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I am new to monad transformers, so sorry easy question.
I have value val :: MaybeT IO String and function fn :: String -> IO [String].
So after binding, I have val >>= liftM fn :: MaybeT IO (IO [String]). How can I remove duplicate IO monad and get result of type MaybeT IO [String]?
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Monad transformers not only make it easier to write getPassphrase but also simplify all the code instances.
liftIO allows us to lift an IO action into a transformer stack that is built on top of IO and it works no matter how deeply nested the stack is.
The Reader monad (also called the Environment monad). Represents a computation, which can read values from a shared environment, pass values from function to function, and execute sub-computations in a modified environment. Using Reader monad for such computations is often clearer and easier than using the State monad.
The state monad is a built in monad in Haskell that allows for chaining of a state variable (which may be arbitrarily complex) through a series of function calls, to simulate stateful code. It is defined as: newtype State s a = State { runState :: (s -> (a,s)) }
Use lift (or liftIO) instead of liftM.
> :t val >>= lift . fn
val >>= lift . fn :: MaybeT IO [String]
liftM is for applying pure functions in a monad. lift and liftIO are for lifting actions into a transformer.
liftM :: Monad m => (a -> b) -> m a -> m b
lift :: (Monad m, MonadTrans t) => m a -> t m a
liftIO :: MonadIO m => IO a -> m a
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