How to recursively go through list of lists and co

2019-12-16 17:17发布

问题:

This is a follow-up to this question.

I have lists or "records" that are appended together into an empty list, that are each in the format (Matthew (AL 21 32)). I am now trying to write a function that would use fetchRecord in order to find the desired record and then multiply the two numbers inside the record. However, my current code works only if I am getting the name on the first record but it returns an empty list for any record after that. Here is my code:

(define (Mult_Num name)
  (cond
    [(empty? db) #f]
    [else (* (car(cdr(car(cdr (fetchRecord name)))))
             (car(cdr(cdr(car(cdr (fetchRecord name)))))))]))

How would I fix this? Also if a certain record has two sets of data like so: (John (AL 25 40) (CA 40 67)) then how would you get both 25*40 and 40*67 etc., and even if it has more than two sets of data? I understand that it would be recursion but am not quite sure how you would set it up.

This is my fetchRecord function:

(define (fetchRecord name)
  (cond
    [(empty? db) #f]
    [(equal? (car (car db)) name) (car db)]
    [else( car (car db)) name (cdr db)]))

This Also may be relevant:

(define db '())

Also I have this but if I have more than two names in here it screws itself up:

 (define(showRec name) ;displays everything following a name. 
  (cond
    [(empty? db) #f]
    [(equal? (car (car db)) name) (cdr (fetchRecord name))]
    [else (cdr (car(fetchRecord name)))])
  )

回答1:

The main problem is that you rely on fetchRecord which not only does not work, but isn't valid Scheme: the else clause should have only 1 argument, not 3.

And you shouldn't be surprised that a function that would use recursion to handle the rest of a list doesn't handle the rest of a list when it doesn't use recursion.