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.

————————————————————————————————————————————————————————————————————————

(cdr x)
(b)

(cdr x)
(b c d)

Without drawing diagrams, the explanation is that (append x y) does not 
mutate x, so (cdr x) is still just (b) as it was before, while 
(append! x y) mutates x so that it now contains the entire list.