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，我尝试将Segment作为Query的一个实例：

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

函数filter和ifilter都采用（Segment a）和函数（a-＆gt; 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)'

我不确定我在这里缺少什么。我确实希望（段a）中的'a'与（a-＆gt; 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的含义没有限制，这意味着应该允许用户为他想要的b选择任何类型。当然，你的实例定义不允许这样做，这就是它被拒绝的原因。

你想说的是“a是在类型b上进行参数化的，那就是我想要使用的类型b”。这就是你怎么说的：

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

然后你写instance Query Segment而不是instance Query (Segment t)，因为类型类定义中的a现在只代表Segment而b代表t {1}}。

为什么GHC说这两种类型不同？我怎样才能让他们变得一样？

1 个答案: