23 hours ago

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)

x)

(define (last-pair x)

(if (null? (cdr x))

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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