我有一個常見模式,其中有一個類型級別列表[*]
,我想將類型* -> *
的構造函數應用於列表中的每個元素。例如,我想將'[Int, Double, Integer]
的類型更改爲'[Maybe Int, Maybe Double, Maybe Integer]
。使用DataKinds的類型級別映射
這裏是我嘗試實現類型級別map
。
{-# LANGUAGE TypeFamilies, MultiParamTypeClasses, FlexibleInstances, FlexibleContexts, TypeOperators, DataKinds, ScopedTypeVariables, GADTs #-}
-- turns a type list '[b1, b2, b3]
-- into the type list '[a b1, a b2, a b3]
class TypeMap (a :: * -> *) (bs :: [*]) where
type Map a bs :: [*]
instance TypeMap a '[b] where
type Map a '[b] = '[a b]
instance TypeMap a (b1 ': b2 ': bs) where
type Map a (b1 ': b2 ': bs) = ((a b1) ': (Map a (b2 ': bs)))
data HList :: [*] -> * where
HNil :: HList '[]
HCons :: a -> HList as -> HList (a ': as)
class Foo as where
toLists :: HList as -> HList (Map [] as)
instance Foo '[a] where
toLists (HCons a HNil) = HCons [a] HNil
instance (Foo (a2 ': as)) => Foo (a1 ': a2 ': as) where
toLists (HCons a as) =
let as' = case (toLists as) of
(HCons a2 as'') -> HCons [head a2] as'' -- ERROR
in HCons [a] as'
這將導致錯誤
Could not deduce (a3 ~ [t0])
from the context (Foo ((':) * a2 as))
bound by the instance declaration at Test.hs:35:10-50
or from ((':) * a1 ((':) * a2 as) ~ (':) * a as1)
bound by a pattern with constructor
HCons :: forall a (as :: [*]).
a -> HList as -> HList ((':) * a as),
in an equation for `toLists'
at Test.hs:36:14-23
or from (Map [] as1 ~ (':) * a3 as2)
bound by a pattern with constructor
HCons :: forall a (as :: [*]).
a -> HList as -> HList ((':) * a as),
in a case alternative
at Test.hs:38:22-34
`a3' is a rigid type variable bound by
a pattern with constructor
HCons :: forall a (as :: [*]).
a -> HList as -> HList ((':) * a as),
in a case alternative
at Test.hs:38:22
Expected type: HList (Map [] ((':) * a2 as))
Actual type: HList ((':) * [t0] as2)
In the return type of a call of `HCons'
In the expression: HCons [head a2] as''
In a case alternative: (HCons a2 as'') -> HCons [head a2] as''
我已經嘗試添加豐富的類型註釋,但錯誤或多或少出來的一樣:GHC甚至不能推斷出的第一個元素HList是一個(正常)列表。我在這裏做些傻事嗎?非法的東西?或者有什麼辦法嗎?
爲什麼你沒有'TypeMap a []'實例? –
@DanielWagner同意,這些實例可能應該是'[]和(a':as)。 – crockeea