[TUHS] Algol68 vs. C at Bell Labs

Peter Jeremy peter at rulingia.com
Sun Jul 3 09:32:17 AEST 2016

On 2016-Jun-30 22:21:27 +0300, Diomidis Spinellis <dds at aueb.gr> wrote:
>First, the 8088 was a 16-bit CPU with an 8-bit data bus in a cheap 
>40-pin package.  This halved the number DRAM chips required and allowed 
>the IBM PC to be easily designed along existing easily-available 8-bit 
>peripherals.  In contrast the 68000 had a 16-bit data bus in a more 
>expensive 64-pin package.  Remember that in the 1980s glue logic was 
>implemented through simple TTL chips, so adopting the 68000 might have 
>doubled the number of chips on the motherboard.

My understanding was that the 8-bit bus was a requirement so IBM could
have a 64KB base model using the then new 64k×1 chips.  IBM also
emasculated the PC so it didn't compete with their existing minis.  The
68008 wasn't available until later (and this would explain why Motorola
pushed the 6809 as a solution).

Both the 8086 and M68k could relatively easily use 8-bit peripherals
(both Intel and Motorola had a range of 8-bit peripherals that they
didn't want to make instantly obsolete).

>In addition, the 8086 architecture was an extension of the 8080 one, 
>which made it easier to make the MS-DOS API compatible with the CP/M 

Since IBM was buying the software, I'm not sure how much of a driver
this was.  Definitely, porting from 8080 to 8086 was easier but writing
from scratch would be far easier on M68k.

Peter Jeremy
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