Path: utzoo!attcan!uunet!lll-winken!ames!eos!shelby!portia.stanford.edu!hobbes
From: hobbes@portia.Stanford.EDU (Andrew Zimmerman)
Newsgroups: comp.sys.handhelds
Subject: Re: Machine code on the 28S (and 48SX)
Message-ID: <1990Jun4.222439.21761@portia.Stanford.EDU>
Date: 4 Jun 90 22:24:39 GMT
References: <55022@microsoft.UUCP>
Organization: AIR, Stanford University
Lines: 55

A couple of questions of the hp48sx assembly language.  My sources for these
questions are notes by Alonzo Gariepy for the hp-28s and the recent note
about the 48sx.

#1 The new document gives XLIB numbers for all of the command.  ie for
RR, it gives a number of 003002.  When I look at the table in 74600, where
it appears that there is a table of commands, the table looks more like

	DB	02	; number of characters in string
	DB	"RR"	; the string
	D3	003	; XLIB# ?
	DB	02	; number of characters in next string
	DB	"SL"	; the next string
	D3	005	; XLIB# ?

(Note: XLIB 4 is skipped, because the table is sorted by the number of
characters, and 004 is RRB)

So, where do the extra 3 nybbles come from in the document.

#2 What is register D1 used for?  The documents say that Register D1 points 
to the end of the stack growing down toward the heap.  I interpret this to
mean that D1 is the stack pointer.  But, in disassembling the code for
peek, we have a sequence of instructions that have
MOVE.A	@D1,A
MOVE.A	A,D0
ADD.a	10,D0
MOVE.A	@D0,C
SWAP.A	C,D0
MOVE.W	@D0,C

from this code, I get the impression that D1 is either a pointer to the 
stack pointer, or the instruction should be MOVE.A D1,A.

#3 If I look at the code for ASR at 1957B, I see D9D2.. which appears to be
D9   MOVE.A	B,C
D2   CLR.A	C
Is this correct?  Seems it is clearing reg C after putting something in it.

#4 Does anyone know where the processor starts execution?  I am guessing that
it starts at 00000.  (Sorry all you motorola people out there :-)) 
If I do an EVAL at 00000, it seems that the hp48 comes up in normal mode.
But, if you look at the code, there is a jump as the second instruction.
This is supposed to be a JUMP.3.  Assuming that was the case, I went to the
address and did an eval.  Didn't seem to work.  But, if I assume that it is
a 4 nybble address, this address is valid, and the machine come up.  Also,
If I do an eval right after the JUMP, assuming 4, the machine does come
up after doing a memory clear.

#5 How was the instruction set determined?  Boy, I can't find any fields in
it, and those it appears to have move around.

Thanks for any help,
Andrew
zimmer@calvin.Stanford.edu