Date: Thu, 15 Jan 1998 13:17:55 +0200 (IST)
From: Eli Zaretskii <eliz AT is DOT elta DOT co DOT il>
To: Nate Eldredge <eldredge AT ap DOT net>
cc: Noam Rotem <nrotem AT johnbryce DOT co DOT il>, djgpp AT delorie DOT com
Subject: Re: Bit fields in djgpp
In-Reply-To: <199801150431.UAA16364@adit.ap.net>
Message-ID: <Pine.SUN.3.91.980115131731.12130P-100000@is>
MIME-Version: 1.0
Content-Type: TEXT/PLAIN; charset=US-ASCII
Precedence: bulk


On Wed, 14 Jan 1998, Nate Eldredge wrote:

> >Sorry, I was wrong.  The ANSI C Standard explicitly says that signed bit 
> >fields of size N can be used to represent values in the range [0, 2^(N-1))
> >so when N is 1, you cannot represent 1.  You need to make it unsigned, as 
> >Nate suggested.
> Oughtn't that to be:
> [0..(2^(N-1))-1] ?
> Because, for instance, a 16-bit signed value can only go up to 32767.
> 
> Also, 2^(N-1) for N=1 equals 2^0 = 1.

Here's what my ANSI C references say (the last column is mine):

       Designation         Minimum range	  For N=1

       int or none	   [0, 2^(N-1))		  [0, 1)  i.e. only 0

       signed or	   (-2^(N-1), 2^(N-1))	  (-1, 1) i.e. only 0
       signed int

       unsigned or	   [0, 2^N)		  [0, 2)  i.e. 0 and 1
       unsigned int

So it seems to me that the only useful way to have a single-bit bit
field is to declare it unsigned.  Of course, the above only specifies
the *minimum* range, so an implementation could behave otherwise, but
if you want to be sure it works, make it unsigned.