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From: friesen@psivax.UUCP (Stanley Friesen)
Newsgroups: net.lang.c
Subject: Re: Byte order (retitled)
Message-ID: <1104@psivax.UUCP>
Date: Wed, 16-Apr-86 16:33:14 EST
Article-I.D.: psivax.1104
Posted: Wed Apr 16 16:33:14 1986
Date-Received: Fri, 18-Apr-86 21:04:29 EST
References: <7046@cca.UUCP>  <7158@cca.UUCP>
Reply-To: friesen@psivax.UUCP (Stanley Friesen)
Organization: Pacesetter Systems Inc., Sylmar, CA
Lines: 25

In article <7158@cca.UUCP> g-rh@cca.UUCP (Richard Harter) writes:
>>
>>Many implementors of C on big-endian machines have had to cope with
>>the pointer conversion problem, which is simpler on little-endians.
>>
>	Color me skeptical.  You may be right, but I would like to
>	see more details.  In the model of pointers that it occurs
>	to me to implement, big-Endian and little-endian come out
>	the same.

	Could you expand on this! Do you mean that if you cast a
pointer to a long to a pointer to a short and dereference you will get
the *high* order portion on a big-endian machine and the *low* order
portion on a little-endian? Clearly a portability problem, and the the
big-endian behavior is counter-intuitive. Or do you intend that the
pointer always point to the low-order byte even on a big-endian
machine? Then you have to index *backwards* to break the item up into
bytes! Really, the only way to get the rigth semantics on a big-endian
machine is to actually convert pointers when a cast is used.
-- 

				Sarima (Stanley Friesen)

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