如何在进入空错误之前中断递归调用

Question

(define l '(* - + 4))
(define (operator? x)
    (or (equal? '+ x) (equal? '- x) (equal? '* x) (equal? '/ x)))
(define (tokes list)
  (if (null? list)(write "empty")
  (if (operator? (car list))
       ((write "operator")
        (tokes (cdr list)))
      (write "other"))))

代码工作得很好，直到（tokes （cdr list）））到达文件末尾。有人可以给我一个提示，告诉我如何防止这种情况。我是 Scheme 的新手，所以如果这个问题很荒谬，我原谅我。

Answer 1

您必须确保在每个情况下推进递归（当列表null时，基本情况除外）。在您的代码中，您不会对(write "other")情况进行递归调用。另外，当有几个条件要测试时，你应该使用cond，让我用一个例子来解释 - 而不是这个：

(if condition1
    exp1
    (if condition2
        exp2
        (if condition3
            exp3
            exp4)))

最好这样写，可读性更强，并且还有一个额外的好处，即您可以在每个条件之后编写多个表达式，而无需使用begin形式：

(cond (condition1 exp1) ; you can write additional expressions after exp1
      (condition2 exp2) ; you can write additional expressions after exp2
      (condition3 exp3) ; you can write additional expressions after exp3
      (else exp4))      ; you can write additional expressions after exp4

。这让我想到了下一点，请注意，您只能为if的每个分支编写一个表达式，如果给定条件在if形式中需要多个表达式，则必须用begin将它们括起来，例如：

(if condition
    ; if the condition is true
    (begin  ; if more than one expression is needed 
      exp1  ; surround them with a begin
      exp2) 
    ; if the condition is false
    (begin  ; if more than one expression is needed 
      exp3  ; surround them with a begin
      exp4))

回到你的问题 - 这是一般的想法，填空：

(define (tokes list)
  (cond ((null? list)
         (write "empty"))
        ((operator? (car list))
         (write "operator")
         <???>)   ; advance on the recursion
        (else
         (write "other")
         <???>))) ; advance on the recursion