具有可变参数长度的调用函数

Question

我有一个函数，它接受变量长度的参数列表。取决于它的长度，我必须调用另一个函数，但参数长度不同。您可以看到下面的功能：

  (define (set-contents . args)
   (define columns-length (length args))
        (cond
          ((= columns-length 1) 
           (send output-list set (empty-list rows-length) (get-nth-item columns-as-list 0)))
          ((= columns-length 2) 
           (send output-list set (empty-list rows-length) (get-nth-item columns-as-list 0) (get-nth-item columns-as-list 1)))
          ((= columns-length 3) 
           (send output-list set (empty-list rows-length) (get-nth-item columns-as-list 0) (get-nth-item columns-as-list 1) (get-nth-item columns-as-list 2)))
          ((= columns-length 4) 
           (send output-list set (empty-list rows-length) (get-nth-item columns-as-list 0) (get-nth-item columns-as-list 1) (get-nth-item columns-as-list 2) (get-nth-item columns-as-list 3)))

这个cond条款还可以再长一些。这是可行的，但绝对不恰当。

有更好的方法来填补这个功能吗？

Answer 1

这个想法的直接翻译如下：

(define (set-contents . args)
  (send/apply output-list set (empty-list rows-length)
                              (for/list ((i (length args)))
                                (get-nth-item columns-as-list i))))

但是，由于您实际上没有使用参数，而是获得参数计数，这似乎很可疑。你想做什么？

Answer 2

使用申请？

(define (set-contents . args)
  (define columns-length (length args))
  (apply send (append (list output-list set '()) 
                      (take column-length columns-as-list))))

(define (take num lst)
  (cond ((< num 1) '())
        ((null? lst) (error "Cannot TAKE from the empty list")
        (else (cons (car lst) (take (- num 1) (cdr lst))))))

当然比做人类编译器强