Scheme统计元音函数报错‘contains?: undefined’求解决方案

Hey there! Let's walk through the issues with your current code and get that vowel counter working properly. The error contains?: undefined is the first red flag, but there are a few other pieces we need to adjust too.

What's Going Wrong #

• contains? isn't a built-in Scheme function: Scheme doesn't have a native contains?—we'll use member instead, which checks if an element exists in a list and returns the sublist starting at that element (or #f if it doesn't exist).
• Undefined variables: You reference vowels and counter but never define them. Plus, Scheme is a functional language—we can't mutate a counter variable like in imperative languages; instead, we'll pass an accumulator through recursion.
• Flawed logic & missing base case: Your current code only handles very specific list scenarios (length 1, or checking the first two elements) and doesn't account for empty lists, which will break recursion. Also, using list as a parameter name is a bad idea—it overrides Scheme's built-in list function!

Corrected Code #

First, let's define our vowels, then build a proper recursive counter:

; Define the list of vowels (include both uppercase and lowercase)
(define vowels '(a e i o u A E I O U))

(define (count-vowels lst)
  ; Helper function that carries our running count (accumulator)
  (define (count-helper remaining current-count)
    (cond
      ; Base case: if the list is empty, return the total count
      ((null? remaining) current-count)
      ; If the first element is a vowel, increment count and recurse
      ((member (car remaining) vowels)
       (count-helper (cdr remaining) (+ current-count 1)))
      ; If not a vowel, just recurse with the same count
      (else
       (count-helper (cdr remaining) current-count))))
  ; Start the recursion with an initial count of 0
  (count-helper lst 0))

How This Works #

1. Vowel List: We explicitly define vowels so the function knows which characters to look for.
2. Helper Function: count-helper takes two arguments: the remaining part of the list to check, and the current number of vowels we've found. This is how we "track" the count without mutable variables.
3. Recursion Cases:
   • When the list is empty (null? remaining), we return the accumulated count—this stops the recursion.
   • If the first element is in the vowels list (checked via member), we add 1 to the count and recurse on the rest of the list.
   • If it's not a vowel, we recurse without changing the count.

Testing It Out #

Try these examples to verify:

(count-vowels '(a b c e i))  ; Returns 3
(count-vowels '())           ; Returns 0 (empty list case)
(count-vowels '(A B o y))    ; Returns 2 (A and o are vowels)

内容的提问来源于stack exchange，提问作者Amir1312