80186 and now AMY chip

From: ajp166 <ajp166_at_bellatlantic.net>
Date: Fri Nov 24 14:13:33 2000

-----Original Message-----
From: Eric Smith <eric_at_brouhaha.com>
To: classiccmp_at_classiccmp.org <classiccmp_at_classiccmp.org>
Date: Friday, November 24, 2000 3:27 AM
Subject: Re: 80186 and now AMY chip


>"Eric J. Korpela" <korpela_at_ellie.ssl.berkeley.edu> wrote:
>> You missed the important one, and the only one that counts
>> when determining the bitness of the processor, the width of
>> the (integer) ALU.
>>
>> Clearly, regardless of the width of the registers and the
>> address bus size, the Z80 is an 8 bit processor, as is the
>> 6502. The 8088 and 8086 are both 16 bitters. The 68000 and
>> 68010 are 16 bitters. The 68020 is a 32 bitter as is the
>> x386.
>
>Actually, using this metric, the 8080 and Z-80 are 4-bit processors,


Where in the world did that come from? The base data paths for z80
and 8080 are 8bits. Key words are "data paths" as the alu on 8080
is basically 8bits (with microcoded 16 bit ops as two sequential 8bit
ops).

Usually the size of the data word is the metric. What confuses the
subject often is the instruction word size (4004 is really 8bit then)
or the ability to manipulate smaller than word size (VAX and pdp-11
byte ops).

>Try asking programmers what width processors are instead of hardware
>engineers. They'll tell you that the 8080 and Z-80 are 8-bitters,
>and that the 68000 and IBM 360 (most models) are 32-bitters.


generally agreed. It when you start playing with math ops and
addressing range that the "size" is noticeable and significant.

However if you look at most cpus there is a basic word size and
alu size that are amoung the defining parameters.

Allison
Received on Fri Nov 24 2000 - 14:13:33 GMT