【SICP练习】121 练习3.52

最新推荐文章于 2019-05-07 10:16:48 发布

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最新推荐文章于 2019-05-07 10:16:48 发布

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原文链接：http://www.cnblogs.com/NoMasp/p/4786079.html

本文探讨了使用Scheme语言实现的序列求和与过滤过程。通过定义一系列函数，包括累加、枚举区间、筛选偶数及能被5整除的数，并利用stream相关操作，展示了如何处理数值序列。此外，还讨论了延迟计算机制的不同实现方式对结果的影响。

练习3-52

原文

Exercise 3.52. Consider the sequence of expressions

(define sum 0) 

(define (accum x)
   (set! sum (+ x sum)) 
   sum) 

(define seq (stream-map accum (stream-enumerate-interval 1 20)))
(define y (stream-filter even? seq)) 
(define z (stream-filter (lambda (x) (= (remainder x 5) 0))                         seq)) 

(stream-ref y 7) 
(display-stream z)

What is the value of sum after each of the above expressions is evaluated? What is the printed response to evaluating the stream-ref and display-stream expressions? Would these responses differ if we had implemented (delay ) simply as (lambda () ) without using the optimization provided by memo-proc ? Explain.

代码

(define seq (stream-map accum (stream-enumerate-interval 1 20)))

sum

;Value: 1

(define y (stream-filter even? seq))

sum

;Value: 6

(define z (stream-filter (lambda (x) (= (remainder x 5) 0)) seq))

(stream-ref y 7)

;Value: 136