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-boun...@classiccmp.org> on behalf of Christian Gauger-Cosgrove <captainkirk...@gmail.com> Sent: Wednesday, December 23, 2015 7:14 PM To: clasyst...@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 <clasyst...@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.
