如何为ListIsomorphic创建Vector.Unboxed实例？

Question

由于 instance for generic vectors似乎是不可能的(或者不是一个好主意)，所以我试图直接为Vector.Unboxed编写一个。

class ListIsomorphic l where
    toList    :: l a -> [a]
    fromList  :: [a] -> l a

instance ListIsomorphic UV.Vector where
    toList   = UV.toList
    fromList = UV.fromList

我得到了以下错误：

test.hs:70:14:
    No instance for (UV.Unbox a) arising from a use of ‘UV.toList’
    Possible fix:
    add (UV.Unbox a) to the context of
        the type signature for toList :: UV.Vector a -> [a]
    In the expression: UV.toList
    In an equation for ‘toList’: toList = UV.toList
    In the instance declaration for ‘ListIsomorphic UV.Vector’

test.hs:71:16:
    No instance for (UV.Unbox a) arising from a use of ‘UV.fromList’
    Possible fix:
    add (UV.Unbox a) to the context of
        the type signature for fromList :: [a] -> UV.Vector a
    In the expression: UV.fromList
    In an equation for ‘fromList’: fromList = UV.fromList
    In the instance declaration for ‘ListIsomorphic UV.Vector’

我试过遵循编译器错误建议，但没有成功。我怎么写呢？

Answer 1

您不能。您的ListIsomorphic类保证fromList可以应用于任何类型的a。但是，未装箱向量要求a是Unbox的一个实例。

您可以编写一个类ListIsomorphicUnboxed，该类包含：

class ListIsomorphicUnboxed l where
    fromListUnboxed :: (Unbox a) => [a] -> l a
    toListUnboxed :: (Unbox a) => l a -> [a]

另一种方法可以是使用约束(您需要添加一些语言扩展)：

{-# LANGUAGE ConstraintKinds, TypeFamilies #-}
import GHC.Exts (Constraint)

class ListIsomorphic l where
    type C l a :: Constraint
    type C l a = () -- default to no constraint

    fromList :: C l a => [a] -> l a
    toList :: C l a => l a -> [a]

instance ListIsomorphic UV.Vector where
    type C UV.Vector a = UV.Unbox a
    toList   = UV.toList
    fromList = UV.fromList

(代码未经测试！)

这样，每个实例都可以有不同的约束。然而，这并不是一个非常普遍的解决方案。