I suspect I’m missing something very basic here, as I just started looking at the manual and tutorial for Chatscript.

I took the basic simpletopic.top that comes with Chatscript (v 2.93 on windows ) and that does the basic questions shown in the tutorial, and tried to replace the “what is your name” rule with this modified one:

#! what is your name $bot = harry
u: ($bot=harry what is your name) My name is Harry Potter—what’s yours?
a: (My name is _*1 _*1 >) $firstninny = ‘_0 $lastnanny = ‘_1 Hello $firstninny $lastnanny .

This syntax is taken out of the manual, page 21, aside from the output. I believe the characters at the start of ‘_0 and ‘_1 are indeed ASCII character 39,  an apostrophe.

In order to run this multiple times I also had to modify the INTRRODUCTIONS topic to add noerase and repeat:

topic: ~INTRODUCTIONS noerase repeat (~emogoodbye ~emohello ~emohowzit name here )

That’s all I changed.  Here’s what I get, on two different machines:

My name             output name
============  =========
a b                 a B
A B                 a B
aaa bbb             AAA B
one two             rule does not fire
zdkjf zdkfjs         passed through correctly
Wade Schuette       rule does not fire
wade schuette       Wade Schutte   (  changed “ue” to “u” in the last name)

I would have expected (as a newbie) that by single quoting ‘_0 and ‘_1 that I would have
bypassed all pre-processing of the first and last name, but this is clearly not true.

What else do I need to do to simply capture, correctly, the two input words, and print them (capitalized or not) without other change in the output stream?

Thank you!

Wade

I had the same issue with that one as well. Wish I could help.

First, you’d have to block all lower level preprocessing done on tokens. In simplecontrol.top,the bot definition of harry has this:
$token = #DO_INTERJECTION_SPLITTING | #DO_SUBSTITUTES   | #DO_NUMBER_MERGE   | #DO_PROPERNAME_MERGE | #DO_SPELLCHECK | #DO_POSTAG | #DO_PARSE

You would need to (for simplicity) make that $token = 0

Then it will not try to merge tokens that are proper names, or correct spellings of input that do not show up in the dictionary.

Ahh,  thank you.
Is $token just a user variable that I can change safely for the duration of one question, say? Or or there other side-effects I should know about?

The following seems to work.  Is this construction fairly safe?

#! what is your name $bot=harry
u: ($bot=harry what is your name) $oldtoken = $token $token = 0 My name is Harry Potter—what’s yours? Token is now $token . Oldtoken is saving value $oldtoken .
a: (My name is _*1 _*1 >) $firstname = ‘_0 $lastname = ‘_1 $token = $oldtoken Hello, $firstname $lastname . Token was restored to $token for the happy case.
      a: ($bot=harry) $token = $oldtoken Greetings, whoever you are.  Token was restored to $token for all other cases.

Yes, $token is just a user variable (albeit a privledged one) that you can change for any duration. In fact, you can change it after the fact.  Because things like user names are OFTEN a problem, when my asks for a user’s name, it alters $token for that volley, as you propose.  But also, if it detects something like the user pattern s: (my name is)  then it changes $token and does a ^retry(SENTENCE) to force a reanalysis of the sentence in the new format.  It can tell on the second pass that it has already done this because $token is now different from the default one.

What a great idea!  Thank you, Bruce!