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，我从编译器中得到以下错误，其含义我无法理解：

/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)’

我可以说，我的想法似乎有意义 - 毕竟，RuleSet只是Rule的容器，应该允许可折叠性，但错误信息不是对我没有任何意义。有人可以澄清我在这里没有掌握的内容吗？

Answer 1

你有没有试过实际上实现这个类？看起来默认定义和类型系列有些奇怪。如果至少定义以下内容，则文件类型将检查：

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

编辑：优雅的前奏，我现在知道我永远不会使用，具有默认实现和包含约束t a ~ mono, a ~ Element (t a)的类型签名。因为我必须在这里三思而后行。 t a ~ RuleSet a0所以t == RuleSet和a == a0。然后a ~ Element (RuleSet a)，这是您在邮件中的确切错误，会建议a ~ Rule a，这是不对的。

Answer 2

澄清默认实现：由于存在大量具有正确多态性的类型 - 因此Functor - MonoFunctor的实例提供了一种简单的方法来实现这些实例MonoFunctor，default method signatures。如果您有Functor，只需声明instance MonoFunctor即可。

在您的情况下，您收到一条令人困惑的错误消息，因为您的类型实际上是Functor，但与您所需的MonoFunctor实例的类型不同。具体来说，根据其形状，RuleSet a是Functor的{{1}}，而您希望它是a。这没有任何问题，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，这不是特定于您的类型：任何不是从Rule a到Functor的简单翻译都需要这项工作。这适用于某些内置实例，例如MonoFunctor和Text。

无法理解为什么我的类型的MonoFoldable不能编译，或者错误消息

2 个答案: