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From: Florian Weimer <fw@deneb.enyo.de>
Newsgroups: comp.lang.ada
Subject: Re: ada vs. cpp
Date: Tue, 16 Oct 2001 22:13:33 +0200
Organization: Enyo's not your organization
Message-ID: <878zebmjgy.fsf@deneb.enyo.de>
References: <slrn9semrm.4l2.whraven@raven.wri> <9q7na102nqn@drn.newsguy.com>
 <slrn9sll0g.i2.lutz@taranis.iks-jena.de> <3BCBF40A.BBDD4FE6@free.fr>
 <slrn9so1pm.hu.lutz@taranis.iks-jena.de> <87669foaf2.fsf@deneb.enyo.de>
 <uheszs7on.fsf@gsfc.nasa.gov>
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Date: 2001-10-16T22:13:33+02:00
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Stephen Leake <stephen.a.leake.1@gsfc.nasa.gov> writes:

> Florian Weimer <fw@deneb.enyo.de> writes:
>
>> lutz@iks-jena.de (Lutz Donnerhacke) writes:
>> 
>> > :3 When applied to an operand that has type char, unsigned char, or signed
>> > :  char (...) the result is 1. When applied to an operant that hast array
>> > :  type, the result is the total number of bytes in the array. ...
>> >
>> > Consequence: No compiler is a conforming one when a char is not a byte?!
>> 
>> Yes, of course.  Unfortunately, this way, a low-level concept (storage
>> unit) and a high-level one (character set) are strongly linked, which
>> is a rather bad idea.
>
> Actually, a C compiler with a 16 bit char is conforming; this
> statement in the C standard simply tells you that a size 1 for this
> compiler means 16 bits.
>
> Remember, there is nothing "standard" about an 8 bit byte.

The revision of POSIX.1 will require 8 bit bytes, I think, but of
course, there's no such requirement in the C language (except that the
size must be greater than 7 bits).