我试图让GHC.Generics中描述的通用二进制编码类的示例实现起作用,但是当我尝试编译它时,我得到一个错误。
import GHC.Generics
class Encode' f where
encode' :: f p -> [Bool]
instance Encode' V1 where
encode' x = undefined
instance Encode' U1 where
encode' U1 = []
instance (Encode' f, Encode' g) => Encode' (f :+: g) where
encode' (L1 x) = False : encode' x
encode' (R1 x) = True : encode' x
instance (Encode' f, Encode' g) => Encode' (f :*: g) where
encode' (x :*: y) = encode' x ++ encode' y
instance (Encode c) => Encode' (K1 i c) where
encode' (K1 x) = encode x
instance (Encode' f) => Encode' (M1 i t f) where
encode' (M1 x) = encode' x
class Encode a where
encode :: a -> [Bool]
default encode :: (Generic a) => a -> [Bool]
encode x = encode' (from x)
GHC抱怨:
Could not deduce (Encode' (Rep a)) arising from a use of ‘encode'’
from the context (Encode a)
bound by the class declaration for ‘Encode’
...
or from (Generic a)
bound by the type signature for encode :: Generic a => a -> [Bool]
...
In the expression: encode' (from x)
In an equation for ‘encode’: encode x = encode' (from x)
我错过了什么?
答案 0 :(得分:5)
并非Generic
的所有内容都可以由encode'
进行编码。只有同时具有Generic
实例的Rep a
且具有代表性Encode'
的内容才能由通用encode'
进行编码。编译器不知道(并且不能知道)Generic
不被Rep
实例覆盖的Encode'
内容Generic
。 Rep
实例的作者可以使用尚未存在的Encode' (Rep a)
类型。
您需要将所请求的default encode
约束添加到default encode :: (Generic a, Encode' (Rep a)) => a -> [Bool]
的上下文中。
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