你将如何（重新）在Haskell中实现迭代？

问题描述

  （++） lambdas，pattern mataching，guards等） 
解决方案
好吧， iterate 构造了一个无限值列表 a 由 f 完成。因此，我将首先编写一个函数，该函数将前缀 a 添加到通过 fa 递归调用迭代构建的列表中： 
 
 
  iterate ::（a  - > a） - > a  - > [a] 
迭代fa = a：iterate f（fa）
  
感谢懒惰评估，只有计算我函数值所需的构造列表的那部分才会被评估。
 
iterate :: (a -> a) -> a -> [a]
(As you probably know) iterate is a function that takes a function and starting value. Then it applies the function to the starting value, then it applies the same function to the last result, and so on. 
Prelude> take 5 $ iterate (^2) 2
[2,4,16,256,65536]
Prelude>
The result is an infinite list. (that's why I use take).My question how would you implement your own iterate' function in Haskell, using only the basics ((:) (++) lambdas, pattern mataching, guards, etc.) ?
(Haskell beginner here)
 解决方案 
Well, iterate constructs an infinite list of values a incremented by f. So I would start by writing a function that prepended some value a to the list constructed by recursively calling iterate with f a:
iterate :: (a -> a) -> a -> [a]
iterate f a = a : iterate f (f a)
Thanks to lazy evaluation, only that portion of the constructed list necessary to compute the value of my function will be evaluated.
                        
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