sicp习题试解 (2.33)

; ======================================================================

;

; Structure and Interpretation of Computer Programs

; (trial answer to excercises)

;

; 计算机程序的构造和解释(习题试解)

;

; created: code17 08/19/05

; modified:

; (保持内容完整不变前提下，可以任意转载)

; ======================================================================

;; SICP No.2.33

(define (map p sequence)

(accumulate (lambda (x y) (cons (p x) y)) () sequence))

(define (append seq1 seq2)

(accumulate cons seq2 seq1))

(define (length sequence)

(accumulate (lambda (x y) (+ y 1)) 0 sequence))

;; Test-it:

;; Welcome to MzScheme version 209, Copyright (c) 2004 PLT Scheme, Inc.

;; > map

;; #<procedure:map>

;; > append

;; #<procedure:append>

;; > length

;; #<procedure:length>

;; ;; 注意到此时map,append,length的原始定义已被新的覆盖，

;; ;; 因为它们现在是procedure而不是primitive

;;

;; > (map sqrt (list 1 9 16 25))

;; (1 3 4 5)

;; > (append (list 1 2 (list 3 4)) (list 5 6))

;; (1 2 (3 4) 5 6)

;; > (length (list 1 2 (list 3 4)))

;; 3