无法理解为什么我的类型的 MonoFoldable 无法编译,或者错误消息

Cannot understand why MonoFoldable for my type doesn't compile, or the error message

我有以下代码:

{-# LANGUAGE NoImplicitPrelude, OverloadedStrings, TypeFamilies #-}

module AI.Analysis.Rules where

import ClassyPrelude

-- Our set of rules

data RuleSet a = RuleSet [Rule a] [Rule a]
  deriving (Eq)

mkRuleSet :: (Ord a) => [Rule a] -> RuleSet a
mkRuleSet rules = uncurry RuleSet (partition isStandard uniques)
  where uniques = ordNub rules
        isStandard x = case x of
          Standard _ _ -> True
          LastResort _ -> False

instance (Show a) => Show (RuleSet a) where
  show (RuleSet s l) = unlines [toLines s, "----", toLines l]
    where toLines = unlines . fmap show

instance (Ord a) => Monoid (RuleSet a) where
  mempty = RuleSet [] []
  mappend (RuleSet s1 l1) (RuleSet s2 l2) = RuleSet (ordNub (s1 ++ s2)) (ordNub (l1 ++ l2))

instance (Ord a) => Semigroup (RuleSet a) where
  (<>) = mappend

type instance Element (RuleSet a) = (Rule a)

instance MonoFoldable (RuleSet a) --this is unhappy

-- A rule in our system
-- For now, we assume rules *individually* are always internally-consistent

data Rule a = Standard [a] a | LastResort a
  deriving (Eq)

mkRule :: (Eq a, Ord a) => [a] -> a -> Rule a
mkRule as c = case as of
  [] -> LastResort c
  _ -> Standard ((sort . ordNub) as) c

-- Last-resort rules and standard rules cannot be compared for consistency
mutuallyConsistent :: (Eq a) => Rule a -> Rule a -> Maybe Bool
mutuallyConsistent (LastResort c1) (LastResort c2) = Just (c1 == c2)
mutuallyConsistent (Standard as1 c1) (Standard as2 c2) = Just ((as1 /= as2) || (c1 == c2))
mutuallyConsistent _ _ = Nothing

instance (Show a) => Show (Rule a) where
  show x = case x of
    Standard as c -> formatAnd as ++ " -> " ++ show c
    LastResort c -> "-> " ++ show c
     where formatAnd = unwords . intersperse "^" . map show . otoList

 -- LastResort rules are always ordered smaller than standard ones
 instance (Ord a) => Ord (Rule a) where
   (<=) (LastResort _) (Standard _ _) = True
   (<=) (Standard _ _) (LastResort _) = False
   (<=) (LastResort c1) (LastResort c2) = c1 <= c2
   (<=) (Standard as1 c1) (Standard as2 c2) = (as1 <= as2) || (c1 <= c2)

但是,我从编译器中收到以下错误,我无法理解其含义:

/home/koz/documents/uni/research/summer-research-2015/clinical/rules-analysis/src/AI/Analysis/Rules.hs:47:10:
    Couldn't match type ‘a’ with ‘Rule a’
      ‘a’ is a rigid type variable bound by
          the instance declaration
          at /home/koz/documents/uni/research/summer-research-2015/clinical/rules-analysis/src/AI/Analysis/Rules.hs:47:10
    Expected type: Element (RuleSet a)
      Actual type: a
    Relevant bindings include
      ofoldMap :: (Element (RuleSet a) -> m) -> RuleSet a -> m
        (bound at /home/koz/documents/uni/research/summer-research-2015/clinical/rules-analysis/src/AI/Analysis/Rules.hs:47:10)
    In the expression:
      mono-traversable-0.10.0.1:Data.MonoTraversable.$gdmofoldMap
    In an equation for ‘ofoldMap’:
        ofoldMap
          = mono-traversable-0.10.0.1:Data.MonoTraversable.$gdmofoldMap
    In the instance declaration for ‘MonoFoldable (RuleSet a)’

据我所知,我的想法似乎有道理 - 毕竟,RuleSet 只是 Rule 的容器,应该允许折叠,但错误消息问题对我来说没有任何意义。有人可以澄清我在这里没有掌握的内容吗?

您是否尝试过实际实施 class?默认定义和您的类型系列似乎有些奇怪。如果您至少定义了以下内容,那么文件类型将检查:

instance MonoFoldable (RuleSet a) where --this is unhappy
    ofoldl1Ex' = undefined
    ofoldr1Ex  = undefined
    ofoldl'    = undefined
    ofoldr     = undefined
    ofoldMap   = undefined

编辑:classy 序曲,我现在知道我永远不会使用它,它具有默认实现和类型签名,其中包括约束 t a ~ mono, a ~ Element (t a)。仔细工作,因为我不得不在这里三思而后行。 t a ~ RuleSet a0 所以 t == RuleSeta == a0。然后 a ~ Element (RuleSet a),这是您在消息中的确切错误,会建议 a ~ Rule a,这是不对的。

澄清默认实现:因为有大量类型是适当多态的 - 因此 Functor 的实例 - MonoFunctor 提供了一种简单的方法来使它们也MonoFunctor 的实例,来自 default method signatures。如果您有 Functor,只需声明 instance MonoFunctor 就足够了。

在您的情况下,您会收到一条令人困惑的错误消息,因为您的类型实际上是 Functor,但与您想要的 MonoFunctor 实例类型不同。具体来说,根据其形状,RuleSet aaFunctor,而您希望它是 Rule a。这没有什么错,它只是与默认实现冲突,因此您需要提供单独的实现。

请注意,这并非特定于您的类型:任何不是从 FunctorMonoFunctor 的简单转换的内容都需要这项工作。这适用于某些 built-in 实例,例如 TextByteString.