Re: (< @X 18) doesn't behave as expected with pilog (SOLVED: short
Hi Eric, On Sun, Nov 13, 2016 at 09:32:20PM +0100, CILz wrote: > I've just created an account on the wiki however I think I can't add > something in the reference part. I think that this how-to could fit very > well here: > > http://software-lab.de/doc/ref.html#pilog > ... Very good examples and explanations! However, I think for the general reference your text is a bit too specialized, more like a tutorial. A (perhaps just short) article in the Wiki would be much better, wouldn't it? ♪♫ Alex -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: (< @X 18) doesn't behave as expected with pilog (SOLVED: short
Hi Eric, > Hum ! I just noticed that I may have been too friendly here, sorry. I wanted > to say "Hi Alexander". My apologies. Best, Eric Not at all! "Alex" is perfectly all right :) ♪♫ Alex -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: (< @X 18) doesn't behave as expected with pilog (SOLVED: short
Hi Alex, I've just created an account on the wiki however I think I can't add something in the reference part. I think that this how-to could fit very well here: http://software-lab.de/doc/ref.html#pilog to illustrate the last part which starts with " Pilog can be called from Lisp and vice versa:" After the last sentence and just before the horizontal rule, it could be word like this: = To illustrate this, let's say that you have those two facts in a Pilog database: (be age (Paul 19) ) (be age (Kate 17) ) and that you want to find the person under 18. In full Prolog you may have written something like this: underage(X) :- age(X,Y), Y < 18. however in Pilog the following rule: (be underage (@X) (age @X @Y) (< @Y 18) ) won't work and the query: (? (underage @X) ) will yield to 'NIL' instead of the expected result '@X=Kate' . The reason is that '<' (less than) is not Pilog function but only a Lisp one in PicoLisp. In order to embed a Lisp expression in a Pilog, you must use '^' operator. It causes the rest of the expression to be taken as Lisp. Then, inside the Lisp code you can in turn access Pilog-bindings with the '->' function. Hence, in our case the Prolog rule above translates as: (be underage (@X) (age @X @Y) (^ @ (< (-> @Y) 18)) ) In '(^ @ (< (-> @Y) 18))', '@' is an anonymous variable used to get the result. If you need to access the result you can bind it to a defined variable like in '(^ @B (+ (-> @A) 7))' where '@B' is now bound to '@A + 7'. You may prefer to define your own Pilog predicate in this particular case. Let's say that to avoid confusion, you want to create a Pilog predicate call 'less_than' to mimic the Lisp function '<': (be less_than (@A @B) (^ @ (< (-> @A) (-> @B) ))) Then the Pilog rule becomes: (be underage_1 (@X) (age @X @Y) (less_than @Y 18) ) and now: (? (underage @X) ) yields to: @X=Kate which is the expected result. = Hope this helps. Best, Eric Le 13/11/2016 à 15:47, Alexander Burger a écrit : You can't be too young, I think. You could write it, and perhaps others may improve it;)
Re: (< @X 18) doesn't behave as expected with pilog (SOLVED: short
Hi Eric, > short how to. I haven't seen such a one on the wiki, so may be it can find > its way there. However I'm too young here to take such a decision ;-) You can't be too young, I think. You could write it, and perhaps others may improve it ;) ♪♫ Alex -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
Re: (< @X 18) doesn't behave as expected with pilog (SOLVED: short
Hi, This is mostly a copy/paste of Alexander's answer below in the form of a short how to. I haven't seen such a one on the wiki, so may be it can find its way there. However I'm too young here to take such a decision ;-) == **How to access a Lisp function from Pilog** Let's say that you have those two facts in a Pilog database: (be age (Paul 19) ) (be age (Kate 17) ) and that you want to find the person under 18. In full Prolog you may have written something like this: underage(X) :- age(X,Y), Y < 18. however in Pilog the following rule: (be underage (@X) (age @X @Y) (< @Y 18) ) won't work and the query: (? (underage @X) ) will yield to 'NIL' instead of the expected result '@X=Kate' . The reason is that '<' (less than) is not Pilog function but only a Lisp one in PicoLisp. In order to embed a Lisp expression in a Pilog, you must use '^' operator. It causes the rest of the expression to be taken as Lisp. Then, inside the Lisp code you can in turn access Pilog-bindings with the '->' function. Hence, in our case the Prolog rule above translates as: (be underage (@X) (age @X @Y) (^ @ (< (-> @Y) 18)) ) In '(^ @ (< (-> @Y) 18))', '@' is an anonymous variable used to get the result. If you need to access the result you can bind it to a defined variable like in '(^ @B (+ (-> @A) 7))' where '@B' is now bound to '@A + 7'. You may prefer to define your own Pilog predicate in this particular case. Let's say that to avoid confusion, you want to create a Pilog predicate call 'less_than' to mimic the Lisp function '<': (be less_than (@A @B) (^ @ (< (-> @A) (-> @B) ))) Then the Pilog rule becomes: (be underage_1 (@X) (age @X @Y) (less_than @Y 18) ) and now: (? (underage @X) ) yields to: @X=Kate which is the expected result. Et voià! == Best, Eric Le 12/11/2016 à 16:27, Alexander Burger a écrit : Hi Eric, (be underage (@X) (age @X @Y) (< @Y 18)) '<' is a Lisp function and not a Pilog rule. To embed a Lisp expression in Pilog, you must use the '^' operator. It causes the rest of the expression to be taken as Lisp, and inside the Lisp code you can in turn access Pilog-bindings with the '->' function. In the case above it should be something like (^ @ (< (-> @Y) 18)) '@' is an anonymous variable here. If you want to bind the result of the Lisp expression to a specific variable, it would be e.g. (^ @X (+ (-> @N) 7)) This binds @X to @N + 7. Of course, if you need '<' more often, you could define your own predicate: : (be < (@A @B) (^ @ (< (-> @A) (-> @B))) ) -> < : (? (< 3 4)) -> T : (? (< 4 2)) -> NIL ♪♫ Alex -- UNSUBSCRIBE: mailto:picolisp@software-lab.de?subject=Unsubscribe
