我有一个共同的模式,我有一个类型级别的[*]
列表,我想将类型* -> *
的类型构造函数应用于列表中的每个元素。例如,我想将类型'[Int, Double, Integer]
更改为'[Maybe Int, Maybe Double, Maybe Integer]
。
这是我尝试实现类型级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'
这会导致错误
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''
我尝试添加大量类型的注释,但错误或多或少都是相同的:GHC甚至不能推断出HList的第一个元素是(正常)列表。我在这做傻事吗?有什么违法的吗?或者有什么办法吗?
答案 0 :(得分:6)
当您编写TypeMap a (b1 ': b2 ': bs)
时,这与您定义Map的递归不一致...当您尝试不是1或2个元素长的TypeMap列表时,这只会导致错误。此外,在你的情况下,只有一个类型系列更清洁。
type family TypeMap (a :: * -> *) (xs :: [*]) :: [*]
type instance TypeMap t '[] = '[]
type instance TypeMap t (x ': xs) = t x ': TypeMap t xs
请注意,这几乎是以下内容的直接翻译:
map f [] = []
map f (x:xs) = f x : map f xs
答案 1 :(得分:4)
使代码编译的最小变化是将[a]
和b1:b2:bs
的实例更改为[]
和b:bs
的实例。
instance TypeMap a '[] where
type Map a '[] = '[]
instance TypeMap a (b ': bs) where
type Map a (b ': bs) = a b ': Map a bs