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