Sorry if this is a common question. I have this simple IO() function:
greeter :: IO()
greeter = do
putStr "What's your name? "
name <- getLine
putStrLn $ "Hi, " ++ name
Now I want to call greeter and at the same time specify a parameter that will pre-fill the getLine, so that I don't actually need to interact. I imagine something like a function
IOwithinputs :: [String] -> IO() -> IO()
then I'd do
IOwithinputs ["Buddy"] greeter
which would produce an IO action requiring no user input that would look something like:
What's your name?
Hi, Buddy
I want to do this without modifying the original IO() function greeter. I also don't want to compile greeter and pipe input from the command line. I don't see anything like IOwithinputs in Hoogle. (withArgs is tantalizingly typed and named, but isn't at all what I want.) Is there an easy way to do this? Or is it impossible for some reason? Is this what Pipes is for?
As others have noted, there's no clean way to "simulate" IO if you're already using things like getLine and putStrLn. You have to modify greeter. You could use the hGetLine and hPutStr versions and mock out IO with a fake Handle, or you could use the Purify Code with Free Monads method.
It's far more complex, but also more general and usually a good fit for this kind of mocking, especially when it gets more complex.. I'll explain it briefly below, though the details are somewhat sophisticated.
The idea is that you will be creating your own "fake IO" monad which can be "interpreted" in multiple ways. The primary interpretation is just to use regular IO. The mocked interpretation replaces getLine with some fake lines and echoes everything to stdout.
We'll use the free package. The first step is to describe your interface with a Functor. The basic notion is that each command is a branch of your functor data type and that the functor's "slot" represents the "next action".
{-# LANGUAGE DeriveFunctor #-}
import Control.Monad.Trans.Free
data FakeIOF next = PutStr String next
| GetLine (String -> next)
deriving Functor
These constructors are almost like the regular IO functions from the point of view of someone building a FakeIOF if you ignore the next action. If we want to PutStr we must provide a String. If we want to GetLine we provide a function with only gives the next action when given a String.
Now we need a little confusing boilerplate. We use the liftF function to turn our functor into a FreeT monad. Notice that we provide () as the next action on PutStr and id as our String -> next function. It turns out that these give us the right "return values" if we think of how our FakeIO Monad will behave.
-- Our FakeIO monad
type FakeIO a = FreeT FakeIOF IO a
fPutStr :: String -> FakeIO ()
fPutStr s = liftF (PutStr s ())
fGetLine :: FakeIO String
fGetLine = liftF (GetLine id)
Using these we can build whatever functionality we like and rewrite greeter with very minimal changes.
fPutStrLn :: String -> FakeIO ()
fPutStrLn s = fPutStr (s ++ "\n")
greeter :: FakeIO ()
greeter = do
fPutStr "What's your name? "
name <- fGetLine
fPutStrLn $ "Hi, " ++ name
This might look a little bit magical---we're using do notation without defining a Monad instance. The trick is that FreeT f m is a Monad for any Monad m and Functorf`.
This completes our "mocked" greeter function. Now we must interpret it somehow as we've implemented almost no functionality so far. To write an interpreter we use the iterT function from Control.Monad.Trans.Free. It's fully general type is as follows
iterT
:: (Monad m, Functor f) => (f (m a) -> m a) -> FreeT f m a -> m a
But when we apply it to our FakeIO monad it looks
iterT
:: (FakeIOF (IO a) -> IO a) -> FakeIO a -> IO a
which is much nicer. We provide it a function that takes FakeIOF functors filled with IO actons in the "next action" position (which is how it gets its name) to a plain IO action and iterT will do the magic of turning FakeIO into real IO.
