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From: akcs.joehorn@hpcvbbs.UUCP (Joseph K. Horn)
Newsgroups: comp.sys.handhelds
Subject: Re: Assembly questions for HP48
Message-ID: <2777de06:1489.1comp.sys.handhelds;1@hpcvbbs.UUCP>
Date: 25 Dec 90 23:40:04 GMT
References: <6488@uqcspe.cs.uq.oz.au>
Lines: 31

Paul Dale (grue) asks:

> Which of the following two pieces of code will run more quickly?
>   clr.a   c
>   move.p2 10, c
> or
>   move.p5 10, c
> 
> The first is shorter by one nibble than the second.
> 
> Also what happens if we change the field in the first code section
> to .p1 or .p3 ??

Unfortunately, your question is in Alonzo mnemonics, and the answer
is found in the HP-71 IDS Volume 1, which is in HP mnemonics, so I
had to "translate" your question before I could answer it...

"clr.a c" is "C=0 A" (Clear C, address field), whose opcode is D2.
This instruction executes in 7 clock cycles.  "move.p2 10,c" is
"LCHEX 10" (Load C with hex 10 starting at the P pointer position),
whose opcode is 3101.  This instruction executes in 5 clock cycles.
Therefore these two instructions take 12 cycles together.

"move.p5 10,c" is "LCHEX 00010", opcode 3401000.  This instruction
executes in 8 clock cycles, and is therefore one nib longer but much
faster than the other two together.

To answer you second question, the execution times of "move.pn hh,c"
instructions, where n is the number of hex digits (hh) to move into
register C, is n+3 cycles.

--  Joseph K. Horn  --  (714) 858-0920  --  Peripheral Vision, Ltd.