haskell - 为什么GHC说这两种类型不一样？我怎样才能让它们相同？

Question

我定义了一个类:

class Query a where
  filter  :: a -> (b -> Bool) -> [b]
  ifilter :: a -> (b -> Bool) -> [(b, Int64)]

我有一个用户定义的类型:

data Segment a = Segment {
  extents :: [Extent],
  array   :: [a]
} deriving(Eq, Show)

最后，我尝试使 Segment 成为 Query 的实例:

instance Query (Segment a) where
  filter  = filterSegment
  ifilter = ifilterSegment

函数 filter 和 ifilter 都接受一个 (Segment a) 和一个函数 (a->Bool) 并返回 [a] 或 [(a,Int64)]。

但是，我收到以下错误:

Couldn't match type `a' with `b'
      `a' is a rigid type variable bound by
          the instance declaration at src/Store/DataColumn.hs:19:10
      `b' is a rigid type variable bound by
          the type signature for filter :: Segment a -> (b -> Bool) -> [b]
          at src/Store/DataColumn.hs:20:3
    Expected type: Segment a -> (b -> Bool) -> [b]
      Actual type: Segment a -> (a -> Bool) -> [a]
    In the expression: filterSegment
    In an equation for `filter': filter = filterSegment
    In the instance declaration for `Query (Segment a)'

Couldn't match type `a' with `b'
      `a' is a rigid type variable bound by
          the instance declaration at src/Store/DataColumn.hs:19:10
      `b' is a rigid type variable bound by
          the type signature for
            ifilter :: Segment a -> (b -> Bool) -> [(b, Int64)]
          at src/Store/DataColumn.hs:21:3
    Expected type: Segment a -> (b -> Bool) -> [(b, Int64)]
      Actual type: Segment a -> (a -> Bool) -> [(a, Int64)]
    In the expression: ifilterSegment
    In an equation for `ifilter': ifilter = ifilterSegment
    In the instance declaration for `Query (Segment a)'

我不确定我在这里错过了什么。我确实希望 (Segment a) 中的 'a' 与 (a->Bool)、[a] 和 [(a, Int64)] 中的 'a' 相同。但是，我无法弄清楚如何做到这一点。任何帮助，将不胜感激。

Answer 1

您对类型类的定义对 b 没有限制。可以，这意味着应该允许用户为 b 选择任何类型。他要。当然，您的实例定义不允许这样做，这就是它被拒绝的原因。

您想说的是“a 参数化了类型 b，这就是我想要使用的类型 b”。你是这样说的:

class Query a where
  filter  :: a b -> (b -> Bool) -> [b]

然后你写 instance Query Segment而不是 instance Query (Segment t)因为 a在类型类定义中现在只代表 Segment和 b代表 t .