Extracting nested monadic result: m (m a) -> m a - haskell

I have a function
parseArgs :: [String] -> StdGen -> IO ()
which selects the function to run. The main looks like
main = parseArgs <$> getArgs <*> getStdGen >>= id
The problem I have, parseArgs <$> getArgs <*> getStdGen is of type IO (IO ()), which I extract using (>>= id) which is of type Monad m => m (m b) -> m b. Is there a way to avoid requiring the "extraction" of the value while having just a single line function?

The easiest way would be with join:
main = join $ parseArgs <$> getArgs <*> getStdGen
Personally, I would prefer the form
main = join $ liftM2 parseArgs getArgs getStdGen
where
join :: Monad m => m (m a) -> m a
liftM2 :: Monad m => (a -> b -> r) -> m a -> m b -> m r
Or just use a do
main = do
args <- getArgs
gen <- getStdGen
parseArgs args gen

You can define an operator for this:
infixl 4 <&>
(<&>) :: Monad m => m (a -> m b) -> m a -> m b
f <&> x = f >>= (x >>=)
If you have a function of type
f :: Monad m => (a1 -> a2 -> ... -> an -> m b) -> m a1 -> m a2 -> ... -> m an -> m b
then you can write
fx :: Monad m => m b
fx = f <$> x1 <*> x2 <*> ... <&> xn
where each xi has type m ai.
In your case it would be simply
parseArgs <$> getArgs <&> getStdGen

You could pair up the arguments and put them through a single bind:
main = uncurry parseArgs =<< (,) <$> getArgs <*> getStdGen
This avoids the extraction from nested IO. Admittedly it's no shorter but I find it easier to think about.
It fits a general pattern of doTheWork =<< getAllTheInputs which might be the way you'd end up arranging things anyway, if the code was more complicated.

Related

combining IO monad with State Monad in example

I created simple evaluator for statements.
I would like to do it using transformers - mix IO monad with State.
Could somebody explain how to do it ? It is something that I can't deal with it - transformers.
execStmt :: Stmt -> State (Map.Map String Int) ()
execStmt s = case s of
SAssigment v e -> get >>= (\s -> put (Map.insert v (fromJust (runReader (evalExpM e) s)) s))
SIf e s1 s2 -> get >>= (\s -> case (fromJust (runReader (evalExpM e) s)) of
0 -> execStmt s2
_ -> execStmt s1
)
SSkip -> return ()
SSequence s1 s2 -> get >>= (\s -> (execStmt s1) >>= (\s' -> execStmt s2))
SWhile e s1 -> get >>= (\s -> case (fromJust (runReader (evalExpM e) s)) of
0 -> return ()
_ -> (execStmt s1) >>= (\s' -> execStmt (SWhile e s1)))
execStmt' :: Stmt -> IO ()
execStmt' stmt = putStrLn $ show $ snd $ runState (execStmt stmt) Map.empty
Here's a basic program outline
newtype StateIO s a = SIO {runSIO :: s -> IO (a, s)}
put :: s -> StateIO s ()
put s' = SIO $ \_s -> return ((), s')
liftIO :: IO a -> StateIO s a
liftIO ia = SIO $ \s -> do
a <- ia
return (a, s)
instance Functor (StateIO s) where
fmap ab (SIO sa) = SIO $ \s -> do
(a, s') <- sa s
let b = ab a
return (b, s')
instance Applicative (StateIO s) where
pure a = SIO $ \s -> return (a, s)
(SIO sab) <*> (SIO sa) = SIO $ \s -> do
(ab, s' ) <- sab s
(a , s'') <- sa s'
let b = ab a
return (b, s')
StateIO s a is something that takes an input state (of type s), and returns an IO action to produce something of type a as well as a new state.
To check for understanding, do the following
Write a value of get :: StateIO s s which retrieves the state.
Write an instance for Monad (StateIO s) (it will be similar to the code above).
Finally, and this is the big step to understanding transformers, is to define newtype StateT m s a = StateT {run :: s -> m (a, s)}, and translate the above code to that (with the constraint Monad m). This will show you how monad transformers work.

