Re: The Apple ][ was cool!

[dempson@actrix.gen.nz (David Empson)]

william strutts <wrstrutts1@nospam.home.com> wrote:

> "Roy Miller" <millers@inebraska.com> wrote in message
> 3B5522F2.662F33B1@inebraska.com">news:3B5522F2.662F33B1@inebraska.com...
> >
> > Huh? The 8088 was a 16 bit CPU, with an 8 bit data bus, and a 20 bit
> > address bus.
> > The 8086 is a 16 bit CPU with a 16 bit data bus, and a 20 bit address bus.
> > Neither was an 8 bit CPU. An 8 bit CPUs have (normally) 16 bit address
> > lines.
>
> History calls it an 8-bit computer due to the 8 bit data bus.  The same
> can be said for the 6809 which was also a 16 bit processor with an 8-bit
> data path.

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.


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