For our default interpreter this is really easy.
interpretNormally :: FakeIO a -> IO a
interpretNormally = iterT go where
go (PutStr s next) = putStr s >> next -- next :: IO a
go (GetLine doNext) = getLine >>= doNext -- doNext :: String -> IO a
But we can also make a mocked interpreter. We'll use the facilities of IO to store some state, in particular a cyclic queue of fake responses.
newQ :: [a] -> IO (IORef [a])
newQ = newIORef . cycle
popQ :: IORef [a] -> IO a
popQ ref = atomicModifyIORef ref (\(a:as) -> (as, a))
interpretMocked :: [String] -> FakeIO a -> IO a
interpretMocked greetings fakeIO = do
queue <- newQ greetings
iterT (go queue) fakeIO
where
go _ (PutStr s next) = putStr s >> next
go q (GetLine getNext) = do
greeting <- popQ q -- first we pop a fresh greeting
putStrLn greeting -- then we print it
getNext greeting -- finally we pass it to the next IO action
and now we can test these functions
λ> interpretNormally greeter
What's your name? Joseph
Hi, Joseph.
λ> interpretMocked ["Jabberwocky", "Frumious"] (greeter >> greeter >> greeter)
What's your name?
Jabberwocky
Hi, Jabberwocky
What's your name?
Frumious
Hi, Frumious
What's your name?
Jabberwocky
Hi, Jabberwocky
I don't think it is easy to do as you ask, but you can do next:
greeter' :: IO String -> IO()
greeter' ioS = do
putStr "What's your name? "
name <- ioS
putStrLn $ "Hi, " ++ name
greeter :: IO ()
greeter = greeter' getLine
ioWithInputs :: Monad m => [a] -> (m a -> m ()) -> m()
ioWithInputs s ioS = mapM_ (ioS.return) s
and test it:
> ioWithInputs ["Buddy","Nick"] greeter'
What's your name? Hi, Buddy
What's your name? Hi, Nick
and even more funny with emulation answer:
> ioWithInputs ["Buddy","Nick"] $ greeter' . (\s -> s >>= putStrLn >> s)
What's your name? Buddy
Hi, Buddy
What's your name? Nick
Hi, Nick
Once you are in IO you can't change the way you get input. To answer your question about pipes, yes it is possible to abstract away the input by defining a Consumer:
import Pipes
import qualified Pipes.Prelude as P
greeter :: Consumer String IO ()
greeter = do
lift $ putStr "What's your name? "
name <- await
lift $ putStrLn $ "Hi, " ++ name
Then you can either specify to use the command line as input:
main = runEffect $ P.stdinLn >-> greeter
... or to use a pure set of strings as input:
main = runEffect $ each ["Buddy"] >-> greeter
Related
I am trying to get a good grip on the do notation in Haskell.
I could use it with Maybe and then print the result. Like this:
maybeAdd :: Maybe Integer
maybeAdd = do one <- maybe1
two <- maybe2
three <- maybe3
return (one + two + three)
main :: IO ()
main = putStr (show $ fromMaybe 0 maybeAdd)
But instead of having a separate function I am trying to use the do notation with the Maybe inside the main function. But I am not having any luck. The various attempts I tried include:
main :: IO ()
main = do one <- maybe1
two <- maybe2
three <- maybe3
putStr (show $ fromMaybe 0 $ return (one + two + three))
main :: IO ()
main = do one <- maybe1
two <- maybe2
three <- maybe3
putStr (show $ fromMaybe 0 $ Just (one + two + three))
main :: IO ()
main = do one <- maybe1
two <- maybe2
three <- maybe3
putStr (show $ (one + two + three))
All of these leads to various types of compilation errors, which unfortunately I failed to decipher to get the correct way to do it.
How do I achieve the above? And perhaps maybe an explanation of why the approaches I tried were wrong also?
Each do block must work within a single monad. If you want to use multiple monads, you could use multiple do blocks. Trying to adapt your code:
main :: IO ()
main = do -- IO block
let x = do -- Maybe block
one <- maybe1
two <- maybe2
three <- maybe3
return (one + two + three)
putStr (show $ fromMaybe 0 x)
You could even use
main = do -- IO block
putStr $ show $ fromMaybe 0 $ do -- Maybe block
one <- maybe1
two <- maybe2
three <- maybe3
return (one + two + three)
-- other IO actions here
but it could be less readable in certain cases.
