Re: The Apple ][ was cool!

["william strutts" <wrstrutts1@nospam.home.com>]

"David Empson" <dempson@actrix.gen.nz> wrote in message
1ewrul3.1lppa9z118k0snN%dempson@actrix.gen.nz">news:1ewrul3.1lppa9z118k0snN%dempson@actrix.gen.nz...
<snip>
> I'd dispute that.  The 6809 has an 8-bit accumulator and ALU, and uses
> an internal 8-bit data path.  It combines 8-bit accumulators to form a
> 16 bit pair, which allow a smaller set of 16-bit operations.  It also
> has the address space typical of an 8-bit processor (64KB).  Even
> Motorola call it an 8-bit micro in the data sheet.
>
> Sounds like an 8-bit micro to me.

Well, I would suppose if we refer to the ALU as the primary
factor. I would have to agree.  I was referring more to the
fact that the processor had built-in 16-bit operations via
register combining.  It was marketed back in the 80's as
an 8/16 bit processor.  The later 68HC11 would further
extend the 16 bit features of the 6809 and move it into
the realm of 1 meg address space and more 16-bit
operations making it a 16-bit computer.

>
> There can be no argument about the 8086 - it is definitely a 16-bit
> micro.

No dispute there.  It was always amazing to me that IBM went
with the more crippled processor to save a few bucks in circuit
board design by using 8-bit components.

>The 8088 is identical internally, but is constrained by an
> external 8-bit data bus.  This may make it an 8-bit micro by some
> definitions, but I'd still call it a 16-bit micro.

The data bus was definitely 8-bit hence we had 8-bit ISA slots
which to me would say that the machine is an 8-bit computer.
With the advent of the 286 AT, we had 16-bit data bus which
lead us to the new standard 16-bit ISA slots.   The perception
of the original XT was that it was an 8-bit compter.  Or more
precisely stated an 8/16 bit computer and the 286 AT was a
16-bit computer.

>
> The 68000 is an interesting case.  It has 32-bit registers and a 16-bit
> external data path, which suggests that it should be classed as 32-bit.
> I don't have a Motorola datasheet handy, but Rockwell second-sourced it,
> and they call it a 16-bit micro.

I think the 16-bit data path is what pegged it as a 16-bit processor
in most people's minds.  It again is a 16/32 bit processor to be more
precise.  I think many companies failed to captialize on the fact that
the 68k processor was 32 bit.  By marketing it as a 16-bit, it made
it on par with the 8086.  Apple would have done a better job if they
marketed the machine as a 32-bit computer.  (That problem would
disappear when the 68020 and 68030 machines appeared.) It would
have then appeared to the public that the Mac was more advanced
than the other 16 bit computers of its day.  I think they were afraid
that the 16-bit bus could back fire on them if they marketed the Mac
as a 32-bit computer.

> I can't see any clear indication, but I expect it has a 16-bit ALU.  For
> data already in registers, it generally takes longer to do 32-bit
> calculations than 16-bit ones, even though the instructions are the same
> length.  It appears to have an internal 32-bit data path, since there is
> no timing difference between 16-bit and 32-bit move operations between
> registers.

Check this link out.  It explains about the ALU's for the 68000

http://bwrc.eecs.berkeley.edu/CIC/archive/cpu_history.html#Sec3Part4

The 68000 had two 16-bit ALU's which were combined to make
a 32-bit ALU which makes for slower 32-bit operations.

--
--
William R. Strutts - wrstrutts1@nospam.home.com - Whatever!

C'est moi!  http://www.facelink.com/wrstrutts

Just hacking away...

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