I'm trying to work with Ubigraph in haskell, but I believe my problem is more generic. I'm trying to compile:
import Graphics.Ubigraph
import Control.Monad
import System.Posix.Unistd
main = do
h <- initHubigraph "http://127.0.0.1:20738/RPC2"
runHubigraph op h
op = do
clear
vs <- mapM (const newVertex) [0..200]
mapM_ (setVAttr (VShape Sphere)) vs
putStrLn "something"
let bind i = zipWithM (\a b -> newEdge (a,b)) vs (drop i vs ++ take i vs)
mapM_ bind [1..15]
mapM_ (removeVertex) vs
return ()
and I am getting
Couldn't match expected type `Control.Monad.Trans.Reader.ReaderT
Ubigraph IO a0'
with actual type `IO ()'
In the return type of a call of `putStrLn'
In a stmt of a 'do' expression: putStrLn "something"
In the expression:
do { clear;
vs <- mapM (const newVertex) [0 .. 200];
mapM_ (setVAttr (VShape Sphere)) vs;
putStrLn "something";
.... }
I can see how the type of op is being implied as something different from the return type of putStrLn, but I am not sure how I would reengineer this code to compile properly. Can I simply change the return type of the op function?
Thanks
Your call to putStrLn buried down in op is in the IO monad. You will need to lift it into the Ubigraph monad, using
liftIO $ putStrLn "foo"
Such lifting functions help you access monadic functions lower in the stack. In this case, Ubigraph is a Reader monad composed with the IO monad, and the IO monad is at the bottom. So things in IO must be lifted.
liftIO is from the MonadIO class, in the transformers package.
Related
Hello community thank you for your time.
I have an error and I am not sure what the error is, but what I think the problem is:
There is no IO transformer from ext-1.2.4.1:Data.Text.Internal.Lazy.Text IO) to Web.Scotty.Internal.Types.ScottyT.
But I wondering why the compiler works with ext-1.2.4.1:Data.Text.Internal.Lazy.Text IO). That's why I am working just with String and I removed all occurrences of {-# LANGUAGE OverloadedStrings #-} but still get the error. On the other hand, this should be IO [String], shouldn't it?
And as you can mention I don't really know what ext-1.2.4.1:Data.Text.Internal.Lazy.Text IO) is.
At another place, I already use liftIO successfully for an a -> IO String function. And I think I use them the same way.
I think I get slowly a feeling for what a monad is, but not quite sure. I don't really know why I have to use a lift function at all.
Error message:
• No instance for (MonadIO
(Web.Scotty.Internal.Types.ScottyT
text-1.2.4.1:Data.Text.Internal.Lazy.Text IO))
arising from a use of ‘liftIO’
• In a stmt of a 'do' block:
paths <- liftIO $ getAllFilePaths2 path
In the expression:
do paths <- liftIO $ getAllFilePaths2 path
pathsToScotty paths
In an equation for ‘pathsToScotty2’:
pathsToScotty2 path
= do paths <- liftIO $ getAllFilePaths2 path
pathsToScotty paths
|
49 | paths <- liftIO $ getAllFilePaths2 path
Where the error occurred:
import Control.Monad.IO.Class
...
pathsToScotty2 :: String -> ScottyM ()
pathsToScotty2 path = do
paths <- liftIO $ getAllFilePaths2 path
pathsToScotty paths
getAllFilePaths2 :: String -> IO [String]
getAllFilePaths2 dir = do
putStrLn dir
isFile <- doesFileExist dir
if isFile
then return [dir]
else do
dirs <- listDirectory dir
foldl foldHelper2 (return []) $ map (\d -> show $ mconcat [dir, "/",d ]) dirs
foldHelper2 :: IO [String] -> String -> IO [String]
foldHelper2 ps path = do
paths <- ps
newPaths <- getAllFilePaths2 path
return (paths ++ newPaths)
Truly understanding monads takes time, practice, and patience, but it shouldn't be too hard to understand the need for liftIO by examining your types.
First off, the type of liftIO is MonadIO m => IO a -> m a. This means that the function can convert any IO action into an action in the monad m so long as m has an instance of MonadIO. In theory, this can only be implemented if m has some way of processing IO actions, so this function is embedding the given action into the m monad.
You're definitely in the right sort of place to use liftIO, so why isn't it working? That is, you have a value getAllFilePaths2 path of type IO [String], and you'd like it to be a value of type ScottyM [String] — this indeed seems like a good place to use liftIO. However, ScottyM is not an instance of MonadIO, as that error message you saw is trying to tell you, so you can't use liftIO.
