Lift a function with a monad parameter into a monad transformer - haskell

I want to lift a function like mask_ :: IO a -> IO a to create a function with this signature: lmask_ :: StateT Bool IO a -> StateT IO a.
My problem is, how to handle the callback/first parameter? Wouldn't the following code be incorrect since it would execute the callback before mask_'s code?
lmask_ :: StateT Bool IO a -> StateT Bool IO a
lmask_ m = do
r <- m
lift (mask_ (return r))
Is there some general way to do this? A helper like lift1 :: MonadTrans t => (m a -> m a) -> (t m a -> t m a)?

If we generalize lmask_ to get rid of the StateT Bool IO, we get something like this:
lift1 :: (Monad m, Monad (t m), MonadTrans t) => (m a -> m a) -> (t m a -> t m a)
lift1 f term = do
x <- term
lift (f (return x))

In general this is not possible without knowing something about the monad transformer. However, there is a way how to do this for all the standard monad transformers. See type class MonadBaseControl. It's superclass MonadBase defines what is the bottom monad in a monad transformer stack (which is IO for all stacks that include IO), and MonadBaseControl defines a way how to embed the monad into the base monad. Its instances are somewhat convoluted, but once they're defined, it's possible to lift all such functions like mask_.
In your case, package lifted-base uses the above construction to re-define the standard IO functions lifted to MonadBaseControl. In particular, there is mask_
mask_ :: MonadBaseControl IO m => m a -> m a
which can be specialized to StateT Bool IO a -> StateT Bool IO a, as StateT s has an instance of MonadBaseControl.
See also Lift a function and its argument to a different monadic context.

Related

Modifying inner reader in a transformer stack

I'm pulling together code from a number of different places, and I'm trying to deal with the following:
Problem
I have a transformer stack with the following simplified type:
action :: m (ReaderT r IO) a
and I'm trying to use the action in the context of a different stack, which has a different reader environment:
desired :: m (ReaderT r' IO) a
I can of course provide
f :: r' -> r
Example
things :: m (ReaderT r' IO) ()
things = do
-- ... some stuff
-- <want to use action here>
action :: m (ReaderT r IO) a -- broken
-- ... more stuff
pure ()
What I've considered
withReaderT :: (r' -> r) -> ReaderT r m a -> ReaderT r' m a
This has the problem that ReaderT is the outer monad, whilst I want to use it on an inner one.
I've also considered that this might be related to MonadBase or MonadTransControl, but I'm not familiar with their workings.
I don't think it's possible to write a function with signature:
changeReaderT :: (MonadTrans m)
=> (r -> r')
-> m (ReaderT r IO) a
-> m (ReaderT r' IO) a
the issue being that the only operation possible, in general, on the second argument is lifting it to t (m (ReaderT r IO)) a for some monad transformer t, which doesn't buy you anything.
That is, the MonadTrans m constraint alone doesn't provide enough structure to do what you want. You either need m to be an instance of a typeclass like MFunctor in the mmorph package that allows you to modify an inner layer of the monad stack in a general way by providing a function like:
hoist :: Monad m => (forall a. m a -> n a) -> t m b -> t n b
(which is what #Juan Pablo Santos was saying), or else you need an ability to dig into the structure of your m monad transformer to partially run and rebuild it (which will be transformer-specific).
The first approach (using hoist from the mmorph package) will be most convenient if your m is already made up of transformers supported by the mmorph package. For example, the following typechecks, and you don't have to write any instances:
type M n = MaybeT (StateT String n)
action :: M (ReaderT Double IO) a
action = undefined
f :: Int -> Double
f = fromIntegral
desired :: M (ReaderT Int IO) a
desired = (hoist $ hoist $ withReaderT fromIntegral) action
You'll need a hoist for each layer in M.
The second approach avoids hoist and requisite MFunctor instances but requires tailoring to your specific M. For the above type , it looks something like:
desired' :: M (ReaderT Int IO) a
desired' = MaybeT $ StateT $ \s ->
(withReaderT fromIntegral . flip runStateT s . runMaybeT) action
You basically need to run the monad down to the ReaderT layer and then rebuild it back up, treating layers like StateT with care. This is exactly what the MFunctor instances in mmorph are doing automatically.

