SICP Exercise 3.14 mystery

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 wafter evaluating this expression. What would be printed as the values of v and w ?

SOLUTION

The code is here:

 Exercise 3.14 mystery

The box-and-pointer diagrams are here:





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