23 Dec
2015
23 Dec
'15
11:02 p.m.
Ok, I think I get it
The acc is one value. Both arg1 and agr0 are init to the second value ( start the same )
I'm not sure how to save a value to the acc on a pdp-8. I know how on my Nicolet 1080
that has a similar ALU.
On my Nicolet:
MEMA ARG0
A+MM ARG0
ANDAM ARG1
A+MA ARG1
ANGA
A+MAM ARG0 ( acc and ARG0 have the xor of the original ARG0 and ARG1 )
I suspect it is similar on a pdp-8 as they have similar ALUs
Dwight
________________________________________
From: cctalk <cctalk-bounces at classiccmp.org> on behalf of dwight <dkelvey at
hotmail.com>
Sent: Wednesday, December 23, 2015 10:38 PM
To: General Discussion: On-Topic and Off-Topic Posts
Subject: Re: Has Anyone Written PDP-8 .XOR. Code Using the MQ Register (Without the EAE)?
I'm not sure how to code it on the machine but I know
how to do it.
Take A AND B = C
2* C = D ( ignore carry )
negate D ( complement and add 1 )
A + B = E ( this could have been done first )
E + D = A XOR B
I think it needs 3 variables?
It works because the carries are the result of the AND of the times 2 ( his RAL )
negating and subtracting is what he want to do, leaving the XOR.
Dwight
________________________________________
From: cctalk <cctalk-bounces at classiccmp.org> on behalf of Christian
Gauger-Cosgrove <captainkirk359 at gmail.com>
Sent: Wednesday, December 23, 2015 7:14 PM
To: CLASystems at gmail.com; General Discussion: On-Topic and Off-Topic Posts
Subject: Re: Has Anyone Written PDP-8 .XOR. Code Using the MQ Register (Without the EAE)?
On 23 December 2015 at 13:44, CLASystems <clasystems at gmail.com> wrote:
Ironically, the shortest and fastest seems to be
avoidance of the MQ
altogether [thus making it work on ANY model].
TAD ARGONE
AND ARGTWO
CLL RAL
CIA
TAD ARGONE
DCA ARGTWO
This works because .XOR. is addition ignoring the carry bits. So, knowing
they will happen, just allow them at first, then remove them.
Hmm, I just tried that in SIMH, and that doesn't XOR at all. I haven't
a clue what it does.
What I have entered:
sim> ie -m 100-105
100: TAD 76
101: AND 77
102: CLL RAL
103: CIA
104: TAD 76
105: DCA 77
Locations 076 and 077 being ARGONE and ARGTWO respectively, at the start:
sim> ie 076-077
76: 1234
77: 4321
After running the above code sample:
sim> ie 076-077
76: 1234
77: 0574
If we "flip" ARGONE and ARGTWO's values (to 4321 and 1234 respectively):
sim> ie 076-077
76: 4321
77: 3661
Neither of those is the expected 5115 of an XOR operation.
Am I missing something blindingly obvious?
Cheers,
Christian
--
Christian M. Gauger-Cosgrove
STCKON08DS0
Contact information available upon request.