Apply a function to a file if it exists

I have a function that apply a function to a file if it exists:
import System.Directory
import Data.Maybe
applyToFile :: (FilePath -> IO a) -> FilePath -> IO (Maybe a)
applyToFile f p = doesFileExist p >>= apply
where
apply True = f p >>= (pure . Just)
apply False = pure Nothing
Usage example:
applyToFile readFile "/tmp/foo"
applyToFile (\p -> writeFile p "bar") "/tmp/foo"
A level of abstraction can be added with:
import System.Directory
import Data.Maybe
applyToFileIf :: (FilePath -> IO Bool) -> (FilePath -> IO a) -> FilePath -> IO (Maybe a)
applyToFileIf f g p = f p >>= apply
where
apply True = g p >>= (pure . Just)
apply False = pure Nothing
applyToFile :: (FilePath -> IO a) -> FilePath -> IO (Maybe a)
applyToFile f p = applyToFileIf doesFileExist f p
That allow usages like:
applyToFileIf (\p -> doesFileExist p >>= (pure . not)) (\p -> writeFile p "baz") "/tmp/baz"
I have the feeling that I just scratched the surface and there is a more generic pattern hiding.
Are there better abstractions or more idiomatic ways to do this?
applyToFileIf can be given a more generic type and a more generic name
applyToIf :: Monad m => (a -> m Bool) -> (a -> m b) -> a -> m (Maybe b)
applyToIf f g p = f p >>= apply
where
apply True = g p >>= (return . Just)
apply False = return Nothing
In the type of applyToIf we see the composition of two Monads
Maybe is a monad ---v
applyToIf :: Monad m => (a -> m Bool) -> (a -> m b) -> a -> m (Maybe b)
^------------- m is a monad -------------^
When we see the composition of two monads, we can expect that it could be replaced with a monad transformer stack and some class describing what that monad transformer adds. The MaybeT transformer replaces m (Maybe a)
newtype MaybeT m a = MaybeT { runMaybeT :: m (Maybe a) }
And adds MonadPlus to what an m can do.
instance (Monad m) => MonadPlus (MaybeT m) where ...
We'll change the type of applyToIf to not have a composition of two monads and instead have a MonadPlus constraint on a single monad
import Control.Monad
applyToIf :: MonadPlus m => (a -> m Bool) -> (a -> m b) -> a -> m b
applyToIf f g p = f p >>= apply
where
apply True = g p
apply False = mzero
This could be rewritten in terms of guard from Control.Monad and given a more generic name.
guardBy :: MonadPlus m => (a -> m Bool) -> (a -> m b) -> a -> m b
guardBy f g p = f p >>= apply
where
apply b = guard b >> g p
The second g argument adds nothing to what guardBy can do. guardBy f g p can be replaced by guardBy f return p >>= g. We will drop the second argument.
guardBy :: MonadPlus m => (a -> m Bool) -> a -> m a
guardBy f p = f p >>= \b -> guard b >> return p
The MaybeT transformer adds possible failure to any computation. We can use it to recreate applyToIf or use it more generally to handle failure through complete programs.
import Control.Monad.Trans.Class
import Control.Monad.Trans.Maybe
applyToIf :: Monad m => (a -> m Bool) -> (a -> m b) -> a -> m (Maybe b)
applyToIf f g = runMaybeT . (>>= lift . g) . guardBy (lift . f)
If you instead rework the program to use monad style classes, it might include a snippet like
import Control.Monad.IO.Class
(MonadPlus m, MonadIO m) =>
...
guardBy (liftIO . doesFileExist) filename >>= liftIO . readFile

