how can you implement a Scheme function has-list recursively, which tests whether a list contains other list as an element. For example (has-list '(1 2 3)) should return false, and (has-list '(1 2 (3 4) 5)) should return true.
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3Spizzy - we aren't just an answer service, we like to see people try things first.zellio– zellio2011-02-27 21:15:58 +00:00Commented Feb 27, 2011 at 21:15
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4 Answers
If your implementation has something like ormap, then:
(define (has-list? l) (ormap list? l))
Using or as in Dan D.'s answer will not work.
2 Comments
ray
ormap is not available before r6rs.
Eli Barzilay
(1) I started it with an "if"; (2) R6RS is Scheme; (3) even in pre-R6RS schemes such a function is very common -- for example, SRFI-1's
any.A list has a list as an element iff it is not the empty list and either its first element is a list or the rest of the list has a list as an element. The translation to Scheme code is left as an exercise for the reader.
Comments
If you need to do it recursively and without using map:
(define (has-list? lst)
(cond
((null? lst) #f)
((list? (car lst)) #t)
(else (has-list? (cdr lst)))))