Apply a function to a value inside IO Maybe

I have a function of type a -> IO (Maybe b) and I want to apply it to IO (Maybe a) and get IO (Maybe b). I wrote a function to do that:
ioMaybeApply :: (a -> IO (Maybe b)) -> IO (Maybe a) -> IO (Maybe b)
ioMaybeApply f ioMaybeA = do
maybeA <- ioMaybeA
maybe (return Nothing) f maybeA
Is there a standard Haskell function to do that? I tried searching with Hoogle but I didn't find anything. If not, is my implementation good, or could it be simpler?
This can be achieved through the MaybeT monad transformer:
GHCi> import Control.Monad.Trans.Maybe
GHCi> :t \f m -> runMaybeT (MaybeT m >>= MaybeT . f)
\f m -> runMaybeT (MaybeT m >>= MaybeT . f)
:: Monad m => (a1 -> m (Maybe a)) -> m (Maybe a1) -> m (Maybe a)
import Control.Monad.Trans.Maybe
-- Making it look like your definition, for the sake of comparison.
ioMaybeApply :: (a -> IO (Maybe b)) -> IO (Maybe a) -> IO (Maybe b)
ioMaybeApply f ioMaybeA = runMaybeT $ do
a <- MaybeT ioMaybeA
MaybeT (f a)
If you are using this pattern in multiple places, it will likely pay off to change your a -> IO (Maybe b) functions to a -> MaybeT IO b -- then you can just use (>>=) and/or seamless do-blocks instead of your special-purpose function. On the other hand, if this is just an one-off you may reasonably think that using MaybeT would be overkill; in that case, your implementation is perfectly fine.
(It is worth mentioning that while there is a general-purpose wrapper for nested functors called Compose that has Functor and Applicative instances, it doesn't have a Monad instance, as nesting two monads doesn't necessarily result in something that can be given a legal Monad instance. That being so, we typically resort to monad transformers tailored for each combination that works.)

MonadBaseControl: how to lift simpleHTTP from Happstack?

How to use MonadBaseControl from monad-control to lift simpleHTTP function defined in happstack-server?
Current type of simpleHTTP:
simpleHTTP :: ToMessage a
=> Conf -> ServerPartT IO a -> IO ()
Expected type of simpleHTTPLifted:
simpleHTTPLifted :: (MonadBaseControl IO m, ToMessage a)
=> Conf -> ServerPartT m a -> m ()
My current attempt (does not compile):
simpleHTTPLifted conf action =
liftBaseWith (\runInBase ->
let
fixTypes :: UnWebT m a -> UnWebT IO a
fixTypes c = runInBase c
in simpleHTTP conf (mapServerPartT fixTypes action)
)
Note that similar puzzle is in my related question: MonadBaseControl: how to lift ThreadGroup
I'd like to understand how to in general lift such functions and what are usual steps taken when presented with such a type puzzle?
EDIT: I guess I need a function of type (StM m a -> a). restoreM is pretty close, but does not make it. I've also found an ugly version of fixTypes:
fixTypes :: UnWebT m a -> UnWebT IO a
fixTypes c = do
x <- newIORef undefined
_ <- runInBase (c >>= liftBase . writeIORef x)
readIORef x
This relies on IO being the base monad which is not an optimal solution.
I don't think you can lift this in general for any MonadBaseControl IO m. There are some ms for which we can.
In General
UnWebT m is isomorphic to WebT m which has a MonadTransControl instance. You can convert to and from WebT with mkWebT :: UnWebT m a -> WebT m a and ununWebT :: WebT m a -> UnWebT m a.
MonadBaseControl is a fancy wrapper around a stack of MonadTransControl transformers that flattens the stack so that running and restoring state happens all the way down the stack and all the way back up it again. You can understand MonadBaseControl by understanding MonadTransControl, which I'll repeat briefly here:
class MonadTrans t => MonadTransControl t where
data StT t :: * -> *
liftWith :: Monad m => (Run t -> m a) -> t m a
restoreT :: Monad m => m (StT t a) -> t m a
type Run t = forall n b. Monad n => t n b -> n (StT t b)
The class says with liftWith, "I'll provide a temporary way to run t ms in m, which you can use to build actions in m, which I will in turn run." The StT type of the result says, "the results of the t m things I run in m for you aren't going to be generally available in t m; I need to save my state somewhere, and you have to give me a chance to restore my state if you want the results."
Another way of saying approximately the same thing is, "I can temporarily unwrap the base monad". The question of implementing fixTypes is reduced to "Given that we can temporarily unwrap a WebT from an m and can temporarily unwrap an m from IO, can we permanently unwrap an m from an IO?" for which the answer, barring the capabilities of IO, is almost certainly "no".
IO Tricks
I suspect that there exist ms such that the "ugly" fixTypes will do horrible things like never call writeIORef and thus return undefined or execute code asynchronously and therefore call writeIORef after readIORef. I'm not sure. This is made complicated to reason about due to the possibility that the action created by liftBaseWith is never used in such degenerate cases.
For Comonads
There should be a way to lift simpleHttp without IO tricks precisely when the state of the monad m is a Comonad and therefore has a function extract :: StM m a -> a. For example, this would be the case for StateT s m which essentially has StM s a ~ (s, a).
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE ScopedTypeVariables #-}
import Happstack.Server.SimpleHTTP
import Control.Comonad
import Control.Monad.Base
import Control.Monad.Trans.Control
simpleHTTPLifted :: forall m a. (MonadBaseControl IO m, Comonad (StM m), ToMessage a)
=> Conf -> ServerPartT m a -> m ()
simpleHTTPLifted conf action =
liftBaseWith (\runInBase ->
let
fixTypes :: UnWebT m b -> UnWebT IO b
fixTypes = fmap extract . runInBase
in simpleHTTP conf (mapServerPartT fixTypes action)
)
In practice this isn't very useful because the newtypes defined in the older versions of monad-control don't have Comonad instances and the type synonymns in the newer versions of monad-control make no effort to have the result as the last type argument. For example, in the newest version of monad-control type StT (StateT s) a = (a, s) .

