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 ?

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It reverses a list, destroying the original.

Specifically it calls loop with the original list and '() as arguments.

(loop x y) returns y if x is empty.

Otherwise it saves the tail of x, alters x to be the original first 
element of x followed by y, and calls loop with the saved tail of x 
and the modified x as arguments.  Since the x that is modified is the 
argument, the original x is destroyed.

This has the effect of creating a new list which contains the original 
elements of x in reversed order, but has the side effect of chopping x 
up into individual elements; a good example of why mutability is evil.