'惯性'的Haskell型不等式 [英] 'idiomatic' Haskell type inequality

问题描述

（从前面的问题编辑，我认为下面的代码不起作用）

我希望实现一个haskell函数f，它有一个限制，使得它的2个参数不能有相同的类型。我已经使用了下面的代码：

$ $ p $ {code> { - ＃LANGUAGE MultiParamTypeClasses，FunctionalDependencies，UndecidableInstances，FlexibleInstances，FlexibleContexts，TypeFamilies，IncoherentInstances＃ - }
data HTrue = HTrue
data HFalse = HFalse

class HEq xyb | x y - > b
实例（b〜HTrue）=> HEq x x b
instance（b〜HFalse）=> HEq xy b


g ::（HEq a b HFalse）=> a - > b - > （）
gxy =（）

现在函数g只接受a和b不同种类。这是在haskell中编码类型不等式的'idiomiatic'方法吗？如果没有，那么它有什么问题？ new

  { - ＃语言DataKinds，PolyKinds，TypeFamilies＃ - } 
 
类型家庭Equal（a :: k）（b :: k）:: Bool 
类型实例其中
等于aa = True 
 Equal ab = False 
  

(edited from previous question where I thought code below doesn't work)

I wish to implement a haskell function f that has a restriction such that its 2 parameters must not have the same type. I have used the following code:

{-# LANGUAGE MultiParamTypeClasses, FunctionalDependencies, UndecidableInstances, FlexibleInstances, FlexibleContexts, TypeFamilies, IncoherentInstances #-}
data HTrue = HTrue
data HFalse = HFalse

class HEq x y b | x y -> b
instance (b ~ HTrue) => HEq x x b
instance (b ~ HFalse) => HEq x y b


g :: (HEq a b HFalse) => a -> b -> ()
g x y = ()

Now the function g only accepts a and b if they have different types. Is this the 'idiomiatic' way to code type inequality in haskell? If not, what are the problems with it?

解决方案

With new features being added to GHC, you'll be able to write:

{-# LANGUAGE DataKinds, PolyKinds, TypeFamilies #-}

type family Equal (a :: k) (b :: k) :: Bool
type instance where
   Equal a a = True
   Equal a b = False

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