The MaybeT monad transformer would come handy in this particular case. MaybeT monad transformer is just a type defined something like;
newtype MaybeT m a = MaybeT {runMaybeT :: m (Maybe a)}
Actually transformers like MaybeT, StateT etc, are readily available in Control.Monad.Trans.Maybe, Control.Monad.Trans.State... For illustration purposes it' Monad instance could be something like shown below;
instance Monad m => Monad (MaybeT m) where
return = MaybeT . return . Just
x >>= f = MaybeT $ runMaybeT x >>= g
where
g Nothing = return Nothing
g (Just x) = runMaybeT $ f x
so as you will notice the monadic f function takes a value that resides in the Maybe monad which itself is in another monad (IO in our case). The f function does it's thing and wraps the result back into MaybeT m a.
Also there is a MonadTrans class where you can have some common functionalities those are used by the transformer types. One such is lift which is used to lift the value into a transformer according to that particular instance's definition. For MaybeT it should look like
instance MonadTrans MaybeT where
lift = MaybeT . (liftM Just)
Lets perform your task with monad transformers.
addInts :: MaybeT IO ()
addInts = do
lift $ putStrLn "Enter two integers.."
i <- lift getLine
guard $ test i
j <- lift getLine
guard $ test j
lift . print $ (read i :: Int) + (read j :: Int)
where
test = and . (map isDigit)
So when called like
λ> runMaybeT addInts
Enter two integers..
1453
1571
3024
Just ()
The catch is, since a monad transformer is also a member of Monad typeclass, one can nest them indefinitelly and still do things under a singe do notation.
Edit: answer gets downvoted but it is unclear to me why. If there is something wrong with the approach please care to elaborate me so that it helps people including me to learn something better.
Taking the opportunity of being on the edit session, i would like to add a better code since i think Char based testing might not be the best idea as it will not take negative Ints into account. So let's try using readMaybe from the Text.Read package while we are doing things with the Maybe type.
import Control.Monad.Trans.Maybe
import Control.Monad.Trans.Class (lift)
import Text.Read (readMaybe)
addInts :: MaybeT IO ()
addInts = do
lift $ putStrLn "Enter two integers.."
i <- lift getLine
MaybeT $ return (readMaybe i :: Maybe Int)
j <- lift getLine
MaybeT $ return (readMaybe j :: Maybe Int)
lift . print $ (read i :: Int) + (read j :: Int)
I guess now it works better...
λ> runMaybeT addInts
Enter two integers..
-400
500
100
Just ()
λ> runMaybeT addInts
Enter two integers..
Not an Integer
Nothing
I'm currently learning about free monads and I was toying with probably the simplest and most common example out there – Teletype:
{-# LANGUAGE DeriveFunctor #-}
import Control.Monad.Free
data TeletypeF a = Put String a
| Get (String -> a)
deriving Functor
type Teletype = Free TeletypeF
Many tutorials interpret Teletype programs in the IO monad. For example:
-- Utilities
get = liftF $ Get id
put s = liftF $ Put s ()
-- Sample programs
echo :: Teletype ()
echo = do word <- get
if word == "\04" -- Ctrl-D
then return ()
else put word >> echo
hello :: Teletype ()
hello = do put "What is your name?"
name <- get
put "What is your age?"
age <- get
put ("Hello, " ++ name ++ "!")
put ("You are " ++ age ++ " years old!")
-- Interpret to IO
interpIO :: Teletype a -> IO a
interpIO = foldFree lift
where
lift (Put s a) = putStrLn s >> return a
lift (Get f) = getLine >>= return . f
I was trying to interpret it in a different monad, namely the RWS monad.
This idea was motivated by the last exercise from this assignment.
I'm using the RWS datatype to fetch input from the Reader part and accumulate output in the State part.
But, unfortunately, I'm not able to get it working. Here is my attempt so far:
import Control.Monad.Trans.RWS.Lazy hiding (get, put)
type TeletypeRWS = RWS [String] () [String]
-- Interpret to TeletypeRWS
interpRWS :: Teletype a -> TeletypeRWS a
interpRWS = foldFree lift
where
lift (Put s a) = state (\t -> ((), t ++ [s])) >> return a
lift (Get f) = reader head >>= local tail . return . f -- This is wrong
mockConsole :: Teletype a -> [String] -> (a, [String])
mockConsole p inp = (a, s)
where
(a, s, _) = runRWS (interpRWS p) inp []
When running the TeletypeRWS "programs", the first value in the environment is not removed:
*Main> mockConsole hello ["john", "18"]
((),["What is your name?","What is your age?","Hello, john!","You are john years old!"])
