需要 #t 但获得第一个元素(方案)

问题描述

我想我几乎已经解决了以前的问题:方案中的文件夹但在代码中是一个小麻烦.我需要#t，但我得到第一个元素，false 就可以了.这是我的代码:

I think I have almost done solution with previous problem: Foldr in scheme but in code is a small trouble. I need #t but I get first element, false is OK. Here is my code:

(define accum
  (lambda (list1 pre?)
    (foldr (lambda (x y)
             (if y
                 (if (or (equal? y #t) (pre? x y))
                     x
                     #f)
                 #f))
           #t
           list1)))

(accum '(1 2 3 4) <=) --> 1 (should be #t)
(accum '(2 2 4 4) <=) --> 2 (should be #t)
(accum '(1 2 5 4) <=) --> #f
(accum '(5 7 2 3) <=) --> #f

如果我写x --> #t"，我总是得到#t，即使是#f.

If I write "x --> #t", I always get #t, even if is #f.

推荐答案

好吧，你总是可以用另一个返回正确类型的过程包装结果:

Well, you can always wrap the result with another procedure that returns the correct type:

(define (accum? list1 pre?)
  (if (accum list1 pre?) #t #f))

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