[英]Why there is no way to derive Applicative Functors in Haskell?
在Haskell中,你可以得到Functor
, Foldable
和Traversable
使用自动deriving
。 但是,无法导出Applicative
。 考虑到有一种明显的方法来定义Applicative
实例(相当于一个压缩应用程序),难道没有任何方法可以deriving Applicative
吗?
不,这根本不明显。 比较以下Applicative
实例:
[]
ZipList
Data.Sequence.Seq
,其Applicative
实例声明长达数百行。IO
(->) r
parsec
, attoparsec
, regex-applicative
。pipes
包中的代理。这里几乎没有统一性,而且大多数情况并不明显。
正如David Young 评论的那样, []
和ZipList
实例“最终都是列表类型的两个不同的、同样有效的Applicative
实例。”
既然DerivingVia
已经发布(GHC-8.6 或更高版本),实际上可以在DeriveGeneric
的帮助下为任何确定性数据类型派生Applicative
! 也就是说,任何只有一个变体的数据类型:
data Foo x = Foo x | Fe -- This is non-deterministic and can't derive Applicative
data Bar x = Bar x x (Bar x) -- This is deterministic and can derive Applicative
data Baz x = Baz (Either Int x) [x] -- This is also ok, since [] and Either Int
-- are both Applicative
data Void x -- This is not ok, since pure would be impossible to define.
要派生Applicative
,我们首先需要定义一个用于通过泛型派生的包装器:
{-# LANGUAGE DerivingVia #-}
{-# LANGUAGE DeriveFunctor #-}
{-# LANGUAGE UndecidableInstances #-}
{-# LANGUAGE DeriveGeneric #-}
module Generically1 where
import GHC.Generics
newtype Generically1 f x = Generically1 { generically1 :: f x }
fromg1 :: Generic1 f => Generically1 f a -> Rep1 f a
fromg1 = from1 . generically1
tog1 :: Generic1 f => Rep1 f x -> Generically1 f x
tog1 = Generically1 . to1
instance (Functor f, Generic1 f, Functor (Rep1 f))
=> Functor (Generically1 f) where
fmap f (Generically1 x) = Generically1 $ fmap f x
instance (Functor f, Generic1 f, Applicative (Rep1 f))
=> Applicative (Generically1 f) where
pure = tog1 . pure
f <*> x = tog1 $ fromg1 f <*> fromg1 x
instance (Functor f, Generic1 f, Monad (Rep1 f)) => Monad (Generically1 f) where
return = pure
m >>= f = tog1 $ fromg1 m >>= fromg1 . f
为了使用它,我们首先为我们的数据类型派生Generic1
,然后通过我们新的Generically1
包装器派生Applicative
:
data Foo x = Foo x (Int -> x) (Foo x)
deriving (Functor, Generic1)
deriving (Applicative, Monad) via Generically1 Foo
data Bar x = Bar x (IO x)
deriving (Functor, Generic1)
deriving (Applicative, Monad) via Generically1 Bar
data Baz f x = Baz (f x) (f x)
deriving (Show, Functor, Generic1)
deriving (Applicative, Monad) via Generically1 (Baz f)
如您所见,我们不仅为我们的数据类型派生了Applicative
,而且还派生了Monad
。
这样做的原因是这些数据类型的Generic1
表示有Applicative
和Monad
实例。 例如参见产品类型 (:*:) 。 然而,对于Sum 类型 (:+:)没有Applicative
实例,这就是为什么我们不能为非确定性类型派生它。
您可以通过编写:kind! Rep1 Foo
来查看数据类型的Generic1
表示:kind! Rep1 Foo
GHCi 中的:kind! Rep1 Foo
。 以下是上述类型表示的简化版本(不包括元数据):
type family Simplify x where
Simplify (M1 i c f) = Simplify f
Simplify (f :+: g) = Simplify f :+: Simplify g
Simplify (f :*: g) = Simplify f :*: Simplify g
Simplify x = x
λ> :kind! Simplify (Rep1 Foo)
Simplify (Rep1 Foo) :: * -> *
= Par1 :*: (Rec1 ((->) Int) :*: Rec1 Foo)
λ> :kind! Simplify (Rep1 Bar)
Simplify (Rep1 Bar) :: * -> *
= Par1 :*: Rec1 IO
λ> :kind! forall f. Simplify (Rep1 (Baz f))
forall f. Simplify (Rep1 (Baz f)) :: k -> *
= forall (f :: k -> *). Rec1 f :*: Rec1 f
编辑: Generically1
包装器也可以在这里找到: https : Generically1
: Genericly1
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