Using fmap with a maybe when a function uses IO

I have a function that I want to use a Maybe val with. Usually I would do func <$> val. But now suppose that func uses the IO monad. func <$> val will return a Maybe (IO ()). So instead I had to define a new operator:
(<$$>) :: Monad m => (a -> m b) -> Maybe a -> m ()
(<$$>) func (Just val) = func val >> return ()
(<$$>) func Nothing = return ()
So now I can write func <$$> val, but is there a better way to do it?
mapM_ from Data.Foldable is probably the best match:
Prelude Data.Foldable> :set -XScopedTypeVariables
Prelude Data.Foldable> :t \f (a :: Maybe a) -> Data.Foldable.mapM_ f a
\f (a :: Maybe a) -> Data.Foldable.mapM_ f a
:: Monad m => (a -> m b) -> Maybe a -> m ()
If you'd like a more specialised type there's also maybe:
Prelude> :t \f -> maybe (return ()) (f $)
\f -> maybe (return ()) (f $)
:: Monad m => (a -> m ()) -> Maybe a -> m ()
Your <$$> is traverse_ from Data.Foldable.
Is a one-liner always better? Here's how purity of undefined can be useful:
(<$$>) g x = maybe (return undefined) g x >> return ()
Example:
Prelude> print <$$> (Just 1)
1
Prelude> print <$$> Nothing
Prelude>
If you have a lot of this in your code, it might be worth employing the MaybeT transformer:
(\func val -> liftIO . func =<< MaybeT (return val) )
:: (a -> IO b) -> Maybe b -> MaybeT IO b
That doesn't immediately bring you any further than plain IO (Maybe ()), but it composes nicely.

How best to type "Any monad transformer stack containing m"

I'd like to write the function
fixProxy :: (Monad m, Proxy p) => (b -> m b) -> b -> () -> p a' a () b m r
fixProxy f a () = runIdentityP $ do
v <- respond a
a' <- lift (f a)
fixProxy f a' v
which works just like you'd think until I try to run the proxy
>>> :t \g -> runRVarT . runWriterT . runProxy $ fixProxy g 0 >-> toListD
(Num a, RandomSource m s, MonadRandom (WriterT [a] (RVarT n)),
Data.Random.Lift.Lift n m) =>
(a -> WriterT [a] (RVarT n) a) -> s -> m (a, [a])
where I use RVarT intentionally to highlight the existence of the Lift class in RVar. Lift represents the existence of a natural transformation n :~> m which ought to encapsulate what I'm looking for, a function like:
fixProxy :: (Monad m, Monad n, Lift m n, Proxy p)
=> (b -> m b) -> b -> () -> p a' a () b n r
Is Lift the right answer (which would require many orphan instances) or is there a more standard natural transformation MPTC to use?
Note the practical solution, as described in comments below, is something like
runRVarT . runWriterT . runProxy
$ hoistK lift (fixProxy (const $ sample StdUniform) 0) >-> toListD

Strict fmap using only Functor, not Monad

One irritation with lazy IO caught to my attention recently
import System.IO
import Control.Applicative
main = withFile "test.txt" ReadMode getLines >>= mapM_ putStrLn
where getLines h = lines <$> hGetContents h
Due to lazy IO, the above program prints nothing. So I imagined this could be solved with a strict version of fmap. And indeed, I did come up with just such a combinator:
forceM :: Monad m => m a -> m a
forceM m = do v <- m; return $! v
(<$!>) :: Monad m => (a -> b) -> m a -> m b
f <$!> m = liftM f (forceM m)
Replacing <$> with <$!> does indeed alleviate the problem. However, I am not satisfied. <$!> has a Monad constraint, which feels too tight; it's companion <$> requires only Functor.
Is there a way to write <$!> without the Monad constraint? If so, how? If not, why not? I've tried throwing strictness all over the place, to no avail (following code does not work as desired):
forceF :: Functor f => f a -> f a
forceF m = fmap (\x -> seq x x) $! m
(<$!>) :: Functor f => (a -> b) -> f a -> f b
f <$!> m = fmap (f $!) $! (forceF $! m)
I don't think it's possible, and also the monadic forceM doesn't work for all monads:
module Force where
import Control.Monad.State.Lazy
forceM :: Monad m => m a -> m a
forceM m = do v <- m; return $! v
(<$!>) :: Monad m => (a -> b) -> m a -> m b
f <$!> m = liftM f (forceM m)
test :: Int
test = evalState (const 1 <$!> undefined) True
And the evaluation:
Prelude Force> test
1
forceM needs a strict enough (>>=) to actually force the result of its argument. Functor doesn't even have a (>>=). I don't see how one could write an effective forceF. (That doesn't prove it's impossible, of course.)

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