[英]Instance of a Functor
I have a following type newtype Arr2 e1 e2 a = Arr2 { getArr2 :: e1 -> e2 -> a }
. 我有以下类型
newtype Arr2 e1 e2 a = Arr2 { getArr2 :: e1 -> e2 -> a }
。 And I got to write Functor instance for it, yet i don't really understand how I tried 我必须为它编写Functor实例,但我真的不明白我的尝试方式
instance Functor (Arr2 e1 e2) where
fmap g (Arr2 a) = Arr2 (g a)
and 和
instance Functor (Arr2 e1 e2) where
fmap g = g . getArr2
which actually results in type 这实际上导致了类型
(a -> b) -> Arr2 e1 e2 a -> b
instead of desired 而不是期望
(a -> b) -> Arr2 e1 e2 a -> Arr2 e1 e2 b
So please, help me 所以,请帮助我
The Functor
class has as definition: Functor
类具有以下定义:
class Functor f where:
fmap :: (a -> b) -> f a -> f b
(<$) :: a -> f b -> f a
The (<$)
has a default implementation: (<$) = fmap . const
(<$)
有一个默认实现: (<$) = fmap . const
(<$) = fmap . const
which works fine. (<$) = fmap . const
工作得很好。
So that means that if we enter a function ( g :: a -> b
) as first argument, and an Arr2
that produces an a
, we have to generate an Arr2
that calls that g
on the outcome of the arrow if it is applied. 这意味着如果我们输入一个函数(
g :: a -> b
)作为第一个参数,并且一个Arr2
产生一个a
,我们必须生成一个Arr2
,如果它被应用,则在箭头的结果上调用g
。
As a result the definition of fmap
for your Arr2
is: 因此,您的
Arr2
的fmap
定义是:
instance Functor (Arr2 e1 e2) where
fmap g (Arr2 a) = Arr2 (
\x y -> g (a x y))
Or more elegantly: 或者更优雅:
instance Functor (Arr2 e1 e2) where
fmap g (Arr2 a) = Arr2 (
\x -> g . (a x))
Or a more elegant version - commented by @Alec : 或更优雅的版本 - 由@Alec评论:
instance Functor (Arr2 e1 e2) where
fmap g (Arr2 a) = Arr2 (
(g .) . a)
(you can convert expressions to pointfree ones using this tool ) (你可以转换使用表达式来pointfree那些该工具 )
The answer provided by Willem Van Onsem is very good, I would just like to suggest the use of a language extension that can easily create Functor
instances for newtypes: DeriveFunctor
. Willem Van Onsem提供的答案非常好,我只想建议使用一种语言扩展,可以轻松地为newtypes创建
Functor
实例: DeriveFunctor
。
At the top of your module you can add: 在模块的顶部,您可以添加:
{-# LANGUAGE DeriveFunctor #-}
Then you can automatically derive your Functor
instance: 然后,您可以自动派生您的
Functor
实例:
newtype Arr2 e1 e2 a = Arr2 { getArr2 :: e1 -> e2 -> a } deriving Functor
Here's is how I would find out the type of fmap
for this instance in GHCi: 以下是我如何在GHCi中找到此实例的
fmap
类型:
λ > :set -XDeriveFunctor
λ > newtype Arr2 e1 e2 a = Arr2 { getArr2 :: e1 -> e2 -> a } deriving Functor
λ > :set -XTypeApplications
λ > :t fmap @(Arr2 _ _)
fmap @(Arr2 _ _) :: (a -> b) -> Arr2 t t1 a -> Arr2 t t1 b
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.