【SICP练习】106 练习3.7

练习3-7

原文

Exercise 3.7. Consider the bank account objects created by make-account, with the password modification described in exercise 3.3. Suppose that our banking system requires the ability to make joint accounts. Define a procedure make-joint that accomplishes this. Make-joint should take three arguments. The first is a password-protected account. The second argument must match the password with which the account was defined in order for the make-joint operation to proceed. The third argument is a new password. Make-joint is to create an additional access to the original account using the new password. For example, if peter-acc is a bank account with password open-sesame, then

(define paul-acc (make-joint peter-acc 'open-sesame 'rosebud))

will allow one to make transactions on peter-acc using the name paul-acc and the password rosebud. You may wish to modify your solution to exercise 3.3 to accommodate this new feature.

分析

make-joint需要有3个参数：
1.有密码保护的帐户名
2.必须与账号的密码匹配的原密码
3.新密码

而其会返回一个过程，因此在此处需要一个lambda表达式，并且其有一个参数mode和一个传入的密码参数。另外在输出错误信息的函数中也需要一个参数，即是它并不使用，只是出于兼容性的考虑，在前面的博客中我们也遇到过这种问题。

代码

(define (make-joint origin-acc old-password new-password)

  (define (display-wrong-message msg)
    (display "Incorrect password"))

  (lambda (given-password mode)
    (if (eq? given-password new-password)
        (origin-acc old-password mode)
        display-wrong-message)))
;Value: make-joint

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