Question

所以我正在尝试为可变长度元组创建一个类型，基本上是Either a (Either (a,b) (Either (a,b,c) ...))和Either (Either (Either ... (x,y,z)) (y,z)) z的更漂亮的版本。

{-# LANGUAGE TypeOperators, TypeFamilies, MultiParamTypeClasses, FlexibleInstances #-}
module Temp where

-- type level addition
data Unit
data Succ n

class Summable n m where
  type Sum n m :: *

instance Summable Unit m where
  type Sum Unit m = Succ m

instance Summable n m => Summable (Succ n) m where
  type Sum (Succ n) m = Succ (Sum n m)

-- variable length tuple, left-to-right
data a :+ b = a :+ Maybe b
infixr 5 :+

class Prependable t r s where
  type Prepend t r s :: *
  prepend :: r -> Maybe s -> Prepend t r s

instance Prependable Unit x y where
  type Prepend Unit x y = x :+ y
  prepend = (:+)

instance Prependable n x y => Prependable (Succ n) (w :+ x) y where
  type Prepend (Succ n) (w :+ x) y = w :+ Prepend n x y
  prepend (w :+ Nothing) _ = w :+ Nothing
  prepend (w :+ Just x) y = w :+ Just (prepend x y)

-- variable length tuple, right-to-left
data a :- b = Maybe a :- b
infixl 5 :-

class Appendable t r s where
  type Append t r s :: *
  append :: Maybe r -> s -> Append t r s

instance Appendable Unit x y where
  type Append Unit x y = x :- y
  append = (:-)

instance Appendable n x y => Appendable (Succ n) x (y :- z) where
  type Append (Succ n) x (y :- z) = Append n x y :- z
  append _ (Nothing :- z) = Nothing :- z
  append x (Just y :- z) = Just (append x y) :- z

但是，编译器似乎无法在递归情况下推断出prepend或append的幻像类型参数：

Temp.hs:32:40:
    Could not deduce (Prepend t1 x y ~ Prepend n x y)
    from the context (Prependable n x y)
      bound by the instance declaration at Temp.hs:29:10-61
    NB: `Prepend' is a type function, and may not be injective
    In the return type of a call of `prepend'
    In the first argument of `Just', namely `(prepend x y)'
    In the second argument of `(:+)', namely `Just (prepend x y)'

Temp.hs:49:34:
    Could not deduce (Append t0 x y ~ Append n x y)
    from the context (Appendable n x y)
      bound by the instance declaration at Temp.hs:46:10-59
    NB: `Append' is a type function, and may not be injective
    In the return type of a call of `append'
    In the first argument of `Just', namely `(append x y)'
    In the first argument of `(:-)', namely `Just (append x y)'

我能做些什么来帮助编译器进行推理吗？

Answer 1

此处错误消息的重要部分是：

NB: `Prepend' is a type function, and may not be injective

这是什么意思？这意味着可能有多个instance Prependable这样type Prepend ... = a，因此，如果您将某些Prepend推断为a，则您不一定知道它属于哪个实例到。

你可以使用data types in type families来解决这个问题，它的优势在于你不处理类型函数，这些函数是完全的，但可能是单射的，而不是类型“关系“，这是双射的（因此每个Prepend类型只能属于一个类型系列，并且每个类型系列都有不同的Prepend类型。”

（如果您希望我在类型系列中显示包含数据类型的解决方案，请发表评论！基本上，只需使用data Prepend代替type Prepend）

Answer 2

我想出的解决方案是添加一个伪参数来将prepend和append绑定到幻像参数：

-- as above, except...

unsucc :: Succ n -> n
unsucc _ = undefined

class Prependable t r s where
  type Prepend t r s :: *
  prepend :: t -> r -> Maybe s -> Prepend t r s

instance Prependable Unit x y where
  type Prepend Unit x y = x :+ y
  prepend _ = (:+)

instance Prependable n x y => Prependable (Succ n) (w :+ x) y where
  type Prepend (Succ n) (w :+ x) y = w :+ Prepend n x y
  prepend _ (w :+ Nothing) _ = w :+ Nothing
  prepend t (w :+ Just x) y = w :+ Just (prepend (unsucc t) x y)

class Appendable t r s where
  type Append t r s :: *
  append :: t -> Maybe r -> s -> Append t r s

instance Appendable Unit x y where
  type Append Unit x y = x :- y
  append _ = (:-)

instance Appendable n x y => Appendable (Succ n) x (y :- z) where
  type Append (Succ n) x (y :- z) = Append n x y :- z
  append _ _ (Nothing :- z) = Nothing :- z
  append t x (Just y :- z) = Just (append (unsucc t) x y) :- z

GHC：无法推断幻像类型参数

2 个答案: