my attempt to do the exercises in sicp.

Sunday, September 6, 2009

sicp exercise 3.12

; 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.

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

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

(define x (list 'a 'b))
(define y (list 'c 'd))
(define z (append x y))

;(display x)(newline)
;(display y)(newline)
;(display z)(newline)

(display (cdr x)) (newline)

(define w (append! x y))
(display w) (newline)

(display (cdr x)) (newline)

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