[英]Simple Applicative Functor Example
I'm reading the Learn You a Haskell book. 我正在读“了解你一本哈斯克尔”一书。 I'm struggling to understand this applicative functor code: 我很难理解这个应用程序的编码器代码:
(*) <$> (+3) <*> (*2) $ 2
This boils down to: (3+2) * (2*2) = 20 这归结为:(3 + 2)*(2 * 2)= 20
I don't follow how. 我不遵循如何。 I can expand the above into the less elegant but more explicit for newbie comprehension version: 我可以将上面的内容扩展为不那么优雅但更明确的新手理解版本:
((fmap (*) (+3)) <*> (*2)) 2
I understand the basics of the <*>
operator. 我理解<*>
运算符的基础知识。 This makes perfect sense: 这很有道理:
class (Functor f) => Applicative f where
pure :: a -> f a
(<*>) :: f (a -> b) -> f a -> f b
But I don't see how the command works? 但我不知道命令是如何工作的? Any tips? 有小费吗?
One method to approach this types of questions is to use substitution. 处理这类问题的一种方法是使用替代。 Take the operator, in this case (<*>)
or function, get its implementation and insert it to the code in question. 在这种情况下,使用运算符(<*>)
或函数,获取其实现并将其插入到相关代码中。
In the case of (*) <$> (+3) <*> (*2) $ 2
you are using the ((->) a)
instance of Applicative
found in the Applicative
module in base , you can find the instance by clicking the source link on the right and searching for "(->": 在的情况下(*) <$> (+3) <*> (*2) $ 2
您所使用的((->) a)
的实例Applicative
中发现的Applicative
在基本模块 ,可以发现实例通过单击右侧的源链接并搜索“( - >”:
instance Applicative ((->) a) where
pure = const
(<*>) f g x = f x (g x)
Using the definition for (<*>)
we can continue substituting: 使用(<*>)
的定义,我们可以继续替换:
((fmap (*) (+3)) <*> (*2)) 2 == (fmap (*) (+3)) 2 ((*2) 2)
== (fmap (*) (+3)) 2 4
Ok now we need the Functor instance for ((->) a)
. 好了,我们需要((->) a)
的Functor实例。 You can find this by going to the haddock info for Functor
, here clicking on the source link on the right and searching for "(->" to find: 您可以通过为黑线鳕信息发现这个Functor
, 在这里点击右边的源链接,并搜索“( - >”中找到:
instance Functor ((->) r) where
fmap = (.)
Now continuing substituting: 现在继续代替:
(fmap (*) (+3)) 2 4 == ((*) . (+3)) 2 4
== (*) ((+3) 2) 4
== (*) 5 4
== 20
Many people report better long term sucess with these types of problems when thinking about them symbolically. 许多人在用象征性思考这些问题时会报告更好的长期成功。 Instead of feeding the 2 value through the problem lets focus instead on (*) <$> (+3) <*> (*2)
and only apply the 2 at the end. 而不是通过问题提供2值,而是将焦点放在(*) <$> (+3) <*> (*2)
并且仅在末尾应用2。
(*) <$> (+3) <*> (*2)
== ((*) . (+3)) <*> (*2)
== (\x -> ((*) . (+3)) x ((*2) x))
== (\x -> ((*) . (+3)) x (x * 2))
== (\x -> (*) (x + 3) (x * 2))
== (\x -> (x + 3) * (x * 2))
== (\x -> 2 * x * x + 6 * x)
Ok now plug in 2 for x 好的,现在插入2 for x
2 * 2 * 2 + 6 * 2
8 + 12
20
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