Question

我想出了一个很好的练习，但是无法使它发挥作用。

我的想法是尝试表达罗马数字，使得类型检查器会告诉我数字是否有效。

    {-# LANGUAGE RankNTypes
               , MultiParamTypeClasses #-}

    data One a b c = One a deriving (Show, Eq)
    data Two a b c = Two (One a b c) (One a b c) deriving (Show, Eq)
    data Three a b c = Three (One a b c) (Two a b c) deriving (Show, Eq)
    data Four a b c = Four (One a b c) (Five a b c) deriving (Show, Eq)
    data Five a b c = Five b deriving (Show, Eq)
    data Six a b c = Six (Five a b c) (One a b c) deriving (Show, Eq)
    data Seven a b c = Seven (Five a b c) (Two a b c) deriving (Show, Eq)
    data Eight a b c = Eight (Five a b c) (Three a b c) deriving (Show, Eq)
    data Nine a b c d e = Nine (One a b c) (One c d e) deriving (Show, Eq)

    data Z = Z deriving (Show, Eq) -- dummy for the last level
    data I = I deriving (Show, Eq)
    data V = V deriving (Show, Eq)
    data X = X deriving (Show, Eq)
    data L = L deriving (Show, Eq)
    data C = C deriving (Show, Eq)
    data D = D deriving (Show, Eq)
    data M = M deriving (Show, Eq)

    i :: One I V X
    i = One I

    v :: Five I V X
    v = Five V

    x :: One X L C
    x = One X

    l :: Five X L C
    l = Five L

    c :: One C D M
    c = One C

    d :: Five C D M
    d = Five D

    m :: One M Z Z
    m = One M

    infixr 4 #

    class RomanJoiner a b c where
      (#) :: a -> b -> c

    instance RomanJoiner (One a b c) (One a b c) (Two a b c) where
      (#) = Two

    instance RomanJoiner (One a b c) (Two a b c) (Three a b c) where
      (#) = Three

    instance RomanJoiner (One a b c) (Five a b c) (Four a b c) where
      (#) = Four

    instance RomanJoiner (Five a b c) (One a b c) (Six a b c) where
      (#) = Six

    instance RomanJoiner (Five a b c) (Two a b c) (Seven a b c) where
      (#) = Seven

    instance RomanJoiner (Five a b c) (Three a b c) (Eight a b c) where
      (#) = Eight

    instance RomanJoiner (One a b c) (One c d e) (Nine a b c d e) where
      (#) = Nine

    main = print $ v # i # i

这可能会有所不同，并且解决方案是不完整的，但是现在我需要理解为什么它抱怨没有RomanJoiner（One IVX）（One IVX）b0的实例，而我认为我声明了这样的木匠。

Answer 1

问题是没有根据实例选择实例只有一个有效：一个扩展FunctionalDependencies有助于获得更多类型推断。启用它，并用| a b -> c说明类型可以从a # b和a的类型推断出b。不幸的是，这不是您必须做的唯一事情，因为您会收到错误Functional dependencies conflict between instance declarations。使用HList中定义的一些类（这些可以在其他任何地方定义），冲突的两个实例可以合并为一个，其中两个（如果计算错误则为3）可能的结果是根据某些类型是否相等来选择的

关于这个解决方案的一些评论是丑陋的：

您不必复制值类型级别的内容再次级别（hCond vs HCond）如果你有更懒的Show实例（比如 instance Show I where show _ = "I"）。

更多现代扩展TypeFamilies许多中间类型变量ba, bb, bc, babc ...可以消除。

{-# LANGUAGE RankNTypes, MultiParamTypeClasses, FunctionalDependencies, ScopedTypeVariables, UndecidableInstances, FlexibleContexts, FlexibleInstances #-}
import Data.HList hiding ((#))
import Data.HList.TypeEqGeneric1
import Data.HList.TypeCastGeneric1
import Unsafe.Coerce

data One a b c = One a deriving (Show, Eq)
data Two a b c = Two (One a b c) (One a b c) deriving (Show, Eq)
data Three a b c = Three (One a b c) (Two a b c) deriving (Show, Eq)
data Four a b c = Four (One a b c) (Five a b c) deriving (Show, Eq)
data Five a b c = Five b deriving (Show, Eq)
data Six a b c = Six (Five a b c) (One a b c) deriving (Show, Eq)
data Seven a b c = Seven (Five a b c) (Two a b c) deriving (Show, Eq)
data Eight a b c = Eight (Five a b c) (Three a b c) deriving (Show, Eq)
data Nine a b c d e = Nine (One a b c) (One c d e) deriving (Show, Eq)

data Z = Z deriving (Show, Eq) -- dummy for the last level
data I = I deriving (Show, Eq)
data V = V deriving (Show, Eq)
data X = X deriving (Show, Eq)
data L = L deriving (Show, Eq)
data C = C deriving (Show, Eq)
data D = D deriving (Show, Eq)
data M = M deriving (Show, Eq)

i :: One I V X
i = One I

v :: Five I V X
v = Five V

x :: One X L C
x = One X

l :: Five X L C
l = Five L

c :: One C D M
c = One C

d :: Five C D M
d = Five D

m :: One M Z Z
m = One M

infixr 4 #

class RomanJoiner a b c | a b -> c where
    (#) :: a -> b -> c


instance RomanJoiner (One a b c) (Two a b c) (Three a b c) where
    (#) = Three

instance RomanJoiner (One a b c) (Five a b c) (Four a b c) where
    (#) = Four

instance RomanJoiner (Five a b c) (One a b c) (Six a b c) where
    (#) = Six

instance RomanJoiner (Five a b c) (Two a b c) (Seven a b c) where
    (#) = Seven

instance RomanJoiner (Five a b c) (Three a b c) (Eight a b c) where
    (#) = Eight

data Error = Error
instance forall a b c a' b' c' ba bb bc bab babc z bn nine.
   (TypeEq a a' ba,
    TypeEq b b' bb,
    TypeEq c c' bc,
    HAnd ba bb bab,
    HAnd bab bc babc,

    TypeEq c a' bn,
    HCond bn (Nine a b c b' c') Error nine,

    HCond babc (Two a b c) nine  z) =>
        RomanJoiner (One a b c) (One a' b' c') z where
    (#) x y = hCond (undefined :: babc)
                (Two (uc x :: One a b c) (uc y :: One a b c)) $
              hCond (undefined :: bn)
                (Nine (uc x :: One a b c) (uc y :: One c b' c'))
                Error
        where uc = unsafeCoerce

main = print $ v # i # i
{-
Prints with ghc 762, HList-0.2.3

*Main> main
Seven (Five V) (Two (One I) (One I)

-}

声明实例的类型推断

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