无法理解为什么我的类型的MonoFoldable不编译，或者错误消息

Question

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，我从编译器那里得到了以下错误，编译器的意思很难理解：

/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)’

就像我能说的那样，我的想法似乎很有意义--毕竟，RuleSet只是Rule的容器，它应该考虑到可折叠性，但是这个错误信息对我来说没有任何意义。有人能澄清一下我在这里没能理解什么吗？

Answer 1

你试过实际实现这个类吗？在默认定义和类型家族中，似乎有一些奇怪之处。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:高级序曲，我现在知道我永远不会使用它，它有默认的实现和类型签名，其中包括约束t a ~ mono, a ~ Element (t a)。工作很仔细，因为我不得不三思而后行。t a ~ RuleSet a0所以t == RuleSet和a == a0。然后a ~ Element (RuleSet a)，这是您在消息中的准确错误，将建议a ~ Rule a，这就是不对的。

Answer 2

为了澄清默认的实现:由于有大量的类型是正确多态的--因此Functor - MonoFunctor的实例提供了一种简单的方法，可以通过默认方法签名来生成MonoFunctor的实例。在有Functor的情况下，只需声明instance MonoFunctor就足够了。

在您的示例中，您会得到一条令人困惑的错误消息，因为您的类型实际上是Functor，但类型与所需的MonoFunctor实例不同。具体来说，就其形状而言，RuleSet a是a的Functor，而您希望它是Rule a。这没有什么问题，它只是与默认实现相冲突，因此您需要提供单独的实现。

请注意，这并不是特定于您的类型:任何不是简单的从Functor到MonoFunctor的转换都需要这项工作。这适用于一些内置实例，如Text和ByteString。