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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，我的想法似乎是有道理的——毕竟，aRuleSet只是Rules 的容器，它应该允许可折叠性，但有问题的错误消息对我来说没有任何意义。有人可以澄清我在这里未能掌握的内容吗？

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2 回答 2

3

澄清默认实现：由于存在大量适当多态的类型 - 因此实例Functor-MonoFunctor提供了一种简单的方法来MonoFunctor通过默认方法签名来制作这些实例。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 a Functor，但类型与您所需的MonoFunctor实例不同。具体来说，根据它的形状，它RuleSet a是一个Functorfor a，而你希望它是Rule a。这没有什么问题，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，这并不特定于您的类型：任何不是从Functorto简单翻译的东西都MonoFunctor需要这项工作。这适用于一些内置实例，例如Text和ByteString。

于 2016-01-13T09:04:12.420 回答
3

您是否尝试过实际实施该课程？默认定义和您的类型系列似乎有些奇怪。如果您至少定义以下内容，则文件类型检查：

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

编辑：我现在知道我永远不会使用的经典前奏具有默认实现和包含约束的类型签名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 == RuleSet和a == a0。然后a ~ Element (RuleSet a)，这是您在消息中的确切错误，会暗示a ~ Rule a这是不正确的。

于 2016-01-13T04:32:12.330 回答