Scheme的"期望一个可以应用于参数的过程"

Question

我用DrRacket.我有这个代码的问题:

          (define (qweqwe n) (
                      (cond 
                        [(< n 10) #t]
                        [(>= (lastnum n) (pochtilastnum n)) (qweqwe (quotient n 10))]
                        [else #f]
                        )
                      )
    )
    (define ( RTY file1 file2 )

     (define out (open-output-file file2 #:mode  'text #:exists 'replace))  
    (define in (open-input-file file1)) 
    (define (printtofile q) (begin
                   (write q out)
                   (display '#\newline out)
                   ))
       (define (next) 
          (define n (read in)) 
(cond 
      [(equal? n eof) #t]
      [else (begin
      ((if (qweqwe n) (printtofile n) #f))
      ) (next)]
      )
)
    (next)   
   (close-input-port in)
   (close-output-port out)) 

但是当我开始时(RTY"in.txt""out.txt")我有一个错误((if(qweqwe n)(printtofile n)#f)):

    application: not a procedure;
    expected a procedure that can be applied to arguments
    given: #f
    arguments...: [none]

有什么问题？

ADD:我将代码更改为:

(cond 
      [(equal? n eof) #t]
      [else
      (if (qweqwe n) (printtofile n) #f)
      (next)]
      )

但问题仍然存在.

Answer 1

有一些不必要的括号,不要这样做:

((if (qweqwe n) (printtofile n) #f))

试试这个:

(if (qweqwe n) (printtofile n) #f)

也在这里:

(define (qweqwe n)
  ((cond [(< n 10) #t]
         [(>= (lastnum n) (pochtilastnum n)) (qweqwe (quotient n 10))]
         [else #f])))

它应该是:

(define (qweqwe n)
  (cond [(< n 10) #t]
        [(>= (lastnum n) (pochtilastnum n)) (qweqwe (quotient n 10))]
        [else #f]))

在这两种情况下,问题是如果你用()表达式包围,这意味着你正在尝试调用一个过程.并且鉴于上述if和cond表达式的结果不返回过程,则会发生错误.此外,begin原始代码中的两个s都是不必要的,a 在每个条件之后cond都有一个隐含的begin,对于过程定义的主体来说是相同的.