如何在这个例子中得到更好的Haskell中的多态类型推断？
我有以下数据类型：
data PValue = IV Int | BV Bool | SV String
deriving (Show, Eq)
我想写一个从Int，Bool或String生成PValue的函数，如下所示：
> 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
这似乎工作，除了诠释：
> 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
我明白这是因为3本身是模糊的类型（可以是Int ， Float等）。 有没有一种方法来强制这种类型推断，而不必在呼叫站点中明确注释它？
在此扩展@ AndrewC的评论。 为了使loadVal 3能够正常工作，请在实例化时进行类型转换：
instance PValues Integer where
loadVal v = IV (fromInteger v)
现在，如果您希望使用Text类型并且不希望用户明确注释它，请同时为String和Text实例：
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你的输入是Text数据类型的场所，它不必经历pack开销。
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