Haskell 如何为函子制作点(合成)和美元(应用)符号?
我为函子做了点符号(○), 但是我的申请(↯) 不工作,我在Haskell 如何为函子制作点(合成)和美元(应用)符号?,haskell,Haskell,我为函子做了点符号(○), 但是我的申请(↯) 不工作,我在test3函数声明中有一个错误 {-# LANGUAGE TypeOperators #-} module Main where import Protolude -- composition of functors, analog of . infixr 9 ○ type (○) f g a = f (g a) -- functor application, analog of $ infixr 0 ↯ type (↯) f
test3
函数声明中有一个错误
{-# LANGUAGE TypeOperators #-}
module Main where
import Protolude
-- composition of functors, analog of .
infixr 9 ○
type (○) f g a = f (g a)
-- functor application, analog of $
infixr 0 ↯
type (↯) f a = f a
test :: [] (Maybe Int)
test = [Just 1]
test2 :: ([] ○ Maybe) Int
test2 = [Just 1]
test3 :: ([] ○ Maybe) ↯ Int -- error here
test3 = [Just 1]
main :: IO ()
main = do
print test
print test2
return ()
我有个错误
[Error]• The type synonym ‘○’ should have 3 arguments, but has been given 2 • In the type signature: test3 :: ([] ○ Maybe) ↯ Int
怎么了
更新 这里是使用newtype的实现,因为不能部分应用
类型同义词(@M.Aroosi)
我不喜欢它,因为我必须一直用数据类型构造函数包装数据
有没有一种方法可以实现它,而不需要始终使用组合
或应用
包装数据?
{-# LANGUAGE TypeOperators #-}
module Main where
import Protolude
-- I can't use `type` here, because type synonyms cannot be partially applied
-- composition of functors, analog of .
infixr 9 ○
newtype (○) f g a = Composition (f (g a)) deriving (Show)
-- functor application, analog of $
infixr 0 ↯
newtype (↯) f a = Apply (f a) deriving (Show)
test :: [] (Maybe Int)
test = [Just 1]
test2 :: ([] ○ Maybe) Int
test2 = Composition [Just 1]
test2' :: [] ○ Maybe ↯ Int
test2' = Apply (Composition [Just 1])
test3 :: ([] ○ Maybe ○ Maybe) Int
test3 = Composition [Composition (Just (Just 1))]
test3' :: [] ○ Maybe ○ Maybe ↯ Int
test3' = Apply (Composition [Composition (Just (Just 1))])
main :: IO ()
main = do
print test
print test2
print test2'
print test3
print test3'
return ()
更新
这可以在idris中轻松完成
module Main
test : List (Maybe Integer)
test = [Just 1]
-- using (.) from prelude
test1 : (List . Maybe) Integer
test1 = [Just 1]
-- using (.) and ($) from prelude
test2 : List . Maybe $ Integer
test2 = [Just 1]
main : IO ()
main = do
print test
print test1
print test2
更新
使用类型的组合也可以在purescript中使用(耶!)
更新
haskell正在努力使这成为可能
因此,根据您的要求,这里是一个涉及类型族的解决方案。它基于软件包背后的思想,并附有一篇解释该思想的文章
在我开始之前,有一点是赞成使用普通数据类型/newtype的:您可以为合成类型定义函子实例,使其作为单个单元,也就是说,您可以定义实例(函子f,函子g)=>函子(合成f g),其中..
