reverse list - scheme

racket, scheme

Solution

The natural way to recur over a list is not the best way to solve this problem. Using `append`, as suggested in the accepted answer pointed by @lancery, is not a good idea either - and anyway if you're learning your way in Scheme it's best if you try to implement the solution yourself, I'll show you what to do, but first a tip - don't use `list` as a parameter name, that's a built-in procedure and you'd be overwriting it. Use other name, say, `lst`.

It's simpler to reverse a list by means of a helper procedure that accumulates the result of consing each element at the head of the result, this will have the effect of reversing the list - incidentally, the helper procedure is tail-recursive. Here's the general idea, fill-in the blanks:

(define (reverse lst)
  (<???> lst '()))                       ; call the helper procedure

(define (reverse-aux lst acc)
  (if <???>                              ; if the list is empty
      <???>                              ; return the accumulator
      (reverse-aux <???>                 ; advance the recursion over the list
                   (cons <???> <???>)))) ; cons current element with accumulator

Of course, in real-life you wouldn't implement `reverse` from scratch, there's a built-in procedure for that.

Problem

I'm trying to reverse a list, here's my code: ``` (define (reverse list) (if (null? list) list (list (reverse (cdr list)) (car list)))) ``` so if i enter (reverse '(1 2 3 4)), I want it to come out as (4 3 2 1), but right now it's not giving me that. What am I doing wrong and how can I fix it?

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