CODEWITHSUNDEEP
schemeracketr5rs
seaweed
seaweed

Reputation: 21

What is happening in each line of this code?

I know the overall code is to return the last nth elements of the list, however, I don't understand the process, like in each line what is happening(and why, if possible)?

(define (last-n lst n)
  (define (help-func lst drop)
    (cond
      ((> drop 0)
       (help-func (cdr lst ) (- drop 1)))
      (else
       (cdr lst ))))
  (if (= (length lst ) n )
      lst
      (help-func lst (- (length lst ) 1 n ))))

Upvotes: 0

Views: 86

Answers (1)

Óscar López
Óscar López

Reputation: 235984

There's a small bug, when n is greater than the length of the list you should return the whole list (or signal an error), I fixed it. Here's a break-down of the code:

(define (last-n lst n)
  (define (help-func lst drop)
    (cond
      ; iterate while there are elements to drop
      ((> drop 0)
       ; advance on the list, and we're one
       ; element closer to reach our target
       (help-func (cdr lst) (- drop 1)))
      (else
       ; else we reached the point we wanted, stop
       (cdr lst))))
  ; if n is at least as big as the list
  (if (>= n (length lst))
      ; return the whole list
      lst
      ; else calculate how many elements
      ; we need to drop and start the loop
      (help-func lst (- (length lst) 1 n))))

FYI, Racket already has this functionality, just use the take-right built-in procedure, it'll even be faster, requiring a single pass over the list (you're calling length a couple of times, and in a clever algorithm that would be unnecessary)

Upvotes: 3

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