How to nest Parser (IO a) while avoiding unsafePerformIO?

While playing around with parsing based on text-icu's BreakIterator, I've got stuck on implementing a function like this
conditionalParser :: (a -> Bool) -> Parser a -> Parser a -> Parser a -> Parser a
conditionalParser f a b c = do
a' <- a
if f a'
then b
else c
but with a type
conditionalParserIO :: (a -> Bool) -> Parser (IO a) -> Parser (IO a) -> Parser (IO a) -> Parser (IO a)
Is it possible without doing unsafePerformIO?
So far I could only get to some nested dos with the final returned type being Parser (IO (Parser (IO a))), but without any idea how to collapse them.
I think what you want is to use ParsecT instead of Parser.
conditionalParserM :: Monad m => (a -> Bool) -> ParsecT s u m a -> ParsecT s u m a -> ParsecT s u m a
conditionalParserM f a b c = do
a' <- a
if f a' then b else c
This function works with all types of Monads, not just IO.
I suppose it's possible to convert from a ParsecT s u IO a to a Parser (IO a) using runParsecT, depending on which Parser (this or this?) you're using. However, I would recommend that you just restructure your code to work with ParsecT instead.
Clarification
conditionalParserM can't be used as a replacement for conditionalParserIO. I'm suggesting that you need to change how your program works, because attempting to do what your doing (without unsafePerformIO, which you should almost never use) is impossible.
You're looking to compose parsers based on the result of an IO operation, which means that the parser itself will perform side effects when it is run. In order to encapsulate this in the type, you need to use a monad transformer instead.
So, to use conditionalParserM, you need to restructure your code to work with ParsecT instead of Parser.
I just wanted to comment on the difference between Parsec s u (IO a) and ParsecT s u IO a.
You correctly observed that trying to implement your function using Parsec (IO a) yields to Parser (IO (Parser (IO a)). Since both Parser and IO are monads, for both of them we have join :: m (m a) -> m a, which allows to collapse double Parser or double IO. However, in our results we have IO and Parser interleaved. What we need is some function of type IO (Parser a) -> Parser (IO a). If we had such a function f and some x :: Parser (IO (Parser (IO a)), we could use it as liftM f x :: Parser (Parser (IO (IO a))) and then use join and liftM join to collapse both parts into desired Parser (IO a).
Unfortunately there is no such general function for swapping two monads. It's not possible to construct such a function without knowing the internals of a monad, and for some monads it's not even possible at all. For example, there is no total function of type (a -> Maybe b) -> Maybe (a -> b) (the first monad being Maybe, the second one the reader monad (->) a).
And this is why we have monad transformers. A monad transformer corresponding to some monad M knows how to interleave M with another monad. For some monads, such as Reader, swapping it with another monad in the above manner is possible and its transformer is doing exactly that. ReaderT r m a is defined as r -> m a and we can construct:
import Control.Monad
import Control.Monad.Trans
import Control.Monad.Trans.Reader
swapReader :: (Monad m) => m (Reader r a) -> Reader r (m a)
swapReader = fromReaderT . join . lift . liftM (toReaderT . liftM return)
where
-- Helpers to convert ReaderT to Reader and back
fromReaderT :: (Monad m) => ReaderT r m a -> Reader r (m a)
fromReaderT = reader . runReaderT
toReaderT :: (Monad m) => Reader r (m a) -> ReaderT r m a
toReaderT = ReaderT . runReader
We convert m (Reader r a) into ReaderT r m (ReaderT r m a) by augmenting both the inner and outer part and then just collapse it using join.
For other monads, such as MaybeT, swapping is impossible (as in the example above with the (->) a monad). So their transformers are defined differently, for example MaybeT m a is defined as m (Maybe a), not Maybe (m a). Therefore ReaderT r Maybe a is isomorphic MaybeT (ReaderT r) a! There is just one sensible way how to combine Reader and Maybe and so both transformers result in the same thing.
Luckily, we don't have to care about this stuff, once somebody defines a transformer for us.
All we need to know is that the laws hold and how to run the transformer stack at the end.
So using ParsecT s u IO a is the proper solution. ParsecT knows how to interleave parsing within another monad and allows you to combine operations from both of them, without having to deal with the internals.

