SICP学习笔记 2.3.1 引号 SICPLispScheme函数式编程

    练习2.53

(a b c)
((george))
((y1 y2))
(y1 y2)
#f
#f
(red shoes blue socks)

 

    练习2.54

(define (equal? a b)
  (cond ((and (null? a) (null? b)) true)
	      ((null? a) false)
	      ((null? b) false)
	      (else (and (eq? (car a) (car b)) (equal? (cdr a) (cdr b))))))

 

    练习2.55

;; 'a 即为 (quote a)

1 ]=> (quote a)
;Value: a

;; 因此 ''abracadabra 等价于 (quote (quote abracadabra))

1 ]=> (quote (quote abracadabra))
;Value: (quote abracadabra)

1 ]=> ''abracadabra
;Value: (quote abracadabra)

1 ]=> (car (quote (quote abracadabra)))
;Value: quote