Scheme 在chez格式中使用匹配

Scheme 在chez格式中使用匹配,scheme,match,chez-scheme,Scheme,Match,Chez Scheme,我正在努力学习如何在scheme中使用match。我有点理解它是如何处理非常短的问题(即:定义长度仅为两行)的,但不理解有多个输入和辅助程序的问题。例如,这里有一种定义联合的流行方式: (define ele? (lambda (ele ls) (cond [(null? ls) #f] [(eq? ele (car ls)) #t] [else (ele? ele (cdr ls))]))) (define union (lambda (ls1 ls2)

我正在努力学习如何在scheme中使用match。我有点理解它是如何处理非常短的问题(即:定义长度仅为两行)的,但不理解有多个输入和辅助程序的问题。例如,这里有一种定义联合的流行方式:

(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))])))

在这两个程序中如何使用match?(或者你甚至需要两个程序吗?

第一个程序可以这样实现:

(define ele?
  (lambda (a b)
    (let ((isa? (lambda (x) (eq? (car x) a))))
      (match b [(? null?) #f]
               [(? isa?) #t]
               [_ (ele? a (cdr b))]))))
那么第二个就容易了

(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))]))))

也许有一个更聪明的方法来避免这一争论(让兰博达斯说吧,但我仍在学习;)

第一个可以这样实现:

(define ele?
  (lambda (a b)
    (let ((isa? (lambda (x) (eq? (car x) a))))
      (match b [(? null?) #f]
               [(? isa?) #t]
               [_ (ele? a (cdr b))]))))
那么第二个就容易了

(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))]))))
也许有一个更聪明的方法来避免这一争论(让兰博达斯说吧,但我仍在学习;)