<=和>运算符作为函数参数
快速排序：
-- First variant:
qsort :: (Ord a) => [a] -> [a]
qsort [] = []
qsort (x:xs) = left x ++ [x] ++ right x
where left n = qsort [m | m <- xs, m <= n]
right n = qsort [m | m <- xs, m > n]
-- λ: qsort [10,2,5,3,1,6,7,4,2,3,4,8,9]
-- [1,2,2,3,3,4,4,5,6,7,8,9,10]我看到left和right函数几乎是相同的。所以我想把它改写得更短一些..。就像这样：
-- Second variant:
qsort' :: (Ord a) => [a] -> [a]
qsort' [] = []
qsort' (x:xs) = (srt <=) ++ [x] ++ (srt >)
where srt f = qsort' [m | m <- xs, m f x]但是，当我试图将它加载到ghci中时，会出现错误
λ: :load temp
[1 of 1] Compiling Main ( temp.hs, interpreted )
temp.hs:34:18:
Couldn't match expected type `[a]'
with actual type `(t0 -> [a]) -> Bool'
Relevant bindings include
srt :: forall t. t -> [a] (bound at temp.hs:35:9)
xs :: [a] (bound at temp.hs:34:11)
x :: a (bound at temp.hs:34:9)
qsort' :: [a] -> [a] (bound at temp.hs:33:1)
In the first argument of `(++)', namely `(srt <=)'
In the expression: (srt <=) ++ [x] ++ (srt >)
In an equation for qsort':
qsort' (x : xs)
= (srt <=) ++ [x] ++ (srt >)
where
srt f = qsort' [m | m <- xs, m f x]
temp.hs:34:37:
Couldn't match expected type `[a]'
with actual type `(t1 -> [a]) -> Bool'
Relevant bindings include
srt :: forall t. t -> [a] (bound at temp.hs:35:9)
xs :: [a] (bound at temp.hs:34:11)
x :: a (bound at temp.hs:34:9)
qsort' :: [a] -> [a] (bound at temp.hs:33:1)
In the second argument of `(++)', namely `(srt >)'
In the second argument of `(++)', namely `[x] ++ (srt >)'
In the expression: (srt <=) ++ [x] ++ (srt >)
temp.hs:35:38:
Could not deduce (a ~ (t -> a -> Bool))
from the context (Ord a)
bound by the type signature for qsort' :: Ord a => [a] -> [a]
at temp.hs:32:11-31
`a' is a rigid type variable bound by
the type signature for qsort' :: Ord a => [a] -> [a]
at temp.hs:32:11
Relevant bindings include
m :: a (bound at temp.hs:35:29)
f :: t (bound at temp.hs:35:13)
srt :: t -> [a] (bound at temp.hs:35:9)
xs :: [a] (bound at temp.hs:34:11)
x :: a (bound at temp.hs:34:9)
qsort' :: [a] -> [a] (bound at temp.hs:33:1)
The function `m' is applied to two arguments,
but its type `a' has none
In the expression: m f x
In a stmt of a list comprehension: m f x
Failed, modules loaded: none.
λ:我读了错误信息，但我还是不明白原因.
回答 1
Stack Overflow用户
发布于 2015-01-05 11:59:00
您不应该使用f作为infix。您可以通过将f放在前面并表示括号(<=)之间的函数来解决这个问题。
-- third variant:
qsort' :: (Ord a) => [a] -> [a]
qsort' [] = []
qsort' (x:xs) = (srt (<=)) ++ [x] ++ (srt (>))
where srt f = qsort' [m | m <- xs, f m x]这主要是因为您基本上想要做的是调用m和x上的 f。现在默认的lambda演算总是先计算左边列出的函数。
Haskell只为操作符提供了一些语法糖:当您编写a+b时，基本上编写的是(+) a b (窗帘后面)。这是Haskell最喜欢的，但是编译器为程序员提供了一些功能。因为编写a*b+c*d比编写(+) ((*) a b) ((*) c d)要容易得多，但是第二个问题实际上是如何用lambda微积分编写这样的东西。
为了将运算符看作函数，将它们写在括号中，因此要获得<=的函数变体，需要编写(<=)。
编辑
正如@Jubobs所指出的，您也可以使用infix，但因此需要使用backticks：
-- fourth variant:
qsort' :: (Ord a) => [a] -> [a]
qsort' [] = []
qsort' (x:xs) = (srt (<=)) ++ [x] ++ (srt (>))
where srt f = qsort' [m | m <- xs, m `f` x]问题主要是您需要通过f传递您的函数，而<=和>不是函数，(<=)和(>)是函数。从技术上讲，这个故事有点复杂，但我想在学习基础知识时，这就足够了。
Haskell通过使用backticks读到：
x `f` y作为：
f x y(请注意，这并不是完全正确的，因为操作符也具有优先级：*比+绑定得更紧，但这些更多地是过程的“细节”)。
在操作符上加上括号是一种相反的效果：
x o y是
(o) x yo是一个操作符。
https://stackoverflow.com/questions/27778999
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