list - 如何在方案中不返回任何内容

Question

我对 Scheme 中分配的编程任务有一个小问题。我们的任务是创建一个函数，该函数仅返回满足给定谓词要求的对结构中的值。它们也将以相同的对结构返回，只需删除违规条目即可。到目前为止，我的代码如下

(define (keep-if-all ia pred ls)
 (cond
  ((null? ls) null)
  ((pair? ls)
   (cons
    (keep-if-all pred (car ls))
    (keep-if-all pred (cdr ls))))
  ((pred ls) ls)
  (else null)))

问题是 else 返回 null 并因此将值替换为 null 而不是删除它。

例如

(keep-if-all odd? (list 1 2 3 (cons 4 4) (list 1 2 3 4 5)))

返回

(1 () 3 (()) (1 () 3 () 5))

而不是想要的

(1 3 () (1 3 5))

朝正确方向戳将不胜感激

Answer 1

只需在此处添加另一个if，

(define (keep-if-all pred ls)
 (cond
  ((null? ls) '())
  ((pair? ls)
     (let ((x (keep-if-all pred (car ls))))
       (if (or (not (null? x))                ; if something's left of (car ls)
               .... )                         ;     or if it was a pair,
        (cons x (keep-if-all pred (cdr ls)))  ;   then, include x in the output
        .... )))                              ;   else, don't include x in the output
  ((pred ls) ls)
  (else '())))

现在它按预期工作：

(keep-if-all odd? (list 1 2 3 (cons 4 4) (list 1 2 3 4 5)))
;Value 15: (1 3 () (1 3 5))