声明实例的类型推断

Question

我想出了一个很好的练习，但做不到。

我们的想法是尝试用这样的方式来表达罗马数字，这样类型检查器就可以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

这可能有不同的方法，而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 (一个I V)(一个I V) b0的实例，而我认为我声明了这样一个合并器。

Answer 1

问题是，选择实例并不是基于唯一有效的实例:一个扩展FunctionalDependencies有助于获得更多的类型推断。启用该功能，并使用| a b -> c表示可以从a和b的类型推断出a # b的类型。不幸的是，这并不是您唯一需要做的事情，因为您将得到错误的Functional dependencies conflict between instance declarations。使用HList中定义的某些类(这些类可以在其他任何地方定义)，冲突的两个实例可以组合成一个实例，其中两个实例(如果计算错误)将根据某些类型是否相等来选择可能的结果。

关于这个解决方案的一些评论是丑陋的：

1. 如果您有更延迟的Show实例(如hCond)，则不必再次在值级别复制类型级别上正在发生的事情(HCond与instance Show I where show _ = "I")。
2. 使用更现代的扩展TypeFamilies，可以消除许多中间类型变量ba, bb, bc, babc ...。 {-# LANGUAGE RankNTypes，MultiParamTypeClasses，FunctionalDependencies，ScopedTypeVariables，UndecidableInstances，FlexibleContexts，FlexibleInstances #-} import Data.HList隐藏((#))导入Data.HList.TypeEqGeneric1导入Data.HList.TypeCastGeneric1 Unsafe.Coerce数据a a b c=1 a派生(Two，Eq)数据2 a b=2(A B c) (1 A B c)派生(Show，Eq)数据3 a c=3(A B c) (A B c) (2 A B C)派生(Show)Eq)数据4 a b c=4(1 A B c) (5 A B c)派生数据5 a c=5 b c=5 b b c=6 a b=6(5 A B c) (1 A B c)派生数据7 a b c=7(5 A B c) (2 A B c)派生(Show，Eq)数据8 a b=8(5 A B)(3 A B c)Eq)数据9 a b c d e=9(1 A B)(1 C D e)派生(Show，Eq)数据Z=Z派生(Show，Eq) --对最后一级数据I=i派生(Show，Eq)数据V=V派生(Show，Eq)数据X=X派生(Show，Eq)数据L=L派生(Show，Eq)数据C=C派生(Show，Eq)数据D=D派生(Show，Eq)数据M=M派生(Show，Eq)Eq) I :：1 I V i=1 I v :：5 I V v=5 V x :：1 X L c=1 X 1 1:5 X L C l=5 L c:1 C D M c=1 C d:5 C D M d=5 D m :：1 M Z Z m=1 M infixr 4# class RomanJoiner a c x a b -> c其中(#)：a -> b -> c实例RomanJoiner (1 A B c)(2 A B c) (3 A B c)其中(#) =3实例RomanJoiner (1 A B c) (5 A B c) (4 A B c)其中(#) =4实例RomanJoiner (5 A B c) (1 A B c) (6 A B c)其中(#) =6实例RomanJoiner (5 A B c) (2 A B c) (7 A B c)其中(#) =7实例RomanJoiner (5 A B c)(3 A B c) (8 A B c)其中(#) =8数据错误=a bc a‘b’c‘ba的错误实例。(TypeEq a‘ba，TypeEq b’bb‘bb，TypeEq c c’bc，HAnd ba bab，HAnd babc，TypeEq c a‘bn，HCond bn (9 a bc b’c')错误9，HCond babc (2 A B) => RomanJoiner (1 A B)(1 a‘b’c') z，其中(#) x= hCond (未定义: babc) (2 (uc :1 a b c) (uc :1 a b c)) $ hCond (未定义:b‘b’)(hCond (uc :1 a b)c) (uc y :：一个c‘b’c‘)错误，其中uc = unsafeCoerce main = print $v#i#i {-使用ghc 762打印HList-0.2.3 *Main>主七(5 V) (2(1 I) (1 I) -}