I am a bit uneasy about updating the Reader, but I don't know how else I can access the next value in the list. The type of TeletypeRWS was chosen based on the exercise mentioned above – so I assume it should be possible to implement interpRWS.
We can't use foldFree: it needs to be parametric in the continuation, so we can't apply local there. In contrast, iterM explicitly gives us the actual continuation without generalization, so this will work.
interpRWS = iterM lift where
lift (Put s a) = modify (\t -> t ++ [s]) >> a
lift (Get f) = reader head >>= local tail . f
EDITED 2015-11-29: see bottom
I'm trying to write an application that has a do-last-action-again button. The command in question can ask for input, and my thought for how to accomplish this was to just rerun the resulting monad with memoized IO.
There are lots of posts on SO with similar questions, but none of the solutions seem to work here.
I lifted the memoIO code from this SO answer, and changed the implementation to run over MonadIO.
-- Memoize an IO function
memoIO :: MonadIO m => m a -> m (m a)
memoIO action = do
ref <- liftIO $ newMVar Nothing
return $ do
x <- maybe action return =<< liftIO (takeMVar ref)
liftIO . putMVar ref $ Just x
return x
I've got a small repro of my app's approach, the only real difference being my app has a big transformer stack instead of just running in IO:
-- Global variable to contain the action we want to repeat
actionToRepeat :: IORef (IO String)
actionToRepeat = unsafePerformIO . newIORef $ return ""
-- Run an action and store it as the action to repeat
repeatable :: IO String -> IO String
repeatable action = do
writeIORef actionToRepeat action
action
-- Run the last action stored by repeatable
doRepeat :: IO String
doRepeat = do
x <- readIORef actionToRepeat
x
The idea being I can store an action with memoized IO in an IORef (via repeatable) when I record what was last done, and then do it again it out with doRepeat.
I test this via:
-- IO function to memoize
getName :: IO String
getName = do
putStr "name> "
getLine
main :: IO ()
main = do
repeatable $ do
memoized <- memoIO getName
name <- memoized
putStr "hello "
putStrLn name
return name
doRepeat
return ()
with expected output:
name> isovector
hello isovector
hello isovector
but actual output:
name> isovector
hello isovector
name> wasnt memoized
hello wasnt memoized
I'm not entirely sure what the issue is, or even how to go about debugging this. Gun to my head, I'd assume lazy evaluation is biting me somewhere, but I can't figure out where.
Thanks in advance!
EDIT 2015-11-29: My intended use case for this is to implement the repeat last change operator in a vim-clone. Each action can perform an arbitrary number of arbitrary IO calls, and I would like it to be able to specify which ones should be memoized (reading a file, probably not. asking the user for input, yes).
the problem is in main you are creating a new memo each time you call the action
you need to move memoized <- memoIO getName up above the action
main :: IO ()
main = do
memoized <- memoIO getName --moved above repeatable $ do
repeatable $ do
--it was here
name <- memoized
putStr "hello "
putStrLn name
return name
doRepeat
return ()
edit: is this acceptable
import Data.IORef
import System.IO.Unsafe
{-# NOINLINE actionToRepeat #-}
actionToRepeat :: IORef (IO String)
actionToRepeat = unsafePerformIO . newIORef $ return ""
type Repeatable a = IO (IO a)
-- Run an action and store the Repeatable part of the action
repeatable :: Repeatable String -> IO String
repeatable action = do
repeatAction <- action
writeIORef actionToRepeat repeatAction
repeatAction
-- Run the last action stored by repeatable
doRepeat :: IO String
doRepeat = do
x <- readIORef actionToRepeat
x
-- everything before (return $ do) is run just once
hello :: Repeatable String
hello = do
putStr "name> "
name <- getLine
return $ do
putStr "hello "
putStrLn name
return name
main :: IO ()
main = do
repeatable hello
doRepeat
return ()
I came up with a solution. It requires wrapping the original monad in a new transformer which records the results of IO and injects them the next time the underlying monad is run.
