80186 and now AMY chip

From: Eric Smith <eric_at_brouhaha.com>
Date: Fri Nov 24 18:17:46 2000

"Eric J. Korpela" <korpela_at_ellie.ssl.berkeley.edu> wrote:
> You missed the important one, and the only one that counts
> when determining the bitness of the processor, the width of
> the (integer) ALU.
>
> Clearly, regardless of the width of the registers and the
> address bus size, the Z80 is an 8 bit processor, as is the

I wrote:
> Actually, using this metric, the 8080 and Z-80 are 4-bit processors,

Allison wrote:
> Where in the world did that come from? The base data paths for z80
> and 8080 are 8bits. Key words are "data paths" as the alu on 8080
> is basically 8bits (with microcoded 16 bit ops as two sequential 8bit
> ops).

Eric said "width of the (integer) ALU", not "data paths". Note that I
said in my message that I thought that was a useless metric.

According to its designers, the ALU on the 8080 is 4 bits wide, and
takes 2 cycles for an 8-bit add or subtract. Presumably it takes at
least 4 cycles for a 16-bit add or subtract. The 8080 takes so many
cycles for *anything* that it's not obvious what it does internally on
any given cycle.

Eric
Received on Fri Nov 24 2000 - 18:17:46 GMT