简单递归中的意外错误(Scheme语言)

Question

我在用球拍学习计划。我做了下面的程序，但它带来了一个违反合同的错误。

预期：(精确-非负-整数？)->。任何/c)：'()

程序在一个间隔内查找所有数字的列表，这些数字可以被3或5整除。

#lang racket

;;Global Definitions
(define upper-bound 10)
(define lower-bound 0)


;;set-bounds: Int, Int -> ()
(define (set-bounds m n)
(set! upper-bound (max m n))
(set! lower-bound (min m n)))

;;get-numbers: () -> (Int)
(define (get-numbers)
    (build-list upper-bound '()))

;;make-list: Int, (Int) -> (Int)
(define (build-list x y)
    (cond
        [(= x lower-bound) y]
        [(= (modulo x 5) 0) (build-list (sub1 x) (cons x y))]
        [(= (modulo x 3) 0) (build-list (sub1 x) (cons x y))]
        [else (build-list (sub1 x) y)]))

编辑:我做了奥斯卡·洛佩兹建议的修改。

Answer 1

另一种方法是使用for/list创建列表：

(define (build-list ub lst)
  (for/list ((i (range lb ub))
             #:when (or (= 0 (modulo i 3))
                        (= 0 (modulo i 5))))
    i))

用法：

(define lb 0)
(build-list 10 '())

输出：

'(0 3 5 6 9)

编辑：

实际上，这里不需要第一步：

(define (build-list ub)
  (for/list ((i (range lb ub))
             #:when (or (= 0 (modulo i 3))
                        (= 0 (modulo i 5))))
    i))

所以你可以打电话：

(build-list 10)

下面是对递归方法的修改(使用'named let')：

(define (build-list2 ub)
  (let loop ((x ub) (lst '()))
    (cond
        [(= x lb) lst]
        [(= (modulo x 5) 0) (loop (sub1 x) (cons x lst))]
        [(= (modulo x 3) 0) (loop (sub1 x) (cons x lst))]
        [else (loop (sub1 x) lst)])))

此外，如果始终必须使用空list '()调用函数，则可以将其作为默认值放在参数列表中：

(build-list x (y '()))

然后您可以使用简化的命令进行调用：

(build-list 10)

Answer 2

您应该首先测试递归停止的条件，即当x等于lower-bound时

(define (build-list x y)
  (cond
    [(= x lower-bound) y]
    [(= (modulo x 5) 0) (build-list (sub1 x) (cons x y))]
    [(= (modulo x 3) 0) (build-list (sub1 x) (cons x y))]
    [else (build-list (sub1 x) y)]))