Hi Cle,
> And till now, I am failing to get a solution, that works for both
> examples. I think the problem may be, that in the first example, the
> free argument '@Var' is unified with the free variable '@C'. But in the
> second example, '@Var' is bound to '(@C @N)'. Those both seem to work
> different.
Yes, looks like. The binding to '@Var' had already happened before
exeuction starts.
> In some Prolog dialects, there is a predicate 'free/1' and 'bound/1'
> that succeed, if the passed variable was free or bound. Unfortunately, I
> found no such check in Pilog.
There is currently no such predicate. Only '->', but this doesn't
distinguish between "value is NIL" and "not bound".
> 1. Is it possible to complete 'findall' more elegantly to work with both
> one as well as with more than one variable?
> 2. Is it possible to check in Pilog whether a certain variable is free
> or already bound?
Thinking about it a little, I came up with a "half" solution:
(be tst (@Var (@Pred . @Args))
(call @Pred . @Args)
(@ unify (-> @Var)) )
: (? (tst @C (ascii @C @N)))
@C=a
@C=b
@C=c
-> NIL
: (? (tst (@C @N) (ascii @C @N)))
@C=a @N=97
@C=b @N=98
@C=c @N=99
-> NIL
At the moment, however, I don't know how to get the bindings into a
single result list. I'm sure you know how to make a recursive predicated
that accomplishes that ;-)
As my Prolog skills are rather poor, I would more easily come up with a
Lisp solution. For the problem at hand, it might be useful to know
that the clauses are stored under the 'T' property of 'ascii':
: (get 'ascii T)
-> (((a 97)) ((b 98)) ((c 99)))
What is the concrete problem? You might be able to solve it directly
(and more efficiently) by accessing that list.
Cheers,
- Alex
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