DrRacket编程

我在一份作业中得到了这一点：

; subst: [Listof Value] [Listof Name] SExp -> SExp
; substitute the corresponding value for each listed variable name in s
; leave all other names in s unmodified

(define (subst vals vars s) ...)

(define s1 '(foo a 29 (bar "z" b)))
(check-expect (subst '(3 "x") '(a b) s1) '(foo 3 29 (bar "z" "x")))

我知道我需要重复这两个列表，但不确定如何进行。

这将起作用：

(define (subst vals vars s)
  (let ((mapping (map vars vals)))
    (let subbing ((s s))
      (if (null? s)`
          '()
           (let ((s1 (car s)))
             (cons (cond ((and (symbol? s1) (assoc s1 mapping)) => cdr)
                         ((list? s1) (subbing s1))
                         (else s1))
                   (subbing (cdr s))))))))

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这看起来像是来自EOPL，在这种情况下，这是本书开头的“手指练习”之一。