Question

我有一个按类型级别列表索引的数据系列，其中列表中的类型对应于数据实例的参数。我想编写具有不同arity和参数的函数，具体取决于数据实例，因此我可以像家庭中的每个数据实例一样使用它。

{-# LANGUAGE KindSignatures, DataKinds, TypeOperators, 
             TypeFamilies, FlexibleInstances, PolyKinds #-}

module Issue where


type family (->>) (l :: [*]) (y :: *) :: * where
    '[]       ->> y = y
    (x ': xs) ->> y = x -> (xs ->> y)

class CVal (f :: [*]) where
    data Val f :: *
    construct :: f ->> Val f

instance CVal '[Int, Float, Bool] where
    data Val '[Int, Float, Bool] = Val2 Int Float Bool
    construct = Val2

编译好。但是当我尝试应用construct函数时：

v :: Val '[Int, Float, Bool]
v = construct 0 0 True

它会产生错误：

Couldn't match expected type `a0
                              -> a1 -> Bool -> Val '[Int, Float, Bool]'
            with actual type `f0 ->> Val f0'
The type variables `f0', `a0', `a1' are ambiguous
The function `construct' is applied to three arguments,
but its type `f0 ->> Val f0' has none
In the expression: construct 0 0 True
In an equation for `v': v = construct 0 0 True

Answer 1

您的代码无法进行类型检查，因为type families are not (necessarily) injective。如果您通过在f中指定f ->> Val f的选项来帮助GHC，那么它会按预期工作：

{-# LANGUAGE KindSignatures, DataKinds, TypeOperators, 
             TypeFamilies, FlexibleInstances, PolyKinds #-}

module Issue where

import Data.Proxy

type family (->>) (l :: [*]) (y :: *) :: * where
    '[]       ->> y = y
    (x ': xs) ->> y = x -> (xs ->> y)

class CVal (f :: [*]) where
    data Val f :: *
    construct :: proxy f -> f ->> Val f

instance CVal '[Int, Float, Bool] where
    data Val '[Int, Float, Bool] = Val2 Int Float Bool deriving Show
    construct _ = Val2

v :: Val '[Int, Float, Bool]
v = construct (Proxy :: Proxy '[Int, Float, Bool]) 0 0 True

关键是将Proxy :: Proxy '[Int, Float, Bool]参数传递给construct，从而修复f的选择。这是因为没有任何东西阻止你让f1和f2类型f1 ->> Val f1 ~ f2 ->> Val f2。

别担心，this shortcoming of the language is being looked at。

将类型级列表'[a，b，c，...]转换为函数a-＆gt; b-＆gt; c-＆gt;

1 个答案: