Author: Eugene Nalimov
Date: 10:34:30 03/10/00
Go up one level in this thread
On March 10, 2000 at 03:21:31, Tom Kerrigan wrote:
>On March 09, 2000 at 23:07:17, Robert Hyatt wrote:
>
>>that's the wrong way to measure this. the 8080 was a 16 bit cpu if you use
>>that logic. The 68000 had a 16 bit bus. I taught hardware design courses at
>
>16 bit external data bus, yeah. But how wide were the register busses? How wide
>was the ALU? I believe the answer to both questions is 32-bit.
add.w d0, d1 - 4 clocks
add.l d0, d1 - 6 clocks
Exactly the same for and/or/xor, sub/cmp, clr, etc. The only exception is mov -
register-to-register is always executed in 4 bytes, regardless of the size.
Eugene
>And BTW, the 286 used similar "register pairs" as the 8080 (AX = AL , AH). Does
>that mean the 286 is 8-bit? You can go either way with your argument logic.
>
>-Tom
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