function - Haskell 中无点函数的先决条件是什么

Question

我一直认为 pointfree 函数的先决条件是将函数参数获取到定义的末尾。例如

-- 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)

我最初看到这个阅读this question 。我们有这个例子:

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成为无点的，而其他两个则不能？

Answer 1

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

\x -> length `map` x 写成自由点就是map length。中缀反引号只是语法糖。 (正如切普纳指出的，如果你真的想要它，你可以使用一个部分，即 (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“容易”。如果你足够努力的话，几乎任何事情都可以成为免费的。在这种情况下，如果我们用 (.) 而不是 ($)< 来编写 agreeLen，则省略 y 参数很容易。/:

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(.)与zipWith (==) x组成来处理它其他函数只是用函数修改值的另一种情况:

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 (==)

这并不是您在实践中真正想要做的事情，但它肯定是可能的。