Haskell:鼓励GHC推断出正确的中间类型

Question

我认为在Haskell中允许任意链式比较是很好的,所以你可以做简单的范围检查,如:

x <= y < z

还有更复杂的东西

x /= y < z == a

以上两者在语义上等同于

x <= y && y < z
x /= y && y < z && z == a

只是看看我是否可以使语法工作.

所以我在那里使用了几个类型类的大部分方式:

{-# LANGUAGE MultiParamTypeClasses, FlexibleInstances #-}
module ChainedOrd where

import Prelude hiding ((<), (<=), (>), (>=), (==), (/=))

class Booly v a where
  truthy :: v -> a
  falsy :: v -> a

instance Booly a Bool where
  truthy = const True
  falsy = const False

instance Booly a (Maybe a) where
  truthy = Just
  falsy = const Nothing

class ChainedOrd a b where
  (<),(>),(<=),(>=),(==),(/=) :: (Booly b c) => a -> b -> c

infixl 4 <
infixl 4 >
infixl 4 <=
infixl 4 >=
infixl 4 ==
infixl 4 /=

instance Ord a => ChainedOrd a a where
  x < y     = case compare x y of LT -> truthy y ; _ -> falsy y
  x > y     = case compare x y of GT -> truthy y ; _ -> falsy y
  x <= y    = case compare x y of GT -> falsy y  ; _ -> truthy y
  x >= y    = case compare x y of LT -> falsy y  ; _ -> truthy y
  x == y    = case compare x y of EQ -> truthy y ; _ -> falsy y
  x /= y    = case compare x y of EQ -> falsy y  ; _ -> truthy y

instance Ord a => ChainedOrd (Maybe a) a where
  Just x < y     = case compare x y of LT -> truthy y ; _ -> falsy y
  Nothing < y    = falsy y
  Just x > y     = case compare x y of GT -> truthy y ; _ -> falsy y
  Nothing > y    = falsy y
  Just x <= y    = case compare x y of GT -> falsy y  ; _ -> truthy y
  Nothing <= y   = falsy y
  Just x >= y    = case compare x y of LT -> falsy y  ; _ -> truthy y
  Nothing >= y   = falsy y
  Just x == y    = case compare x y of EQ -> truthy y ; _ -> falsy y
  Nothing == y   = falsy y
  Just x /= y    = case compare x y of EQ -> falsy y  ; _ -> truthy y
  Nothing /= y   = falsy y

由于中间类型的问题,编译很好,但似乎不允许链接.

-- works
checkRange1 :: Ord a => a -> a -> a -> Bool
checkRange1 x y z = x `lem` y <= z
  where lem :: Ord a => a -> a -> Maybe a
        lem = (<=)

-- works
checkRange2 :: Ord a => a -> a -> a -> Bool
checkRange2 x y z = (x <= y) `leb` z
  where leb :: Ord a => Maybe a -> a -> Bool
        leb = (<=)

checkRange1并且checkRange2工作正常,因为它们都对中间类型施加约束(作为第一次比较的结果,或作为第二次比较的参数).

-- error
checkRange3 :: Ord a => a -> a -> a -> Bool
checkRange3 x y z = (x <= y) <= z

当我试图让编译器推断出中间类型时,它会咆哮我.

ChainedOrd.hs:64:30:
    Ambiguous type variable `a0' in the constraints:
      (ChainedOrd a0 a) arising from a use of `<='
                        at ChainedOrd.hs:64:30-31
      (Booly a a0) arising from a use of `<=' at ChainedOrd.hs:64:24-25
    Probable fix: add a type signature that fixes these type variable(s)
    In the expression: (x <= y) <= z
    In an equation for `checkRange3': checkRange3 x y z = (x <= y) <= z

有没有什么方法可以说服编译器它应该Maybe a用作中间类型a0令人满意Booly a a0, ChainedOrd a0 a,因为这是它知道的唯一实例？

如果不这样做,还有另一种方法可以进行任意比较链接工作吗？

Answer 1

infixl 4 ==?

class ChainedEq a b where
  (==?) :: a -> b -> Maybe b

instance (Eq a) => ChainedEq (Maybe a) a where
  x ==? y = if x == Just y
    then x
    else Nothing

instance (Eq a) => ChainedEq a a where
  x ==? y = if x == y
    then Just x
    else Nothing

unChain :: Maybe a -> Bool
unChain Nothing = False
unChain (Just _) = True

test :: Int -> Int -> Int -> Bool
test x y z = unChain $ x ==? y ==? z