Haskell中无点功能的先决条件是什么 [英] What are the prerequisites for a point-free function in Haskell

问题描述

我一直认为，无点函数的前提条件是使函数参数到达定义的末尾.例如

I always thought that the prerequisites for a pointfree function were to get the function arguments to the end of the definition. E.g.

-- This can be made pointfree quite easily:    
let lengths x = map length x    
let lengths' = map length

-- However this cannot:
let lengthz x = length `map` x
-- let lengthz' = length `map` (parse error)

我最初遇到的是

I originally came across this reading this question. There we have this example:

agreeLen :: (Eq a) => [a] -> [a] -> Int
agreeLen x y = length $ takeWhile id $ zipWith (==) x y
-- This may look like it can easily be made pointfree, however it cannot
-- agreeLen' :: (Eq a) => [a] -> [a] -> Int
-- agreeLen' = length $ takeWhile id $ zipWith (==) (zipWith is applied to too few arguments)

那么为什么我的第一个示例可以变得没有意义，而其他两个却不能?

So why can my first example be made pointfree, but the other two cannot?

推荐答案

-- However this cannot:
let lengthz x = length `map` x
-- let lengthz' = length `map` (parse error)

\ x->长度`map` x 的免费点简直就是 map length .中缀反引号只是语法糖.(正如chepner指出的，如果您确实想要它，则可以使用一个部分，即(长度为"map"的长度).)

\x -> length `map` x written point free is simply map length. The infix backticks are just syntactical sugar. (As chepner points out, if you really want it you can use a section, i.e. (length `map`).)

agreeLen :: (Eq a) => [a] -> [a] -> Int
agreeLen x y = length $ takeWhile id $ zipWith (==) x y
-- This may look like it can easily be made pointfree, however it cannot

这里的关键词是轻松".如果您足够努力，几乎所有东西都可以变得毫无意义.在这种情况下，如果我们用(.)而不是($):

The key word here is "easily". Pretty much anything can be made point-free if you try hard enough. In this case, omitting the y parameter is easy if we write agreeLen in terms of (.) rather than ($):

agreeLen x y = (length . takeWhile id . zipWith (==) x) y
agreeLen x = length . takeWhile id . zipWith (==) x

对于 x ，我们可以通过使用(.)将 zipWith(==)x 与其他功能只是使用功能修改值的另一种情况:

As for x, we can handle it by treating the use of (.) to compose zipWith (==) x with the other functions as just another case of a value being modified with a function:

agreeLen x = (.) (length . takeWhile id) (zipWith (==) x)
agreeLen x = ((length . takeWhile id) .) (zipWith (==) x) -- cosmetical change
agreeLen x = (((length . takeWhile id) .) . zipWith (==)) x
agreeLen = ((length . takeWhile id) .) . zipWith (==)

这不是您实际上在实践中想要做的事情，但是肯定有可能.

It is not something you'd actually want to do in practice, but it is certainly possible.

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