Re: Validating a Scheme list

[Orm Finnendahl]

Hi Urs,

 it doesn't make too much sense to first collect the whole list and
then check for membership of #f. It'd be more reasonable to stop
iteration once any of the preconditions isn't met. You could achieve
this by using a recursive function like below (you could make it
more terse by using if instead of cond but in recursive functions cond
dispatch is more common and I find it more readable...):

 #(define-scheme-function (parser location lst) (list?)
     (letrec ((test (lambda (x) 
              (cond ((null? x) #t)
                    (#t (and (list? (car x)) 
                             (= (length (car x)) 2)
                             (test (cdr x))))))))
      (if (test lst)
            (display "invalid")
            (display "valid"))
      (newline)))

Am Donnerstag, den 15. Mai 2014 um 10:09:10 Uhr (+0200) schrieb Urs Liska:
> Hi all,
> 
> I am working on a Scheme function and would like to check if I have found the 
> best solution for a specific subpart. Somehow it looks more complicated than 
> necessary.
> 
> The function needs to test if each element of a given list is a (sub)list 
> with exactly two elements. So
> 
> '((1 2)(3 4))
> 
> would return #t while
> 
> '((5 6)(7 8 9))
> 
> would return #f.
> 
> My solution is
> 
> \version "2.19.6"
> 
> validate =
> 
> #(define-scheme-function (parser location lst) (list?)
> 
>    (if (memv #f (map (lambda sig
> 
>                        (and (list? (car sig))
> 
>                             (= 2 (length (car sig))))) lst))
> 
>        (display "invalid")
> 
>        (display "valid"))
> 
>    (newline))
> 
> {
> 
>   \validate #'((1 2)(3 4))
> 
>   \validate #'((5 6)(7 8 9))
> 
> }
> 
> The "framework" doesn't matter, it's just a compilable example. My question 
> is only about the "if" expression.
> 
> What it does is:
> - go through the elements of lst
> - produce a list of boolean values,
>   - #t if we have a two element list,
>   - #f if not
> - check if this intermediate list contains at least one #f
> 
> Somehow this looks clumsy to me, and I'd like to know (and learn) if there's 
> a better solution for this.
> 
> TIA
> Urs
> 

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