Lisp 错误语法（标识符后有多个表达式）_Lisp_Scheme_Racket

Lisp 错误语法（标识符后有多个表达式）

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Lisp 错误语法（标识符后有多个表达式）,lisp,scheme,racket,Lisp,Scheme,Racket,我正在编写一个函数，尝试弱提示，从大量响应中选择一个响应。该程序本质上是与用户的对话 (define try-weak-cues (lambda (sentence context) (define helper (lambda(list-of-pairs) (define helper2 (lambda(list-of-pairs context) (cond((nul

我正在编写一个函数，尝试弱提示，从大量响应中选择一个响应。该程序本质上是与用户的对话

(define try-weak-cues
        (lambda (sentence context)
        (define helper
           (lambda(list-of-pairs)
          (define helper2
            (lambda(list-of-pairs context)
              (cond((null? list-of-pairs)
                  (cond((null? list-of-pairs) '())
                       ((any-good-fragments?(cue-part(car list-of-pairs))sentence) (helper2(cdr(car list-of-pairs))context))
                        (else(helper(cdr list-of-pairs)))))))))
                       (helper *weak-cues*))))

下面是函数应该从中提取的响应列表：

 (define *weak-cues*
  '( ( ((who) (whos) (who is))
       ((first-base)
           ((thats right) (exactly) (you got it)
        (right on) (now youve got it)))
       ((second-base third-base)
           ((no whos on first) (whos on first) (first base))) )
     ( ((what) (whats) (what is))
       ((first-base third-base)
       ((hes on second) (i told you whats on second)))
       ((second-base)
       ((right) (sure) (you got it right))) )
     ( ((whats the name))
       ((first-base third-base)
       ((no whats the name of the guy on second)
        (whats the name of the second baseman)))
       ((second-base)
    ((now youre talking) (you got it))))
   ))

错误：

define:语法错误标识符后面有多个表达式in:define helper lambda对列表定义helper2 lambda对列表上下文条件为空？配对列表是否为空？成对列表引用一些好的片段？球杆零件车对列表句子帮助器2 cdr车辆对列表上下文帮助器弱提示对的cdr列表

问题是，当您在lambda主体内定义内部过程时，必须在定义之后编写表达式，通常调用内部过程。例如，这是错误的：

(define f
  (lambda (x)
    (define g
      (lambda (y)
        <body 1>))))

此外，如果避免这种类型的过程定义，则代码将更易于调试：

(define f
  (lambda (x)
    <body>))

它将变得更短更清晰，如下所示：

(define (f x)
  <body>)

(define (try-weak-cues sentence context)
  (define (helper list-of-pairs)
    (define (helper2 list-of-pairs context)
      (cond ((null? list-of-pairs)
             (cond ((null? list-of-pairs) '())
                   ((any-good-fragments? (cue-part (car list-of-pairs)) sentence)
                    (helper2 (cdr (car list-of-pairs)) context))
                   (else (helper (cdr list-of-pairs)))))))
    <missing body>)
  (helper *weak-cues*))

现在，回到你的代码。修复格式并切换到较短的过程定义语法后，代码将如下所示：

(define (f x)
  <body>)

(define (try-weak-cues sentence context)
  (define (helper list-of-pairs)
    (define (helper2 list-of-pairs context)
      (cond ((null? list-of-pairs)
             (cond ((null? list-of-pairs) '())
                   ((any-good-fragments? (cue-part (car list-of-pairs)) sentence)
                    (helper2 (cdr (car list-of-pairs)) context))
                   (else (helper (cdr list-of-pairs)))))))
    <missing body>)
  (helper *weak-cues*))

现在很明显，helper的主体丢失了，在helper2的内部定义之后，您没有编写任何内容。可能您当时打算打电话给helper2，但您的代码已经够混乱的了，我猜不出在缺少的正文中应该写什么，这取决于您。

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