SICP exercise 1.43

cfmonkey

于 2007-06-13 17:28:00 发布

阅读量488

点赞数

CC 4.0 BY-SA版权

文章标签： application function integer lambda

本文链接：https://blog.youkuaiyun.com/cfmonkey/article/details/1650836

本文介绍了一个过程，通过定义一个过程来实现数值函数f的n次重复应用。此过程接受一个计算f的过程和两个正整数作为输入，并返回计算第n次重复应用f的过程。文章还提供了一个具体的实现示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

书的网页提供的Complete Code中没有. 所以加上

Exercise 1.43.
If f is a numerical function and n is a positive integer, then we can form the nth repeated
application of f, which is defined to be the function whose value at x is f(f(...(f(x))...)). For example, if
f is the function x ->x + 1, then the nth repeated application of f is the function x ->x + n. If f is the
operation of squaring a number, then the nth repeated application of f is the function that raises its
n th power. Write a procedure that takes as inputs a procedure that computes f and a argument to the 2 positive integer n and returns the procedure that computes the nth repeated application of f. Your procedure should be able to be used as follows:
((repeated square 2) 5)
625
Hint: You may find it convenient to use compose from exercise 1.42

solvation:

(define (repeated fn n)
(lambda (x)
    (if (= n 1)
        (fn x)
        (fn ((repeated fn (- n 1)) x)))))

(define (square x)
(* x x))