Implementing last-non-zero without continuations

Question

last-non-zero takes a list of numbers and return the last cdr whose car is 0.

So, I can implement it using continuations, but how do I do this with natural recursion.

(define last-non-zero
  (lambda (ls)
    (let/cc return
      (letrec
          ((lnz
            (lambda (ls)
              (cond
                ((null? ls) '())
                ((zero? (car ls))      ;; jump out when we get to last 0.
                   (return (lnz (cdr ls))))
                (else 
                   (cons (car ls) (lnz (cdr ls))))))))
        (lnz ls)))))

Answer 1

Here's an obvious version which is not tail-recursive:

(define (last-non-zero l)
  ;; Return the last cdr of l which does not contain zero
  ;; or #f if there is none
  (cond
    ((null? l)
     #f)
    ((zero? (car l))
     (let ((lnzc (last-non-zero (cdr l))))
       ;; This is (or lnzc (cdr l)) but that makes me feel bad
       (if lnzc
           lnzc
           (cdr l))))
    (else
     (last-non-zero (cdr l)))))

Here is that version turned into a tail-recursive equivalent with also the zero test moved around a bit.

(define (last-non-zero l)
  (let lnzl ([lt l]
             [r #f])
    (if (null? lt)
        r
      (lnzl (cdr lt) (if (zero? (car lt)) (cdr lt) r)))))

It's much clearer in this last version that the list is traversed exactly once.