Scheme function to reverse a list

Question

For my programming languages class I\'m supposed to write a function in Scheme to reverse a list without using the pre-made reverse function. So far what I got was

Answer 1

You are half way there. The order of the elements in your result is correct, only the structure needs fixing.

What you want is to perform this transformation:

 (((() . c) . b) . a)           ; input
 --------------------
 (((() . c) . b) . a)   ()      ; trans-
  ((() . c) . b)       (a)      ;      for-
   (() . c)          (b a)      ;         mation
    ()             (c b a)      ;               steps
      --------------------
                   (c b a)      ;               result

This is easy to code. The car and cdr of the interim value are immediately available to us. At each step, the next interim-result is constructed by (cons (cdr interim-value) interim-result), and interim-result starts up as an empty list, because this is what we construct here - a list:

(define (transform-rev input)
   (let step  ( (interim-value    input)       ; initial set-up of 
                (interim-result    '() ) )     ;  the two loop variables
     (if (null? interim-value)
       interim-result                  ; return it in the end, or else
       (step  (car interim-value)      ; go on with the next interim value
              (cons                    ;      and the next interim result
                  (... what goes here? ...)
                  interim-result )))))

interim-result serves as an accumulator. This is what's known as "accumulator technique". step represents a loop's step coded with "named-let" syntax.

So overall reverse is

(define (my-reverse lst)
   (transform-rev 
       (reverseList lst)))

Can you tweak transform-rev so that it is able to accept the original list as an input, and thus skip the reverseList call? You only need to change the data-access parts, i.e. how you get the next interim value, and what you add into the interim result.