,这是使用下面的方法无法做到的。
可能有一个库允许您使用类型列表而不仅仅是2(因此类似于Compose[Maybe,[],或Int]a
),但我现在似乎找不到它,因此如果有人知道它,它可能是一个比我下面介绍的更好的解决方案(在我看来)
首先,我们需要一些语言扩展:
{-# LANGUAGE
TypeFamilies,
TypeInType,
TypeOperators
#-}
我们还包括类型的数据.Kind
import Data.Kind (Type)
让我们定义一个类型expa
,它将表示a
我们还将定义一个类型族Eval
,它将执行繁重的工作,它将需要Exp a
并给我们一个a
type Exp a = a -> Type
type family Eval (e :: Exp a) :: a
现在我们可以定义运算符(○)代码>和(↯)
(我更喜欢在这里使用更容易键入的运算符,比如#和$,但我会使用您为这个答案选择的运算符)。
我们将它们定义为空数据类型。这就是TypeInType
的作用(和TypeOperators
但这是因为我们使用的是运算符)
请注意,对于他们来说,最后一种是expa
,这允许我们为他们提供Eval
type instance Eval ((○) f g a) = f (Eval (g a))
type instance Eval ((↯) f a) = Eval (f a)
现在您可能想知道“(○)
的第二个参数是类a->Exp b
,但我想给它一些类似的东西,它有类*->*
!”,这就是我们有3种解决方案的地方:
添加另一个操作符,例如(%)
,与(○)
但采用第二种类型的参数a->b
,而不是a->Exp b
。这只需要替换最右边的合成运算符
“提升”类a->b
到a->Exp b
,我将使用名为lift
的数据类型。这只需要对组合中最右边的类型执行
提供一种“不做任何事情”的数据类型a->Exp b
,我将其称为纯
以下是用Haskell编写的三种解决方案:
infixr 9 %
data (%) :: (b -> c) -> (a -> b) -> a -> Exp c
type instance Eval ((%) f g a) = f (g a)
data Lift :: (a -> b) -> a -> Exp b
type instance Eval (Lift f a) = f a
data Pure :: a -> Exp a
type instance Eval (Pure a) = a
使用这个设置,我们可以做的另一件事是创建一个类型级别的函数datatype,我们称之为“Compose”,它将获取一个类型列表并生成它们的组合
data Compose :: [a -> a] -> a -> Exp a
type instance Eval (Compose '[] a) = a
type instance Eval (Compose (x:xs) a) = x (Eval (Compose xs a))
现在我们可以通过一些测试和一个只打印测试值的main
来制作程序:
{-# LANGUAGE
TypeFamilies,
TypeInType,
TypeOperators
#-}
module Main where
import Data.Kind (Type)
type Exp a = a -> Type
type family Eval (e :: Exp a) :: a
infixr 9 ○
data (○) :: (b -> c) -> (a -> Exp b) -> a -> Exp c
infixr 0 ↯
data (↯) :: (a -> Exp b) -> a -> Exp b
type instance Eval ((○) f g a) = f (Eval (g a))
type instance Eval ((↯) f a) = Eval (f a)
infixr 9 %
data (%) :: (b -> c) -> (a -> b) -> a -> Exp c
type instance Eval ((%) f g a) = f (g a)
data Lift :: (a -> b) -> a -> Exp b
type instance Eval (Lift f a) = f a
data Pure :: a -> Exp a
type instance Eval (Pure a) = a
data Compose :: [a -> a] -> a -> Exp a
type instance Eval (Compose '[] a) = a
type instance Eval (Compose (x:xs) a) = x (Eval (Compose xs a))
test :: [] (Maybe Int)
test = [Just 1]
-- using %
test2 :: Eval (([] % Maybe) Int)
test2 = [Just 1]
test2' :: Eval ([] % Maybe ↯ Int)
test2' = [Just 1]
-- works for longer types too
test3 :: Eval (([] ○ Maybe % Maybe) Int)
test3 = [Just (Just 1)]
test3' :: Eval ([] ○ Maybe % Maybe ↯ Int)
test3' = [Just (Just 1)]
-- we can instead Lift the rightmost type
test4 :: Eval (([] ○ Maybe ○ Lift Maybe) Int)
test4 = [Just (Just 1)]
test4' :: Eval ([] ○ Maybe ○ Lift Maybe ↯ Int)
test4' = [Just (Just 1)]
-- an even longer type, with definition "matching" the type declaration
test5 :: Eval ([] ○ Maybe ○ Either Bool % Maybe ↯ Int)
test5 = (:[]) . Just . Right . Just $ 1
-- Same as above, but instead let's use Pure so we don't need to lift the Maybe or use %
test6 :: Eval ([] ○ Maybe ○ Either Bool ○ Maybe ○ Pure ↯ Int)
