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From: dewar@cs.nyu.edu (Robert Dewar)
Newsgroups: comp.lang.ada
Subject: Re: Ada95 - Determining Machine Byte Order
Date: 5 Feb 1995 22:34:18 -0500
Organization: Courant Institute of Mathematical Sciences
Message-ID: <3h45bq$fev@gnat.cs.nyu.edu>
References: <3gpo58$g48@ixnews3.ix.netcom.com>
 <3gvudv$k4b@liberator.et.tudelft.nl>
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Date: 1995-02-05T22:34:18-05:00
List-Id: <comp.lang.ada>

Not only is Mark's solution:

        int n = 1, little_endian = (*((char *)&n) == n);

not very clear (though if you study it for a second and can find your
way around * and & in C, you can see how it works), but more significantly
there is no guarantee that this will work, it is equivalent to the use of
unchecked conversion in Ada, and equally problematic. Mark if you don't
believe me, try seeing what this gives on a Cray, or if you want to take
the theoretical approach, tell me where in the ANSI standard it is written
that this will do something well defined.

All in all I think that the Ada way of doing things:

   Little_Endian : constant Boolean :=
        System.Default_Bit_Order = System.Low_Order_First;

maybe just a *little* bit clearer :-)