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From: glenn@gla-aux.uucp (Glenn Austin)
Newsgroups: comp.sys.mac.misc,comp.windows.ms
Subject: Re: Mac Vs. Windows? (sorry)
Message-ID: <0E010021.a4dwre@gla-aux.uucp>
Date: 6 Jun 91 04:54:51 GMT
Reply-To: glenn%gla-aux.uucp@skinner.cs.uoregon.edu
Organization: The Pit Lane
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In article <1991Jun4.154854.19649@dbase.A-T.COM>, tomr@dbase.A-T.COM (Tom Rombouts) writes:
> In article <0E010021.e0mxxc@gla-aux.uucp> glenn%gla-aux.uucp@skinner.cs.uoregon.edu writes:
> >
> >In article <1991May18.050842.5732@cs.uoregon.edu>, akm@obelix.cs.uoregon.edu (Anant Kartik Mithal) writes:
> >> So, my question is: if the Mac is faster than a "comparable" pc in
> >> execution, what is the reason? Is it ROM? (I don't think so for the
> >> reasons listed above.) What other things can make a difference?
> >
> >Two things -- processor effeciency and levels of routines.
> 
> [ rest of post deleted ]
> 
> >2) The Motorola (and for that matter, any linear-addressing processor) is
> >faster than segmented memory.
> 
> Correct me if I am wrong, but in small (or tiny) memory model apps,
> the 80x86 is essentially a linear-addressing processor.  I would tend
> to think it would be faster shuffling 16-bit addresses (since the
> segment registers do not change) than a Motorola with (is it?) 32-bit
> addresses.

It would be -- if they dealt with 16-bit (or wider) fetches, but because
of the roots of the 80x86 in the 8080, and the fact that extensions beyond
the 8086 generally have a single prefix byte (which must be fetched before
it can determine that it IS an extension to the 8086 instruction set), the
80x86 tend to be inefficient in memory accesses.  This is (once again) why
zero-wait-state RAM has more benefits on the 80x86 than it does on the Motorola
680x0 processors (which do long word and word fetches).

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| Glenn L. Austin                | "Turn too soon, run out of room.           |
| Macintosh Wizard and           |    Turn too late, much better fate."       |
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