test6= (:[]) . Just . Right . Just $ 1
-- same as above, uses Compose
test7 :: Eval (Compose [[], Maybe, Either Bool, Maybe] Int)
test7= (:[]) . Just . Right . Just $ 1
main :: IO ()
main = do
print test
print test2
print test2'
print test3
print test3'
print test4
print test4'
print test5
print test6
print test7
return ()
@M.Aroosi(我看到你删除了你的评论)非常糟糕,类型同义词不能部分应用
,每次我都必须将我的对象包装在数据类型构造函数组合
和应用
中时,代码看起来很难看(发布了答案).有没有一种方法可以实现无需数据
或新类型
的函子合成?有没有一种方法可以实现函子合成,而无需始终包装在数据构造函数合成
或应用
中?例如,在idris中是否可以使用类型同义词?有一种使用类型族之类的解决方案,但它涉及使用类型签名,如Eval([]○ 大概○ 也许可以举起来↯ Int)
,或者如果您希望为最后一个(好吧,因为它与右侧关联,所以是第一个)组合引入另一个运算符:Eval([]○ 也许%也许↯ Int)
如果这符合您的要求,请告诉我,我会将其作为答案发布。@M.Aroosi是的,请发布一个答案:)我想补充的一点是,Fcf
包已经定义了其中的大部分内容,您只需要在答案中添加一些内容:合成操作符,应用程序操作符,如果您想要unicode lightning(在Fcf
中已经有一个,他们只需使用($)
)在Fcf
中,他们将我命名的Lift
称为Pure1
,您也可以使用fish操作符对常见问题的意外答案tnx
data Compose :: [a -> a] -> a -> Exp a
type instance Eval (Compose '[] a) = a
type instance Eval (Compose (x:xs) a) = x (Eval (Compose xs a))
{-# LANGUAGE
TypeFamilies,
TypeInType,
TypeOperators
#-}
module Main where
import Data.Kind (Type)
type Exp a = a -> Type
type family Eval (e :: Exp a) :: a
infixr 9 ○
data (○) :: (b -> c) -> (a -> Exp b) -> a -> Exp c
infixr 0 ↯
data (↯) :: (a -> Exp b) -> a -> Exp b
type instance Eval ((○) f g a) = f (Eval (g a))
type instance Eval ((↯) f a) = Eval (f a)
infixr 9 %
data (%) :: (b -> c) -> (a -> b) -> a -> Exp c
type instance Eval ((%) f g a) = f (g a)
data Lift :: (a -> b) -> a -> Exp b
type instance Eval (Lift f a) = f a
data Pure :: a -> Exp a
type instance Eval (Pure a) = a
data Compose :: [a -> a] -> a -> Exp a
type instance Eval (Compose '[] a) = a
type instance Eval (Compose (x:xs) a) = x (Eval (Compose xs a))
test :: [] (Maybe Int)
test = [Just 1]
-- using %
test2 :: Eval (([] % Maybe) Int)
test2 = [Just 1]
test2' :: Eval ([] % Maybe ↯ Int)
test2' = [Just 1]
-- works for longer types too
test3 :: Eval (([] ○ Maybe % Maybe) Int)
test3 = [Just (Just 1)]
test3' :: Eval ([] ○ Maybe % Maybe ↯ Int)
test3' = [Just (Just 1)]
-- we can instead Lift the rightmost type
test4 :: Eval (([] ○ Maybe ○ Lift Maybe) Int)
test4 = [Just (Just 1)]
test4' :: Eval ([] ○ Maybe ○ Lift Maybe ↯ Int)
test4' = [Just (Just 1)]
-- an even longer type, with definition "matching" the type declaration
test5 :: Eval ([] ○ Maybe ○ Either Bool % Maybe ↯ Int)
test5 = (:[]) . Just . Right . Just $ 1
-- Same as above, but instead let's use Pure so we don't need to lift the Maybe or use %
test6 :: Eval ([] ○ Maybe ○ Either Bool ○ Maybe ○ Pure ↯ Int)
test6= (:[]) . Just . Right . Just $ 1
-- same as above, uses Compose
test7 :: Eval (Compose [[], Maybe, Either Bool, Maybe] Int)
test7= (:[]) . Just . Right . Just $ 1
main :: IO ()
main = do
print test
print test2
print test2'
print test3
print test3'
print test4
print test4'
print test5
print test6
print test7
return ()