我正在尝试学习如何在计划中使用匹配。我有点理解它如何与真正的短问题一起工作(即:定义长度只有两行)但不存在有多个输入和辅助程序的问题。例如,这是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))])))
你如何在两个程序中使用匹配? (或者你甚至需要两个程序?)
答案 0 :(得分:1)
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))]))))
然后第二个很容易
(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))]))))
也许有一种更聪明的方法来避免这些争论让lambdas但我还在学习;)