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问题描述

我正在阅读教程

  import System.Environment 

main :: IO()
main = getArgs>> = print。 haqify。头部

haqify s =Haq!++ s

这个程序在HLint下出现以下错误;

  ./ Haq.hs:11:1:警告:Eta减少
找到:
haqify s =Haq!++ s
为什么不是:
haqify =(Haq!++)

p>

有人可以从中了解Eta Reduce的含义吗?


Eta裁减正在转向 \ x - >>

解决方案

fx 转换为 f ,只要 f 没有<$ $的空闲出现次数C $ C> X 。

要检查它们是否相同,请将它们应用于某些值 y

 (\ x  - > fx)y === f'y  - (其中f'是通过用y代替所有x得到的) )
=== fy - 因为f没有任何空闲的x

您的定义 haqify 被视为 \s - > Haq!++ s ,它是 \s->的语法糖。 (++)Haq!s 。反过来,可以将eta简化为(++)Haq!,或者等价地,对运算符使用分段符号,(Haq !++)


I'm looking at the tutorial http://haskell.org/haskellwiki/How_to_write_a_Haskell_program

import System.Environment

main :: IO ()
main = getArgs >>= print . haqify . head

haqify s = "Haq! " ++ s

When running this program under HLint it gives the following error;

./Haq.hs:11:1: Warning: Eta reduce
Found:
  haqify s = "Haq! " ++ s
Why not:
  haqify = ("Haq! " ++ )

Can someone shed some light on what exactly "Eta Reduce" means in this context?

解决方案

Eta reduction is turning \x -> f x into f as long as f doesn't have a free occurence of x.

To check that they're the same, apply them to some value y:

(\x -> f x) y === f' y -- (where f' is obtained from f by substituting all x's by y)
              === f y  -- since f has no free occurrences of x

Your definition of haqify is seen as \s -> "Haq! " ++ s, which is syntactic sugar for \s -> (++) "Haq! " s. That, in turn can be eta-reduced to (++) "Haq! ", or equivalently, using section notation for operators, ("Haq! " ++).

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10-11 01:20