Question

我正在尝试使用这个库构建一个slackbot：https://hackage.haskell.org/package/slack-api，只是为了学习更多的haskell，并希望最终了解monad -_-。

然后我有以下类型：

data BotState = BotState
  { 
    _appState :: AppState
  }

makeLenses ''BotState


type AppState = HM.Map String ChannelState

emptyState :: AppState
emptyState = HM.empty

data ChannelState = ChannelState
{ _counter :: Int}

type Bot = Slack.Slack BotState

我用我的机器人运行：

initApp = lookupEnv "SLACK_API_TOKEN" >>=
  \apiToken -> case apiToken of
    Nothing -> throwM ApiTokenMissingException
    Just t -> void $ Slack.runBot (Slack.SlackConfig t) runApp $ BotState emptyState

其中：

runApp :: Slack.Event -> Bot ()
runApp m@(Slack.Message cid uid body _ _ _) = sendMessage cid "GAH I CAN HAZ CHZBURGHER!"

这样运行正常，现在我希望添加更新系统状态的功能（通过递增计数器或其他方式）。

所以我将一个modifyState函数添加到我的Bot：

modifyState :: (AppState -> AppState) -> Bot ()
modifyState f = uses Slack.userState $ view appState >>=
  \state -> modifying Slack.userState $ set appState $ f state

这打破了：

 No instance for (Control.Monad.State.Class.MonadState
                           (Slack.SlackState BotState) ((->) BotState))
          arising from a use of ‘modifying’
        In the expression: modifying Slack.userState
        In the expression:
          modifying Slack.userState $ set appState $ f state
        In the second argument of ‘(>>=)’, namely
          ‘\ state -> modifying Slack.userState $ set appState $ f state’

鉴于modifying的签名：

，这是有道理的

modifying :: MonadState s m => ASetter s s a b -> (a -> b) -> m ()

但是，查看Slack.userState的文档：

userState :: forall s s. Lens (SlackState s) (SlackState s) s s Source

然后：

data SlackState s

 ... Constructor ...
    Instances
Show s => Show (SlackState s)Source  
MonadState (SlackState s) (Slack s)Source

那么为什么BotState已经不是MonadState的实例呢？我怎么能解决这个问题？

Answer 1

$运算符的固定性为0，而>>=的固定性为1，因此这样的代码可以正常工作：

main :: IO ()
main = do
  putStrLn "hello world" >>= \_ -> putStrLn "hi"

但不是这一个：

main :: IO ()
main = do
  putStrLn $ "hello world" >>= \_ -> putStrLn "hi"

它被解释为：

main :: IO ()
main = do
  putStrLn ("hello world" >>= \_ -> putStrLn "hi")

要查看固定信息，请使用ghci＆＃39; :info命令：

 :info $
($) ::
  forall (r :: ghc-prim-0.5.0.0:GHC.Types.RuntimeRep) a (b :: TYPE
                                                                r).
  (a -> b) -> a -> b
    -- Defined in ‘GHC.Base’
infixr 0 $
 :info >>=
class Applicative m => Monad (m :: * -> *) where
  (>>=) :: m a -> (a -> m b) -> m b
  ...
    -- Defined in ‘GHC.Base’
infixl 1 >>=

另外，如果您不确定，那么老的括号总是在这里拯救：）

缺少Monadstate实例

1 个答案: