strangest systems I've sent email from
turing at shaw.ca
Tue May 24 15:53:25 CDT 2016
Without an MMU or a segmentation scheme, 16-bits = 64K.
The 68000 is not a 16-bit processor, it's 32-bit, and exposed (ISTR) a 24-bit address.
20-bits = 1M addresses, 24-bits = 16M addresses.
You're confusing data bus width (8-bit) with address bus width (16-bit).
----- Original Message -----
From: "Liam Proven" <lproven at gmail.com>
To: "General Discussion: On-Topic and Off-Topic Posts" <cctalk at classiccmp.org>
Sent: Tuesday, May 24, 2016 1:29:48 PM
Subject: Re: strangest systems I've sent email from
On 22 May 2016 at 04:52, Guy Sotomayor Jr <ggs at shiresoft.com> wrote:
> Because the 808x was a 16-bit processor with 1MB physical addressing. I
> would argue that for the time 808x was brilliant in that most other 16-bit
> micros only allowed for 64KB physical.
Er, hang on. I'm not sure if my knowledge isn't good enough or if that's a typo.
AFAIK most *8* bits only supported 64 kB physical. Most *16* bits
(e.g. 68000, 65816, 80286, 80386SX) supported 16MB physical RAM.
Am I missing something here?
I always considered the 8088/8086 as a sort of hybrid 8/16-bit processor.
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