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Haskell:Applicative Functor

[英]Haskell : Applicative Functor

I am new to Haskell. 我是Haskell的新手。 I've made a type Maybe3. 我做了一个类型Maybe3。

data Maybe3 a= Just3 a| Unknown3 | Missing3 deriving (Show, Eq, Ord)
eq3 :: Eq a => Maybe3 a-> Maybe3 a-> Bool3
eq3 Unknown3 _ = Unk3
eq3 Missing3 _ = False3
eq3 _ Missing3 = False3
eq3 _ Unknown3 = Unk3 
eq3 (Just3 a) (Just3 b)=if a==b then True3 else False3

How to make Maybe3 an applicative functor? 如何使Maybe3成为一个应用函子? And how to make it a Monad? 如何让它成为Monad?

Key idea 关键的想法

My understanding is that Missing3 and Unknown3 work a bit like Nothing , except they give a little more feedback about why there's no answer, so might behave slightly differently to each other. 我的理解是Missing3Unknown3工作有点像Nothing ,除了他们提供了一些关于为什么没有答案的反馈,所以可能表现得彼此略有不同。 Certainly I think Missing3 should behave like Nothing . 当然我认为Missing3应该表现得像Nothing

Let's look at how these are defined for Maybe : 让我们来看看这些是如何为Maybe定义的:

Functor 函子

Here's the Functor instance for Maybe : 这是Maybe的Functor实例:

instance  Functor Maybe  where
    fmap _ Nothing       = Nothing
    fmap f (Just a)      = Just (f a)

I think it's clear how to deal with Missing3 and Unknown3 here. 我认为这里很清楚如何处理Missing3Unknown3

Monad 单子

instance  Monad Maybe  where
    (Just x) >>= k      = k x
    Nothing  >>= _      = Nothing

    (Just _) >>  k      = k
    Nothing  >>  _      = Nothing

    return              = Just
    fail _              = Nothing

You can't help but do the same here with >>= for Missing3 and Unknown3 , since you don't have a value to bind with. 你不能不在这里使用>>= for Missing3Unknown3 ,因为你没有要绑定的值。 The only question is do you fail with Unknown3 or Missing3 ? 唯一的问题是你用Unknown3Missing3失败了吗?

Applicative 合用的

Here's where there's a little more digging: 这里有更多挖掘的地方:

instance Applicative Maybe where
    pure = return
    (<*>) = ap

ap                :: (Monad m) => m (a -> b) -> m a -> m b
ap                =  liftM2 id

liftM2  :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
liftM2 f m1 m2          = do { x1 <- m1; x2 <- m2; return (f x1 x2) }

Now that translates to 现在转换为

mf <*> mx = do
    f <- mf
    x <- mx
    return (f x)

Which you can use all the time to turn a Monad into an Applicative. 您可以随时使用Monad将Monad变为Applicative。

Aside: Applicative is great. 旁白:适用性很好。

In fact, whenever you find yourself writing 事实上,每当你发现自己写作时

this thing = do
    something <- some monadic thing
    more <- some other thing
    yetmore <- another thing too
    return (combine something more yetmore)

you should rewrite it using applicative notation: 你应该使用applicative notation重写它:

this thing = combine <$> some monadic thing 
                     <*> some other thing 
                     <*> another thing too

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