Posting it here so my answer is complete.
{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE GeneralizedNewtypeDeriving #-}
{-# LANGUAGE LambdaCase #-}
import Control.Applicative (Applicative(..))
import Data.Dynamic
import Data.Maybe (fromJust)
import Control.Monad.RWS
-- | A monad transformer adding the ability to record the results
-- of IO actions and later replay them.
newtype ReplayT m a =
ReplayT { runReplayT :: RWST () [Dynamic] [Dynamic] m a }
deriving ( Functor
, Applicative
, Monad
, MonadIO
, MonadState [Dynamic]
, MonadWriter [Dynamic]
, MonadTrans
)
-- | Removes the first element from a list State and returns it.
dequeue :: MonadState [r] m
=> m (Maybe r)
dequeue = do
get >>= \case
[] -> return Nothing
(x:xs) -> do
put xs
return $ Just x
-- | Marks an IO action to be memoized after its first invocation.
sample :: ( MonadIO m
, Typeable r)
=> IO r
-> ReplayT m r
sample action = do
a <- dequeue >>= \case
Just x -> return . fromJust $ fromDynamic x
Nothing -> liftIO action
tell [toDyn a]
return a
-- | Runs an action and records all of its sampled IO. Returns a
-- action which when invoked will use the recorded IO.
record :: Monad m
=> ReplayT m a
-> m (m a)
record action = do
(a, w) <- evalRWST (runReplayT action) () []
return $ do
evalRWST (runReplayT action) () w
return a
Could someone give a super simple (few lines) monad transformer example, which is non-trivial (i.e. not using the Identity monad - that I understand).
For example, how would someone create a monad that does IO and can handle failure (Maybe)?
What would be the simplest example that would demonstrate this?
I have skimmed through a few monad transformer tutorials and they all seem to use State Monad or Parsers or something complicated (for a newbee). I would like to see something simpler than that. I think IO+Maybe would be simple, but I don't really know how to do that myself.
How could I use an IO+Maybe monad stack?
What would be on top? What would be on bottom? Why?
In what kind of use case would one want to use the IO+Maybe monad or the Maybe+IO monad? Would that make sense to create such a composite monad at all? If yes, when, and why?
This is available here as a .lhs file.
The MaybeT transformer will allow us to break out of a monad computation much like throwing an exception.
I'll first quickly go over some preliminaries. Skip down to Adding Maybe powers to IO for a worked example.
First some imports:
import Control.Monad
import Control.Monad.Trans
import Control.Monad.Trans.Maybe
Rules of thumb:
In a monad stack IO is always on the bottom.
Other IO-like monads will also, as a rule, always appear on the bottom, e.g. the state transformer monad ST.
MaybeT m is a new monad type which adds the power of the Maybe monad to the monad m - e.g. MaybeT IO.
We'll get into what that power is later. For now, get used to thinking of MaybeT IO as the maybe+IO monad stack.
Just like IO Int is a monad expression returning an Int, MaybeT IO Int is a MaybeT IO expression returning an Int.
Getting used to reading compound type signatures is half the battle to understanding monad transformers.
Every expression in a do block must be from the same monad.
I.e. this works because each statement is in the IO-monad:
greet :: IO () -- type:
greet = do putStr "What is your name? " -- IO ()
n <- getLine -- IO String
putStrLn $ "Hello, " ++ n -- IO ()
This will not work because putStr is not in the MaybeT IO monad:
mgreet :: MaybeT IO ()
mgreet = do putStr "What is your name? " -- IO monad - need MaybeT IO here
...
Fortunately there is a way to fix this.
To transform an IO expression into a MaybeT IO expression use liftIO.
liftIO is polymorphic, but in our case it has the type:
liftIO :: IO a -> MaybeT IO a
mgreet :: MaybeT IO () -- types:
mgreet = do liftIO $ putStr "What is your name? " -- MaybeT IO ()
n <- liftIO getLine -- MaybeT IO String
liftIO $ putStrLn $ "Hello, " ++ n -- MaybeT IO ()
Now all of the statement in mgreet are from the MaybeT IO monad.
Every monad transformer has a "run" function.
The run function "runs" the top-most layer of a monad stack returning
a value from the inside layer.
