[REBOL] Re: Can I define an anti function?
From: andreas:bolka:gmx at: 24-Feb-2004 13:53
Monday, February 23, 2004, 8:55:15 PM, Ladislav wrote:
> Question: Is the guess about Rebol correct? (I will post my answer
> later)
I'll second Gabriele's answer: The reasoning seems to be intriguing at
first glance, but it's (at least partly) wrong.
A (naive) demonstration:
anti: func [ f ] [ func first :f compose [ not (second :f) ] ]
f: func [ a b ] [ = a b ]
anti-f: anti :f
f 1 2
; == false
anti-f 1 2
; == true
The above obviously does not cope with functions where 'return is
used, however I'm not going into more detail here, as I'm really
looking forward to Ladislav's answer (which will be more correct than
mine could ever be, anyway).
But let me explain why I weakened my above statement with 'at least
partly'. In Scheme the following is perfectly legal:
(define (anti f) (lambda args (not (apply f args))))
(define (f a b) (equal? a b))
((anti f) 1 2)
; == #t
In REBOL however, my naive approach given above would not give the
desired results in this case (anti :f 1 2), as anti :f does not return
a "loaded" function, i.e. we'd have to force it's application by using
'do:
>> do anti :f 1 2
== true
>> do anti :f 1 1
== false
In my opinion, that is the perfect REBOL equivalent to the special
Scheme form used above. However, Scheme purists may argue with me on
that point :) Nevertheless, even in Scheme, I imagine the "typical"
use of anti would be:
(define anti-f (anti f))
--
Best regards,
Andreas
