16 v. 32 bit

[Dave Lewis]

In article <730 at brl-tgr.ARPA> bader at b.psy.cmu.edu (Miles Bader) writes:
>Well, the 8088 *IS* a 16 bit processor internally-- i.e., 16 bit data paths,
>16 bit alu.  But the 68000 is ALSO a 16 bit processor internally-- 16 bit
>data path and alu.  It just happens to have operations on longs built in...
>
>Not that I like the 8088, of course.
>
>[processor wars!]
>				-Miles

 [foot in mouth, you mean]

  The Motorola 68000 family of microprocessors ALL implement full 32-bit
internal architecture. That means 32-bit registers, 32-bit data paths, and
a 32-bit ALU. I recommend that you obtain FACTS next time instead of posting
hearsay to the net. Try reading the Motorola MICROPROCESSOR COMPONENTS DATA
MANUAL (or Hitachi's -- they're a second source) and then say "32 bits"
until you're convinced.

  The 8088 is indeed a 16-bit machine (very strangely put together, it's true)
which is then squished down onto an 8-bit data bus. Seems Intel wanted to make
it `compatible' with the old 8-bit machines.

  An aside -- once some people (no names) with an Intel fixation tried to build
a high-performance processor board. Two 8086's and an 8087 didn't do the job
but a 68000 all by itself did. Tell you anything?

-------------------------------
		Dave Lewis    Loral Instrumentation   San Diego

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