sicp习题试解 (2.4)

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

;

; 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 04/23/05

; modified:

; (保持内容完整不变前提下，可以任意转载)

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

;; SICP No.2.4

;; cdr可定义为

(define (cdr z)

(z (lambda (p q) q)))

;; 验证

;; (car (cons x y))

;; => (car (lambda (m) (m x y)))

;; => ((lambda (m) (m x y)) (lambda (p q) p))

;; => ((lambda (p q) p) x y)

;; => x

;;

;; cdr同理

;; Test-it:

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

;; > (define test12 (cons 1 2))

;; > (car test12)

;; 1

;; > (cdr test12)

;; 2

;; > (define test123 (cons test12 3))

;; > (cdr test123)

;; 3

;; > (car test123)

;; #<procedure:STDIN::21>

;; > (cdr (car test123))

;; 2