如何使用bind与嵌套monads？ [英] How to use bind with nested monads?

问题描述

我有两个函数，一个试图从web服务获取令牌并可能失败，另一个尝试使用此令牌获取用户名并可能失败。

  getToken :: IO（可能令牌）
 getUsername :: Token  - > IO（Maybe String）
  

我想将getToken的结果提供给 getUsername 。如果只有 IO 或 Maybe ，我可以简单地使用bind，但由于存在嵌套的monads，所以我可以T。我怎样才能写出一些相当于 getToken>> = getUsername :: IO（Maybe String）？

更一般地说，什么函数具有类型 m1 m2 a - > （a - > m1 m2 b） - > m1 m2 b ？

额外问题：我将如何使用 IO中的符号 useToken context


解决方案显示你的用例：
$ b $ pre $ 类型Token = String

getToken :: IO（也许令牌）
getToken = undefined

getUsername :: Token - > IO（可能是字符串）
getUsername = undefined

useToken :: IO（可能是String）
useToken = do
令牌< - getToken
case token
只需x - > getUsername x
Nothing - > return Nothing


如果您不想使用 do
pre $ use $ or $ useToken2 :: IO（可能是字符串）
useToken2 = getToken> ;> = \ token - >也许（没有返回）getUsername令牌


或者使用monad变换器，你的代码将变得更简单：

  import Control.Monad.Trans.Maybe 
类型Token = String 
 
 getToken :: MaybeT IO令牌
 getToken =未定义
 
 getUsername ::令牌 - > MaybeT IO字符串
 getUsername =未定义
 
 useToken :: MaybeT IO字符串
 useToken = do 
令牌<  -  getToken 
 getUsername令牌
  

请注意，您也可以直接提升monad变换器中的IO操作。正如@Robedino所指出的那样，现在代码将更加简洁，不需要记号：

  useToken :: MaybeT IO字符串
 useToken = getToken>> = getUsername 
  

I have two functions, one that tries to get a token from a webservice and may fail, and one that tries to use this token to get the username and may fail.

getToken :: IO (Maybe Token)
getUsername :: Token -> IO (Maybe String)

I would like to take the result of getToken and feed it to getUsername. If there was only IO or Maybe, I could simply use bind, but since there are down nested monads, I can't. How can I write something equivalent to getToken >>= getUsername :: IO (Maybe String) ?

More generally, what function has type m1 m2 a -> (a -> m1 m2 b) -> m1 m2 b?

Bonus question: how would I do that using the do notation in an IO context?

解决方案

I have defined a function useToken showing your use case:

type Token = String

getToken :: IO (Maybe Token)
getToken = undefined

getUsername :: Token -> IO (Maybe String)
getUsername = undefined

useToken :: IO (Maybe String)
useToken = do
  token <- getToken
  case token of
    Just x -> getUsername x
    Nothing -> return Nothing

If you don't want to use do notation, then you can use:

useToken2 :: IO (Maybe String)
useToken2 = getToken >>= \token -> maybe (return Nothing) getUsername token

Or using monad transformers, your code will become simpler:

import Control.Monad.Trans.Maybe
type Token = String

getToken :: MaybeT IO Token
getToken = undefined

getUsername :: Token -> MaybeT IO String
getUsername = undefined

useToken :: MaybeT IO String 
useToken = do
  token <- getToken
  getUsername token

Note that, you can also directly lift IO operations inside the monad transformer. As @Robedino points out, now the code will be more concise without do notation:

useToken :: MaybeT IO String 
useToken = getToken >>= getUsername

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