This may seem crazy—can you really not embed IO actions into ScottyM?—but there's actually a good reason for this. What happens if the IO action throws an error? Does your whole web app crash? It would if you naively used liftIO. Instead, scotty provides the function liftAndCatchIO, which, as the docs describe, is "Like liftIO, but catch any IO exceptions and turn them into Scotty exceptions." This is the preferred way to embed IO actions into Scotty.
And here comes the final gotcha: Note that liftAndCatchIO actually produces values of type ActionM a, not ScottyM a. Additionally, there's no way to take a value in the ActionM monad and get it into the ScottyM monad. Instead, you need to use that value as an action. So, I'm not sure what pathsToScotty does, but it's very likely that you'll need to rewrite it.
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
All About Monads explains sequence_:
The sequence_ function (notice the underscore) has the same behavior as sequence but does not return a list of results. It is useful when only the side-effects of the monadic computations are important.
Then, looking at TestSequence.hs:
import Control.Monad
f :: String -> IO ()
f x = print x
run :: [String] -> IO ()
run xs = sequence_ . map f $ xs
I can run it:
λ: run ["foo", "bar"]
"foo"
"bar"
Is sequence_ calling unsafePerformIO on each IO () to get the result, i.e. the ()?
And, is sequence_ discouraged? Or is it, for the IO Monad, simply used "at the end of the world" to run a list of IO actions?
No, it is not calling unsafePerformIO on each IO () action. In fact, its type is not even specific to IO:
sequence_ :: (Monad m, Foldable t) => t (m a) -> m ()
In the old libraries, when it was specific to lists (rather than generic over all Foldables), it was implemented in the following perfectly readable way:
sequence_ [] = return ()
sequence_ (x:xs) = x >> sequence_ xs
It is absolutely not discouraged; sequence_ (and its big brother, mapM_) are extremely useful, to the point that it is even one of my motivating examples for why Monads as an abstraction are useful.
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 the following code:
import Control.Monad
import Control.Monad.Trans
import Control.Monad.Trans.State
type T = StateT Int IO Int
someMaybe = Just 3
f :: T
f = do
x <- get
val <- lift $ do
val <- someMaybe
-- more code in Maybe monad
-- return 4
return 3
When I use do notation inside to work in Maybe monad it fails. From the error it gives it looks like type signature for this do doesn't match. However I have no idea how to fix it. I tried some lift combinations, but none of them worked and I don't want to guess anymore.
The problem is that Maybe is not part of your transformer stack. If your transformer only knows about StateT Int and IO, it does not know anything about how to lift Maybe.
You can fix this by changing your type T to something like:
type T = StateT Int (MaybeT IO) Int
(You'll need to import Control.Monad.Trans.Maybe.)
You will also need to change your inner do to work with MaybeT rather than Maybe. This means wrapping raw Maybe a values with MaybeT . return:
f :: T
f = do
x <- get
val <- lift $ do
val <- MaybeT $ return someMaybe
-- more code in Maybe monad
return 4
return 3
This is a little awkward, so you probably want to write a function like liftMaybe:
liftMaybe = MaybeT . return
If you used lift to lift IO a values in other parts of your code, this will now break because you have three levels in your transformer stack now. You will get an error that looks like this:
Couldn't match expected type `MaybeT IO t0'
with actual type `IO String'
To fix this, you should use liftIO for all your raw IO a values. This uses a typeclass to life IO actions through any number of transformer layers.
In response to your comment: if you only have a bit of code depending on Maybe, it would be easier just to put the result of the do notation into a variable and match against that:
let maybeVal = do val <- someMaybe
-- more Maybe code
return 4
case maybeVal of
Just res -> ...
Nothing -> ...
This means that the Maybe code will not be able to do an IO. You can also naturally use a function like fromMaybe instead of case.
If you want to run the code in the inner do purely in the Maybe monad, you will not have access to the StateT Int or IO monads (which might be a good thing). Doing so will return a Maybe value, which you will have to scrutinize:
import Control.Monad
import Control.Monad.Trans
import Control.Monad.Trans.State
type T = StateT Int IO Int
someMaybe = Just 3
f :: T
f = do
x <- get
-- no need to use bind
let mval = do
-- this code is purely in the Maybe monad
val <- someMaybe
-- more code in Maybe monad
return 4
-- scrutinize the resulting Maybe value now we are back in the StateT monad
case mval of
Just val -> liftIO . putStrLn $ "I got " ++ show val
Nothing -> liftIO . putStrLn $ "I got a rock"
return 3