SICP Exercise 3.12 append and append!

Exercise 3.12.  The following procedure for appending lists was introduced in section 2.2.1:

(define (append x y)
  (if (null? x)
      y
      (cons (car x) (append (cdr x) y))))

Append forms a new list by successively consing the elements of x onto y. The procedure append! is similar to append, but it is a mutator rather than a constructor. It appends the lists by splicing them together, modifying the final pair of x so that its cdr is now y. (It is an error to call append! with an empty x.)

(define (append! x y)
  (set-cdr! (last-pair x) y)
  x)

Here last-pair is a procedure that returns the last pair in its argument:

(define (last-pair x)
  (if (null? (cdr x))
      x
      (last-pair (cdr x))))

Consider the interaction

(define x (list 'a 'b))
(define y (list 'c 'd))
(define z (append x y))
z
(a b c d)
(cdr x)
<response>
(define w (append! x y))
w
(a b c d)
(cdr x)
<response>

What are the missing <response>s? Draw box-and-pointer diagrams to explain your answer.

 SOLUTION

The code is here:

 Exercise 3.12 append and append!

The box-and-pointer diagrams are here:

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