# 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 ;-)

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**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

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