Xref: utzoo comp.sys.mac.misc:12952 comp.windows.ms:13579
Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!think.com!sdd.hp.com!mips!apple!voder!nsc!amdahl!rtech!dosbears!mikel
From: mikel@dosbears.UUCP (Mike Lipsie)
Newsgroups: comp.sys.mac.misc,comp.windows.ms
Subject: Re: Mac Vs. Windows? (sorry)
Message-ID: <412@dosbears>
Date: 6 Jun 91 14:03:24 GMT
References: <0E010021.e0mxxc@gla-aux.uucp> <1991Jun4.154854.19649@dbase.A-T.COM> <20@ryptyde.UUCP>
Reply-To: dosbears!mikel@pyramid.com (Mike Lipsie)
Organization: DOS Bears
Lines: 29

In article <20@ryptyde.UUCP> dant@ryptyde.UUCP (Daniel Tracy) writes:
>
>Well, not really. I'm assuming that even though the segment "window" doesn't
>have to be moved, the accessing routine is still slower. The window doesn't
>change, but it's still there. In order to request a memory address, a segmented
>chip must do an extra multiply (at least) every time. That is, it multiplies
>the address location within any segment by the segment being used, so it doesn't 
>act as linear-addressing at all. 

Er.  This explanation makes me worry about the rest of your logic.  It does
NOT do a multiply; it does an ADD.  The segment register multiplied by 16
is added to the offset register.  In 8086 mode this means the _equivalent_ 
of a four bit shift to the segment register.

>                                   The 386+ DOES have a fairly good linear
>memory mode, but this isn't used by any OS's except for Unix. Also, there are
>many other factors in 680x0 vs 80x86. Intel's line doesn't even come close
>in most cases to MC's chip line. Also, because the 80x86 began as 16-bit chips
>internally, many kludges had to be performed in order to maintain compatibility
>across the line, and thus the 680x0 line is also accelerating past the 80x86
>at a faster clip because it isn't being held back by mode up mode implimentation. 

You can't be "better" and then "pass the competition".  Make up your mind.


--
Mike Lipsie
dosbears!mikel@pyramid.com
mikel%dosbears.uucp@ingres.com