haskell - 对于这个例子，如何在 Haskell 中获得更好的多态类型推断？

Question

我有以下数据类型:

data PValue = IV Int | BV Bool | SV String
            deriving (Show, Eq)

我想编写一个函数，从 Int、Bool 或 String 生成 PValue，例如:

> loadVal 3
IV 3

> loadVal True
BV Bool

> loadVal "Ha"
SV "Ha"

由于 loadVal 的参数是多态的，我尝试创建一个类:

class PValues v where
  loadVal :: v -> PValue

instance PValues Int where
  loadVal v = IV v

instance PValues Bool where
  loadVal v = BV v

instance PValues String where
  loadVal s = SV s

这似乎有效，除了 Int:

> loadVal "Abc"
SV "Abc"
> loadVal False
BV False
> loadVal 3

<interactive>:8:1:
    No instance for (PValues v0) arising from a use of `loadVal'
    The type variable `v0' is ambiguous
    Note: there are several potential instances:
      instance PValues String -- Defined at Types.hs:22:10
      instance PValues Bool -- Defined at Types.hs:19:10
      instance PValues Int -- Defined at Types.hs:16:10
    In the expression: loadVal 3
    In an equation for `it': it = loadVal 3

<interactive>:8:9:
    No instance for (Num v0) arising from the literal `3'
    The type variable `v0' is ambiguous
    Note: there are several potential instances:
      instance Num Double -- Defined in `GHC.Float'
      instance Num Float -- Defined in `GHC.Float'
      instance Integral a => Num (GHC.Real.Ratio a)
        -- Defined in `GHC.Real'
      ...plus 8 others
    In the first argument of `loadVal', namely `3'
    In the expression: loadVal 3
    In an equation for `it': it = loadVal 3

据我了解，这是因为 3 本身的类型不明确(可能是 Int、Float 等)。有没有办法强制进行这种类型推断，而无需在调用站点中显式注释它？

Answer 1

在此处扩展@AndrewC 的评论。为了使 loadVal 3 工作，请执行以下操作 实例化时的类型转换:

instance PValues Integer where
  loadVal v = IV (fromInteger v)

现在，如果您想让它与 Text 类型一起使用并且不希望您的 用户要显式注释它，请提供 String 的两个实例 以及文本:

data PValue = IV Int | BV Bool | SV Text
            deriving (Show, Eq)

instance PValues String where
  loadVal s = SV (pack s)

instance PValues Text where
  loadVal s = SV s

对于编译器能够推断出您的输入的地方 Text 数据类型，它不必经过 pack 开销。