sicp exercise 3.14

; Exercise 3.14.  The following procedure is quite useful, although obscure:
;
; (define (mystery x)
;   (define (loop x y)
;     (if (null? x)
;         y
;         (let ((temp (cdr x)))
;           (set-cdr! x y)
;           (loop temp x))))
;   (loop x '()))
;
; Loop uses the ``temporary'' variable temp to hold the old value of the cdr of x, since the set-cdr! on the next line destroys the cdr. Explain what mystery does in general. Suppose v is defined by (define v (list 'a 'b 'c 'd)). Draw the box-and-pointer diagram that represents the list to which v is bound. Suppose that we now evaluate (define w (mystery v)). Draw box-and-pointer diagrams that show the structures v and w after evaluating this expression. What would be printed as the values of v and w ?

(define (mystery x)
  (define (loop x y)
    (if (null? x)
        y
        (let ((temp (cdr x)))
          (set-cdr! x y)
          (loop temp x))))
  (loop x '()))

(define v (list 'a 'b 'c 'd))
(display v) (newline)
(define w (mystery v))
(display v) (newline)
(display w) (newline)

; It reverses the list