For MaybeT IO, the run function is:
runMaybeT :: MaybeT IO a -> IO (Maybe a)
Example:
ghci> :t runMaybeT mgreet
mgreet :: IO (Maybe ())
ghci> runMaybeT mgreet
What is your name? user5402
Hello, user5402
Just ()
Also try running:
runMaybeT (forever mgreet)
You'll need to use Ctrl-C to break out of the loop.
So far mgreet doesn't do anything more than what we could do in IO.
Now we'll work on an example which demonstrates the power of mixing
the Maybe monad with IO.
Adding Maybe powers to IO
We'll start with a program which asks some questions:
askfor :: String -> IO String
askfor prompt = do
putStr $ "What is your " ++ prompt ++ "? "
getLine
survey :: IO (String,String)
survey = do n <- askfor "name"
c <- askfor "favorite color"
return (n,c)
Now suppose we want to give the user the ability to end the survey
early by typing END in response to a question. We might handle it
this way:
askfor1 :: String -> IO (Maybe String)
askfor1 prompt = do
putStr $ "What is your " ++ prompt ++ " (type END to quit)? "
r <- getLine
if r == "END"
then return Nothing
else return (Just r)
survey1 :: IO (Maybe (String, String))
survey1 = do
ma <- askfor1 "name"
case ma of
Nothing -> return Nothing
Just n -> do mc <- askfor1 "favorite color"
case mc of
Nothing -> return Nothing
Just c -> return (Just (n,c))
The problem is that survey1 has the familiar staircasing issue which
doesn't scale if we add more questions.
We can use the MaybeT monad transformer to help us here.
askfor2 :: String -> MaybeT IO String
askfor2 prompt = do
liftIO $ putStr $ "What is your " ++ prompt ++ " (type END to quit)? "
r <- liftIO getLine
if r == "END"
then MaybeT (return Nothing) -- has type: MaybeT IO String
else MaybeT (return (Just r)) -- has type: MaybeT IO String
Note how all of the statemens in askfor2 have the same monad type.
We've used a new function:
MaybeT :: IO (Maybe a) -> MaybeT IO a
Here is how the types work out:
Nothing :: Maybe String
return Nothing :: IO (Maybe String)
MaybeT (return Nothing) :: MaybeT IO String
Just "foo" :: Maybe String
return (Just "foo") :: IO (Maybe String)
MaybeT (return (Just "foo")) :: MaybeT IO String
Here return is from the IO-monad.
Now we can write our survey function like this:
survey2 :: IO (Maybe (String,String))
survey2 =
runMaybeT $ do a <- askfor2 "name"
b <- askfor2 "favorite color"
return (a,b)
Try running survey2 and ending the questions early by typing END as a response to either question.
Short-cuts
I know I'll get comments from people if I don't mention the following short-cuts.
The expression:
MaybeT (return (Just r)) -- return is from the IO monad
may also be written simply as:
return r -- return is from the MaybeT IO monad
Also, another way of writing MaybeT (return Nothing) is:
mzero
Furthermore, two consecutive liftIO statements may always combined into a single liftIO, e.g.:
do liftIO $ statement1
liftIO $ statement2
is the same as:
liftIO $ do statement1
statement2
With these changes our askfor2 function may be written:
askfor2 prompt = do
r <- liftIO $ do
putStr $ "What is your " ++ prompt ++ " (type END to quit)?"
getLine
if r == "END"
then mzero -- break out of the monad
else return r -- continue, returning r
In a sense, mzero becomes a way of breaking out of the monad - like throwing an exception.
Another example
Consider this simple password asking loop:
loop1 = do putStr "Password:"
p <- getLine
if p == "SECRET"
then return ()
else loop1
This is a (tail) recursive function and works just fine.
In a conventional language we might write this as a infinite while loop with a break statement:
def loop():
while True:
p = raw_prompt("Password: ")
if p == "SECRET":
break
With MaybeT we can write the loop in the same manner as the Python code:
loop2 :: IO (Maybe ())
loop2 = runMaybeT $
forever $
do liftIO $ putStr "Password: "
p <- liftIO $ getLine
if p == "SECRET"
then mzero -- break out of the loop
else return ()
The last return () continues execution, and since we are in a forever loop, control passes back to the top of the do block. Note that the only value that loop2 can return is Nothing which corresponds to breaking out of the loop.
