I'm trying to learn how to use match in scheme. I sort of understand how it works with really short problems (ie: defining length is just two lines) but not with problems where there's more than one input, and helper programs. For example, here's a popular way of defining union:
(define ele?
(lambda (ele ls)
(cond
[(null? ls) #f]
[(eq? ele (car ls)) #t]
[else (ele? ele (cdr ls))])))
(define union
(lambda (ls1 ls2)
(cond
[(null? ls2) ls1]
[(ele? (car ls2) ls1) (union ls1 (cdr ls2))]
[else (union (cons (car ls2) ls1) (cdr ls2))])))
How do you do this using match in both programs? (or would you even need two programs?)
the first one could be implemented like that:
(define ele?
(lambda (a b)
(let ((isa? (lambda (x) (eq? (car x) a))))
(match b [(? null?) #f]
[(? isa?) #t]
[_ (ele? a (cdr b))]))))
then the second is easy
(define uni
(lambda (ls1 ls2)
(let ((carinls2? (lambda (x) (ele? (car x) ls1))))
(match ls2 [(? null?) ls1]
[(? carinls2?) (uni ls1 (cdr ls2))]
[_ (uni (cons (car ls2) ls1) (cdr ls2))]))))
maybe there is a smarter way to avoid these one argument let lambdas but i'm still learning ;)