目标是消除所有内括号。
(flatten '(a (bc) d)) 变成 '(abcd)
这是我在球拍中的代码
; if slist is null, return empty
; otherwise, if it is a pair, recursively solve car and cdr and concat them
; if it is a symbol, return the symbol
(define flatten
(lambda (slist)
(cond
[ (null? slist) '()]
[ (pair? slist)
(cons ((flatten (car slist)) (flatten (cdr slist))))]
[ (symbol? slist) slist])))
它在抱怨
procedure application: expected procedure, given: c; arguments were: ()
这意味着我正在尝试访问一个car
空cdr
列表。
I did the trace:
> (flatten '(a (b c) d))
pair?-car-cdr
a
((b c) d)
symbol?
a
pair?-car-cdr
(b c)
(d)
pair?-car-cdr
b
(c)
symbol?
b
pair?-car-cdr
c
()
symbol?
c
(stops here)
跟踪代码很简单——一堆显示。
(define flatten
(lambda (slist)
(cond
[ (null? slist) '()]
[ (pair? slist)
(display 'pair?-car-cdr)
(newline)
(display (car slist))
(newline)
(display (cdr slist))
(newline)
(cons ((flatten (car slist)) (flatten (cdr slist))))]
[ (symbol? slist)
(display 'symbol?)
(newline)
(display slist)
(newline)
slist])))
我不明白的是为什么第一个条件(null? slist)
没有抓住空列表?我有两个递归调用。如果它确实捕获了空列表,它将进入下一个递归,即 list {d}
。
我的递归逻辑有什么问题?
更新版本
(define flatten
(lambda (slist)
(cond
[ (null? slist) '()]
[ (pair? slist)
(cons (flatten (car slist)) (flatten (cdr slist)))]
[ (symbol? slist) slist])))
(display (equal? (flatten '(a (b a) b a c (a b) c (e f (b a)))) '(a b a b a c a b c e f b a)))
(newline)
(display (equal? (flatten '(a b c)) '(a b c)))
(newline)
(display (equal? (flatten '(a (b c))) '(a b c)))
(newline)
(display (equal? (flatten '((a)(b)(c) d)) '(a b c d)))
(newline)
(display (equal? (flatten '(a (b) ((c)) (((d))) ((((e (f g))))))) '(a b c d e f g )))
(newline)
(display (equal? (flatten '()) '()))
(newline)
(display (equal? (flatten '(a b () ())) '(a b)))
(newline)
正如 Ross Larson 建议的那样, append 将使程序正常工作。但是为了学习,如果有人有空的话,我的测试结果只显示通过了基本案例(第二个和空列表)
我想过写一个包装函数来调用(cons (flatten slist) empty)