Depending on the situation you might find it easier to write loop2 rather than the recursive loop1.
Suppose you have to work with IO values that "may fail" in some sense, like foo :: IO (Maybe a), func1 :: a -> IO (Maybe b) and func2 :: b -> IO (Maybe c).
Manually checking for the presence of errors in a chain of binds quickly produces the dreaded "staircase of doom":
do
ma <- foo
case ma of
Nothing -> return Nothing
Just a -> do
mb <- func1 a
case mb of
Nothing -> return Nothing
Just b -> func2 b
How to "automate" this in some way? Perhaps we could devise a newtype around IO (Maybe a) with a bind function that automatically checks if the first argument is a Nothing inside IO, saving us the trouble of checking it ourselves. Something like
newtype MaybeOverIO a = MaybeOverIO { runMaybeOverIO :: IO (Maybe a) }
With the bind function:
betterBind :: MaybeOverIO a -> (a -> MaybeOverIO b) -> MaybeOverIO b
betterBind mia mf = MaybeOverIO $ do
ma <- runMaybeOverIO mia
case ma of
Nothing -> return Nothing
Just a -> runMaybeOverIO (mf a)
This works! And, looking at it more closely, we realize that we aren't using any particular functions exclusive to the IO monad. Generalizing the newtype a little, we could make this work for any underlying monad!
newtype MaybeOverM m a = MaybeOverM { runMaybeOverM :: m (Maybe a) }
And this is, in essence, how the MaybeT transformer works. I have left out a few details, like how to implement return for the transformer, and how to "lift" IO values into MaybeOverM IO values.
Notice that MaybeOverIO has kind * -> * while MaybeOverM has kind (* -> *) -> * -> * (because its first "type argument" is a monad type constructor, that itself requires a "type argument").
Sure, the MaybeT monad transformer is:
newtype MaybeT m a = MaybeT {unMaybeT :: m (Maybe a)}
We can implement its monad instance as so:
instance (Monad m) => Monad (MaybeT m) where
return a = MaybeT (return (Just a))
(MaybeT mmv) >>= f = MaybeT $ do
mv <- mmv
case mv of
Nothing -> return Nothing
Just a -> unMaybeT (f a)
This will allow us to perform IO with the option of failing gracefully in certain circumstances.
For instance, imagine we had a function like this:
getDatabaseResult :: String -> IO (Maybe String)
We can manipulate the monads independently with the result of that function, but if we compose it as so:
MaybeT . getDatabaseResult :: String -> MaybeT IO String
We can forget about that extra monadic layer, and just treat it as a normal monad.
I have code in the Reader Monad, so as to pass a file handle as an invisible parameter down the Reader chain.
In writeMail, I am trying to create a Reader, which, when run using runReader, produces an IO () output which is itself the result of a chain of IO monads
writeMail :: Reader Handle (IO ())
writeMail mail = do
wmh <- writeMailHeaders mail
wmb <- writeMailBody mail
return $ wmh >>= \_ -> wmb
However I am finding that only the last in the IO chain i.e. wmb, prints at the console.
Can anyone see what I should be doing to get wmh, then wmb to print?
With simpler example:
module Read where
import Data.Functor.Identity
write :: Monad m => m (IO ())
write = do
a <- return $ putStrLn "foo"
b <- return $ putStrLn "bar"
return $ a >> b
main :: IO ()
main = runIdentity write
main prints both "foo" and "bar". So I suspect the error is in writeMailHeaders.
What you need is not just a reader, but a ReaderT monad transformer with IO as a base monad.
Since your example was incomplete, I made some changes to show your options:
import Control.Monad.Reader
writeMail :: ReaderT Handle IO ()
writeMail = do
-- Here's how you get your handle to further do something to it:
handle <- ask
-- Here's how you do the IO actions.
-- Notice the `lift` function,
-- which allows us to run actions of the base monad,
-- which in that case is `IO`.
lift $ do
print "bla bla"
print "bla"