
comp.lang.{c,c++}, and questions about gcc in gnu.gcc.*.  I don't keep
saying this because it bugs me (I'll end up reading 'em anyway); I say
it because you're more likely to get a good answer if you ask in the
appropriate forum.