Haskell/GHC 泛型文档示例缺少类型类约束?
Haskell/GHC Generics documentation example missing a typeclass constraint?
我正在按照 GHC.Generics
的 hackage documentation 数据类型构建通用 "encode" 函数的教程进行操作。我已将代码复制粘贴到(包括)"The wrapper and generic default" 部分,但出现以下错误:
Main.hs:35:14:
Could not deduce (Encode' (Rep a)) arising from a use of ‘encode'’
from the context (Encode a)
bound by the class declaration for ‘Encode’
at Main.hs:(32,1)-(35,29)
or from (Generic a)
bound by the type signature for encode :: Generic a => a -> [Bool]
at Main.hs:33:13-23
In the expression: encode' (from x)
In an equation for ‘encode’: encode x = encode' (from x)
Failed, modules loaded: none.
我复制的代码如下:
{-# LANGUAGE DeriveGeneric #-}
{-# LANGUAGE TypeOperators #-}
{-# LANGUAGE DefaultSignatures #-}
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE DeriveAnyClass #-}
module Main where
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)
我认为问题出在最后的 class 声明 (class Encode a where ...
)。我通过添加一个额外的约束来修复它,得到这个:
class Encode a where
encode :: a -> [Bool]
default encode :: (Generic a, Encode' (Rep a)) => a -> [Bool]
encode x = encode' (from x)
这似乎像宣传的那样工作 - 我可以声明新的数据类型并使它们可以用 DeriveAnyClass
和 deriving
编码。但是,我不太确定为什么我的修复是必要的。我的问题是:
- 文档有误吗?
Encode' (Rep a)
约束是否应该出现在 hackage 的示例代码中?
- 如果没有,我应该添加什么才能使代码正常工作?
是的,我认为文档是错误的。约束是必需的。
我正在按照 GHC.Generics
的 hackage documentation 数据类型构建通用 "encode" 函数的教程进行操作。我已将代码复制粘贴到(包括)"The wrapper and generic default" 部分,但出现以下错误:
Main.hs:35:14:
Could not deduce (Encode' (Rep a)) arising from a use of ‘encode'’
from the context (Encode a)
bound by the class declaration for ‘Encode’
at Main.hs:(32,1)-(35,29)
or from (Generic a)
bound by the type signature for encode :: Generic a => a -> [Bool]
at Main.hs:33:13-23
In the expression: encode' (from x)
In an equation for ‘encode’: encode x = encode' (from x)
Failed, modules loaded: none.
我复制的代码如下:
{-# LANGUAGE DeriveGeneric #-}
{-# LANGUAGE TypeOperators #-}
{-# LANGUAGE DefaultSignatures #-}
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE DeriveAnyClass #-}
module Main where
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)
我认为问题出在最后的 class 声明 (class Encode a where ...
)。我通过添加一个额外的约束来修复它,得到这个:
class Encode a where
encode :: a -> [Bool]
default encode :: (Generic a, Encode' (Rep a)) => a -> [Bool]
encode x = encode' (from x)
这似乎像宣传的那样工作 - 我可以声明新的数据类型并使它们可以用 DeriveAnyClass
和 deriving
编码。但是,我不太确定为什么我的修复是必要的。我的问题是:
- 文档有误吗?
Encode' (Rep a)
约束是否应该出现在 hackage 的示例代码中?- 如果没有,我应该添加什么才能使代码正常工作?
是的,我认为文档是错误的。约束是必需的。