Adding type constraints to the context of instance declarations in Haskell

Question

I am trying to represent weighted edges. I eventually want to have OutE to be an instance of Eq and Ord, with the constraint that etype is an instance of Eq and Ord. Assume I have following file as temp.hs:

data (Ord etype)=> OutE vtype etype = OutE {destVertex:: vtype, edgeValue::etype}

applyFunBy accessor ordfun = (\x y -> (ordfun (accessor x) (accessor y)))

instance Eq (OutE vtype etype) where
    --(==) :: Ord etype => (OutE vtype etype) -> (OutE vtype etype) -> Bool
    --(/=) :: Ord etype => (OutE vtype etype) -> (OutE vtype etype) -> Bool
    (==) = applyFunBy edgeValue (==)
    (/=) = applyFunBy edgeValue (/=)

when I load this in ghci, I get the following errors:

temp.hs:10:19:
    Could not deduce (Ord etype)
      from the context (Eq (OutE vtype etype))
      arising from a use of `edgeValue' at temp.hs:10:19-27
    Possible fix:
      add (Ord etype) to the context of the instance declaration
    In the first argument of `applyFunBy', namely `edgeValue'
    In the expression: applyFunBy edgeValue (==)
    In the definition of `==': == = applyFunBy edgeValue (==)

temp.hs:11:19:
    Could not deduce (Ord etype)
      from the context (Eq (OutE vtype etype))
      arising from a use of `edgeValue' at temp.hs:11:19-27
    Possible fix:
      add (Ord etype) to the context of the instance declaration
    In the first argument of `applyFunBy', namely `edgeValue'
    In the expression: applyFunBy edgeValue (/=)
    In the definition of `/=': /= = applyFunBy edgeValue (/=)
Failed, modules loaded: none.

If include the lines for the type signatures for (==) and (\=), I get:

temp.hs:6:1:
    Misplaced type signature:
    == ::
      (Ord etype) => (OutE vtype etype) -> (OutE vtype etype) -> Bool

temp.hs:7:1:
    Misplaced type signature:
    /= ::
      (Ord etype) => (OutE vtype etype) -> (OutE vtype etype) -> Bool

Answer 1

你限制etype是Ord在把定义OutE:

data (Ord etype) => OutE vtype etype = ...

但在Eq实例中,您实际上是在尝试为任何 etype无限制的实例定义实例.

instance Eq (OutE vtype etype) where

当然这不起作用,因为OutE它本身只是为Ord etypes 定义的,因此你也必须将类型类约束添加到实例定义中.

instance (Ord etype) => Eq (OutE vtype etype) where

请注意,一个==或多个定义/=足以使类型类工作.

请注意,它通常更容易,因此被认为是更好的样式,不要在data-types 上有类型类约束,而只是在实际需要类型类功能的实例/方法上.

在许多情况下,一个人不需要约束,最终会得到不必要的笨拙类型签名.

以某些有序地图类型为例Ord key => Map key value.

如果我们只想列出所有密钥怎么办？或者获得元素的数量？我们不需要这些键Ord,所以为什么不简单地让地图不受限制

getKeys :: Map key value -> [key]
getLength :: Map key value -> Int

只需在我们真正需要它的时候添加类型类

insert :: Ord key => key -> value -> Map key value