CODEWITHSUNDEEP
haskellmonadsstate-monad
Abraham P
Abraham P

Reputation: 15471

Missing Monadstate instance

I am attempting to build a slackbot using this library: https://hackage.haskell.org/package/slack-api, just to learn a little bit more haskell, and hopefully, finally understand monads -_-.

I then have the following types:

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

and I run my bot with:

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

where:

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

This runs fine, now I wish to add the ability to update the system state (by incrementing the counter, or in other ways).

so I add a modifyState function to my Bot:

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

This breaks with:

 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’

Which makes sense given the signature for modifying:

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

However, upon looking at the documentation for Slack.userState:

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

And then:

data SlackState s

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

So then why isn't the BotState already an instance of MonadState? How could I fix this?

Upvotes: 0

Views: 185

Answers (1)

Kostia R
Kostia R

Reputation: 2565

$ operator has fixity 0, while >>= has fixity 1, so code like this would work:

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

But not this one:

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

It's being interpreted as:

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

To see fixity info, use ghci's :info command:

 :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 >>=

Also, if you're not sure, good old parentheses are always here for the rescue :)

Upvotes: 2

Related Questions