Haskell Convert from IO a to m a

I want to convert a monadic value into a monadic value of a other Monad class.
Lets say I have a instance declaration:
instance ClassM TypeT where
funcX = abc >>= \x -> return (x)
ClassM : is a own monad defined class
TypeT : is a own defined type/data with a monad instance implementation
abc : is of type IO a
how do i convert the monadic value of abc::IO a
to a monadic value of classM m => m a
aka m1 a -> m2 a (where m1 is not m2)(example: IO 5 -> Just 5)
my implementation of funcX is obviously not correct. If it is possible, what should be the correct implementation?
Thank you for your time and help.
Monad the class represents all things which are monads---it's an adjective more than a noun. If you'd like to convert a particular monad into a value which is generic in all monads that type would look like one of these
Monad m => IO a -> m a
Monad m => Maybe a -> m a
Monad m => [a] -> m a
Monad m => Either e a -> m a
and it's in general impossible, though one type of very special monads has this property.
Another thing you might do is use a Monad transformer which IO at the bottom. This means that you layer another monad "on top of" IO. This lets you have the general operation
lift :: MonadTrans t => m a -> t m a -- notice that t takes two parameters
lift :: IO a -> MyTransformer IO a -- specializing the generic function
and, depending on what MyTransformer is, the specific operation
runMyTransformer :: MyTransformer m a -> m a
runMyTransformer :: MyTransformer IO a -> IO a -- specialized
For instance, the very simplest MonadTrans is IdT.
newtype IdT m a = IdT { runIdT :: m a } deriving Functor
instance Monad m => Monad (IdT m) where
return a = IdT (return a)
IdT ma >>= f = IdT (ma >>= runIdT . f)
instance MonadTrans IdT where
lift ma = IdT ma
Giving us the operations
lift :: IO a -> IdT IO a
runIdT :: IdT IO a -> IO a
which are just inverses of one another, in this case. In general the behavior can be much more complex.
Monad "class" it is not an "class" from objects.
You are asking about Comonads->Monad transformation: Comonad package
convert :: (Comonad w, Monad m) => w a -> m a
convert = return . extract
In any case you can't have comonad from IO
You could use some hackery using unsafePerormIO

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