在Haskell中如何以及为什么将ap定义为liftM2 id [英] How and why is ap defined as liftM2 id in Haskell

问题描述

在试图更好地理解Applicative的同时，我研究了< *>的定义，它倾向于被定义为ap，后者又定义为：

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

查看liftM2和id的类型签名，即：

  liftM2 ::（Monad m）=> （a1→a2→r）→> m a1  - > m a2  - > m r 
 id :: a  - > a 
  

我不明白如何通过传入id，类型签名的相关部分似乎从（a1 - > a2 - > r） - > （a - > b）。我在这里错过了什么？

解决方案

类型变量 a from id 可以以任何类型实例化，在这种情况下，该类型是 a - > B'/ code>。因此我们在（a - > b） - >>处实例化 id （a - > b）。现在来自 liftM2 的类型变量 a1 正在实例化为（a - > b） ， a2 正在实例化为 a ， r 正在被实例化为 b 。


把它们放在一起， liftM2 在（（a - > b ）→>（a→b））→> m（a - > b） - > m a - > m b 和 liftM2 id :: m（a - > b） - > m a - > m b 。

Whilst trying to better understand Applicative, I looked at the definition of <*>, which tends to be defined as ap, which in turn is defined as:

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

Looking at the type signatures for liftM2 and id, namely:

liftM2  :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r
id                      :: a -> a

I fail to understand how just by passing in id, the relevant part of the type signature seems to transform from (a1 -> a2 -> r) -> m a1 to m (a -> b). What am I missing here?

解决方案

The type variable a from id can be instantiated at any type, and in this case that type is a -> b.

So we are instantiating id at (a -> b) -> (a -> b). Now the type variable a1 from liftM2 is being instantiated at (a -> b), a2 is being instantiated at a, and r is being instantiated at b.

Putting it all together, liftM2 is instantiated at ((a -> b) -> (a -> b)) -> m (a -> b) -> m a -> m b, and liftM2 id :: m (a -> b) -> m a -> m b.

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