嵌套元组链上的多态函数

Question

假设我有一个类型列表，类型 [*]:

let Ts = '[Int, Bool, Char]

我想将其转换为元组链：

type family Tupled (ts :: [*]) z :: *
type instance Tupled (t ': ts) z = (t, Tupled ts z)
type instance Tupled '[] z = z

目前一切顺利：

> :kind! Tupled Ts ()
Tupled Ts () :: *
= (Int, (Bool, (Char, ())))

现在我希望能够编写一个类型 Fun 来表示该链“底部”的多态函数。例如，Fun Ts Ts 应该适用于以下任一类型：

(Int, (Bool, (Char, (String, ()))))
(Int, (Bool, (Char, (Word, (ByteString, ())))))

我试过这个：

newtype Fun as bs = Fun
  { unfun :: forall z. Tupled as z -> Tupled bs z }

但它无法进行类型检查：

Couldn't match type ‘Tupled bs z’ with ‘Tupled bs z0’
NB: ‘Tupled’ is a type function, and may not be injective
The type variable ‘z0’ is ambiguous
Expected type: Tupled as z -> Tupled bs z
  Actual type: Tupled as z0 -> Tupled bs z0
In the ambiguity check for the type of the constructor ‘Fun’:
  Fun :: forall z. Tupled as z -> Tupled bs z
To defer the ambiguity check to use sites, enable AllowAmbiguousTypes
In the definition of data constructor ‘Fun’
In the newtype declaration for ‘Fun’

我看到了使用数据族来避免单射性问题的建议：

data family Tupled (ts :: [*]) z :: *
data instance Tupled (t ': ts) z = Cons t (Tupled ts z)
data instance Tupled '[] z = Nil

确实这使得 Fun 可以编译，但看起来这让我在 Cons 和 Nil 的土地上“卡住了”，当我想要使用元组时，像这样：

Fun $ \ (i, (b, (c, z))) -> (succ i, (not b, (pred c, z)))

我能以某种方式解决这个问题吗？

Answer 1

启用AllowAmbiguousTypes。从 GHC 8 开始，歧义检查是完全多余的，因为任何（基本可解决的）歧义都可以通过类型应用程序解决。此外，您的案例似乎只是歧义检查的误报，因为即使没有类型应用程序，我们也可以清楚地使用 Fun。

嵌套元组链上的多态函数

Polymorphic functions over chains of nested tuples

haskell

fold

type-level-computation