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| From: Alex Grosholz
Date: 2000-07-25
Subject: RE: Interrupts
|
-> On Tue, 25 Jul 2000, Zsolt Prievara wrote:
->
-> > > Read-modify-write instructions write the unmodified value
-> back first. So,
-> > > the ASL $ff09 will first write $82, then $04.
-> >
-> > This is very interesting. ASL $xxxx is a 6 cycles long
-> instruction. How do
-> > the
-> > instuctions work? What happen exactly in the cycles?
-> > Is there any recipe for the instruction execution?
->
-> Read the 64doc file advertised earlier today. :-) The first two
-> cycles of
-> all instructions are the same: fetch from PC and PC+1. ASL $ff09 goes
-> like this, assuming that $ff09 contains $82:
->
-> # addr R/W data
-> 1 PC R 0e
-> 2 PC+1 R 09
-> 3 PC+2 R ff
-> 4 ff09 R $82
-> 5 ff09 W $82
-> 6 ff09 W $04
->
-> Marko
true....true...
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