From owner-freebsd-current  Tue Oct 10 13:09:54 1995
Return-Path: owner-current
Received: (from root@localhost)
          by freefall.freebsd.org (8.6.12/8.6.6) id NAA02243
          for current-outgoing; Tue, 10 Oct 1995 13:09:54 -0700
Received: from rocky.sri.MT.net (sri.MT.net [204.94.231.129])
          by freefall.freebsd.org (8.6.12/8.6.6) with ESMTP id NAA02231
          for <current@freebsd.org>; Tue, 10 Oct 1995 13:09:50 -0700
Received: (from nate@localhost) by rocky.sri.MT.net (8.6.12/8.6.12) id OAA22172; Tue, 10 Oct 1995 14:12:01 -0600
Date: Tue, 10 Oct 1995 14:12:01 -0600
From: Nate Williams <nate@rocky.sri.MT.net>
Message-Id: <199510102012.OAA22172@rocky.sri.MT.net>
To: "Garrett A. Wollman" <wollman@lcs.mit.edu>
Cc: Nate Williams <nate@rocky.sri.MT.net>, current@freebsd.org
Subject: Re: tail dumps core
In-Reply-To: <9510102001.AA10553@halloran-eldar.lcs.mit.edu>
References: <9510101735.AA10192@halloran-eldar.lcs.mit.edu>
	<199510101836.LAA10827@phaeton.artisoft.com>
	<9510101901.AA10410@halloran-eldar.lcs.mit.edu>
	<199510101943.NAA22092@rocky.sri.MT.net>
	<9510102001.AA10553@halloran-eldar.lcs.mit.edu>
Sender: owner-current@freebsd.org
Precedence: bulk

> > So, what's your point?  Other than some comments in this set of lines
> 
> > which implies that memset doesn't have to set an integer to all zeros,
> > what's the problem?  How does the C standard allow for memseting an int
> > to be non-zero?
> 
> The same way it allows for memsetting a pointer to be non-NULL: it
> makes no guarantees whatsoever about the underlying bitwise
> representation of its data types in memory.

You're implying that a 'int i = 0' doesn't assume that the memory
location doesn't contain the bytes '0x0000', correct?

The prototype of memset() implies that it will indeed set all of the
bytes of the underlying data type to the byte value specified in the
second arguement.

> My point in saying this is not to say that for all practical purposes
> you can't depend on this, but rather to say that arguments suggesting
> that `calloc' is somehow ``more portable'' are completely specious.

I disagree.  It is common to assume that sizeof(char) == 8 bits, but
this isn't specified anywhere either but most code written would break
if this were not true.

Using 'calloc()' is much more portable than using bzero(), simply
because bzero doesn't exist on many OS's while memset() does.


Nate