SICP_3.7-3.8

第一次写的：

 1 ;(define (make-account balance secret-password-list)
 2
 3 ;        (define (withdraw amount)
 4 ;        (if (>= balance amount)
 5 ;            (begin (set! balance (- balance amount))
 6 ;                  balance)
 7 ;            "Insufficient funds"))
 8 ;
 9 ;             (define (deposit amount)
10 ;               (set! balance (+ balance amount))
11 ;               balance)
12 ;
13 ;             (define (dispatch m)
14 ;               (cond
15 ;                     ((eq? m ‘withdraw) withdraw)
16 ;                     ((eq? m ‘deposit) deposit)
17 ;                     (else (error "Unknow request -- MAKE-ACCOUNT"
18 ;                                  m))))
19 ;  (lambda (p m)
20 ;    (if (in? p secret-password-list) (dispatch m)
21 ;        (lambda (x) "Incorrect password"))))
22
23
24 ;(define in?
25 ;  (lambda (x list)
26 ;    (cond ((null? list) #f)
27 ;          ((eq? x (car list)) #t)
28 ;          (else (in? x (cdr list))))))
29
30 ;(define (make-joint account old-pa new-pa)
31 ;  (set! account (make-account ((account (car old-pa) ‘withdraw) 0)
32 ;                              (cons new-pa old-pa))))

我尝试用建立列表的方法来做，但是这就导致了不同的账户可以用不同的密码来操作。

同时我写的这个过程还有bug，我用set！去 把一个新过程赋给一个旧过程，但是结果

上，Paul 可以通过 Peter 的密码来操作账户，Peter可以用Paul的来操作。可是，这两个

账户却是独立的，并没有关联起来，Paul对账户的操作，不会影响Peter账户，反之亦然。

所以set！到底做了什么，会出现这种情况？

思路相同的答案

换一个思路后：

 1 (define (make-account balance secret-password)
 2
 3         (define (withdraw amount)
 4         (if (>= balance amount)
 5             (begin (set! balance (- balance amount))
 6                    balance)
 7             "Insufficient funds"))
 8
 9              (define (deposit amount)
10                (set! balance (+ balance amount))
11                balance)
12
13              (define (dispatch m)
14                (cond
15                      ((eq? m ‘withdraw) withdraw)
16                      ((eq? m ‘deposit) deposit)
17                      (else (error "Unknow request -- MAKE-ACCOUNT"
18                                   m))))
19   (lambda (p m)
20     (if (eq? p secret-password) (dispatch m)
21         (lambda (x) "Incorrect password"))))
22
23
24 (define (make-joint account old-password new-password)
25   (lambda (password mode)
26     (if (eq? password new-password)
27         (account old-password mode)
28         "Incorrect password")))
29
30 (define peter-acc (make-account 100 ‘open-sesame))
31
32 (define paul-acc
33   (make-joint peter-acc ‘open-sesame ‘rosebud))
34
35 ((paul-acc ‘rosebud ‘withdraw) 10)
36 ((peter-acc ‘open-sesame ‘withdraw) 20)
37 ((paul-acc ‘rosebud ‘deposit) 10)
38 ((peter-acc ‘rosehud ‘deposit) 10)

这个是在make-joint 过程中对密码进行判断，如果给的密码和新密码相同就通过旧密码去访问账户

不同的账户只能通过，自己的密码来访问账户

3.8：

 1 (define f
 2   (let ((called #f))
 3     (lambda (x)
 4       (if called
 5           0
 6           (begin (set! called #t)
 7           x)))))
 8
 9 (+ (f 1) (f 0))
10 (+ (f 0) (f 1))