【SICP练习】107 练习3.8

2015/09/08 21:47
阅读数 55

练习3-8

原文

Exercise 3.8. When we defined the evaluation model in section 1.1.3, we said that the first step in evaluating an expression is to evaluate its subexpressions. But we never specified the order in which the subexpressions should be evaluated (e.g., left to right or right to left). When we introduce assignment, the order in which the arguments to a procedure are evaluated can make a difference to the result. Define a simple procedure f such that evaluating

(+ (f 0) (f 1)) 

will return 0 if the arguments to + are evaluated from left to right but will return 1 if the arguments are evaluated from right to left.

分析

题目中已经说明了求解一个表达式的第一步就是求值其中的子表达式,而对于

(+ (f 0) (f 1)) 

意味着分别求解(f 0)和(f 1),可以猜测着其中有2个lambda表达式。又因为从左往右和从右往左的求值结果不同,则意味着2个lambda是嵌套的关系。

代码

(define f (lambda (first-value) (set! f (lambda (second-valua) 0)) first-value))
;Value: 

测试

(define f (lambda (first-value) (set! f (lambda (second-valua) 0)) first-value))
;Value: f

(+ (f 0) (f 1))
;Value: 1

(+ (f 1) (f 0))
;Value: 0

总结

当我们调用(f n)时,n为实参将会代替第四行的first-value,并进一步传入第二行的lambda表达式。返回的结果则是n,但与此同时又将

(lambda (second-value) 0)

赋值给过程f。而当下一次调用(f m)时,不论m为何值,都会返回0。也就是说第一次传入的n,(f n)返回值为n,以后传入的m,(f m)返回0,并且无论传入多少次m,返回值均为0。当然了,m的值并未改变。

(f 1)
;Value: 1

(f 1)
;Value: 0

(f 1)
;Value: 0

(f 1)
;Value: 0

(define m 2)
;Value: m

(f m)
;Value: 0

m
;Value: 2

同时也可以得出结论,MIT-Scheme对子表达式的求值顺序是从右至左。

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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