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" <lpro...@gmail.com> To: "General Discussion: On-Topic and Off-Topic Posts" <cctalk@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 <g...@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. -- Liam Proven • Profile: http://lproven.livejournal.com/profile Email: lpro...@cix.co.uk • GMail/G+/Twitter/Flickr/Facebook: lproven MSN: lpro...@hotmail.com • Skype/AIM/Yahoo/LinkedIn: liamproven Cell/Mobiles: +44 7939-087884 (UK) • +420 702 829 053 (ČR)
