Haskell websockets header - haskell

I am trying to read websockets header in haskell.
Current code is as follows
import Network
import Network.Socket.Internal
import Text.Printf
import Control.Monad
import Control.Monad.IO.Class
import Data.Maybe
import Data.List
import Data.Digest.Pure.SHA (bytestringDigest, sha1)
import Text.Regex
import System.IO
port :: Int
port = 8080
bufferLength :: Int
bufferLength = 2048
keyRegex :: String
keyRegex = "Sec-WebSocket-Key: (.+)(\r)"
response :: String
response = "HTTP/1.1 101 Web Socket Protocol Handshake\r\nUpgrade: WebSocket\r\nConnection: Upgrade\r\nsec-websocket-accept: changethis\r\n\r\n"
guidString :: String
guidString = "258EAFA5-E914-47DA-95CA-C5AB0DC85B11"
talk :: MonadIO m => Handle -> m String
talk handle = do liftIO $ hGetContents handle
main :: IO()
main = withSocketsDo $ do
sock <- listenOn (PortNumber (fromIntegral port))
printf "Listening on port %d\n" port
forever $ do
(handle, host, port) <- accept sock
printf "Accepted connection from %s: %s\n" host (show port)
hSetBuffering handle NoBuffering
putStrLn $ talk handle
The code throws error
Main.hs:38:16:
Couldn't match type `[Char]' with `Char'
Expected type: String
Actual type: [String]
In the return type of a call of `talk'
In the second argument of `($)', namely `talk handle'
In a stmt of a 'do' block: putStrLn $ talk handle
How to print header string? Talk function should get the header string and return it to be printed.

Return type of talk handle is (MonadIO m => m String).
putStrLn has type String -> IO () or [Char] -> IO ().
[] is a monad.
putStrLn expects a list of chars, but it gets a value of (MonadIO m => m String).
m becomes a list, and putStrLn gets a value of type [String] and this is error.
As #Zeta suggests, use talk handle >>= putStrLn instead:
>>= is operator which allows to continue computation with the result from another computation. In your case, you can read talk handle >>= putStrLn as follows:
talk handle -- first compute it
>>= putStrLn -- then take string from result of `talk handle` and print it
There won't be any error such was with putStrLn $ talk handle:
(>>=) :: Monad m => m a -> (a -> m b) -> m b
talk handle :: MonadIO m => m String
(>>=) (talk handle) :: MonadIO m => (String -> m b) -> m b
putStrLn :: String -> IO ()
(>>=) (talk handle) putStrLn :: IO () -- `m` becomes `IO`
Edited:
Here you can find introduction to IO monad: LYHFGG, Input and Output
Here you can find introduction to monads in general: LYHFGG, A Fistful of Monads.
You should know about Functor and Applicative in order to read A Fistful of Monads. You can find it in book's contents anyway.

Related

Maintaining state during interactive command line prompt

I want to write a toy program that has an interactive prompt and that can save and display all previous inputs. This is my first attempt, but does not compile (using ghc):
import System.IO
import Control.Monad.State
data ProgramState = ProgramState
{ events :: [Int] } -- Placeholder for now
parse_input :: String -> State ProgramState Bool
parse_input prompt = do
putStr prompt
hFlush stdout
current_state <- get
str <- getLine
case str of
"c" -> do
put (current_state { events = [1,2,3] } ) -- this should become actual appending
return True
"l" -> return True
"q" -> return False
"quit" -> return False
"h" -> return True
_ -> do
putStrLn "Invalid input."
parse_input prompt
main :: IO ()
main = do
should_continue <- parse_input "Enter your command."
if should_continue then main else return ()
main.hs:9:5: error:
• Couldn't match type ‘IO’
with ‘StateT ProgramState Data.Functor.Identity.Identity’
Expected type: StateT
ProgramState Data.Functor.Identity.Identity ()
Actual type: IO ()
Note: line 9 is putStr prompt
The same error is given for lines 10, 12, 22, 27.
I have since thought of doing the recursion purely inside parse_input, in which case I don't seem to need the state monad. But I am still curious why I get the compilation error. Any help is appreciated, I am very new to Haskell.
You seem to be mixing values of type State s a with values of type IO a. In your main action, you call parse_input in a context expecting IO. In parse_input, you call putStr and so on in a context expecting State. That's not going to work!
The usual way to do this kind of thing is to switch from State to StateT, and import Control.Monad.IO.Class. Now, you can use
evalStateT :: StateT s m a -> s -> m a
to "lower" your loop to IO, and
-- liftIO :: IO a -> StateT s IO a
liftIO :: MonadIO m => IO a -> m a
to "lift" the IO actions to StateT within the loop. Now (untested code ahead):
-- Needed for flexible use of
-- the MonadState class.
{-# LANGUAGE FlexibleContexts #-}
import System.IO
-- You almost always want the "strict"
-- version of `StateT`; the lazy one is weird.
import Control.Monad.State.Strict
import Control.Monad.IO.Class
data ProgramState = ProgramState
{ events :: [Int] } -- Placeholder for now
-- Renaming your function to follow convention.
parseInput
:: (MonadState ProgramState m, MonadIO m)
=> String -> m Bool
parseInput prompt = do
str <- liftIO $ do
putStr prompt
hFlush stdout
getLine
current_state <- get
case str of
"c" -> do
put (current_state { events = [1,2,3] } ) -- this should become actual appending
return True
"l" -> return True
"q" -> return False
"quit" -> return False
"h" -> return True
_ -> do
liftIO $ putStrLn "Invalid input."
parseInput prompt
main :: IO ()
main = do
-- You need to supply the initial state; I've just guessed here.
should_continue <- evalStateT (parseInput "Enter your command.") (ProgramState [])
if should_continue then main else return ()
As Daniel Wagner points out, this will not preserve the state from one main run to the next. If that's your intention, you can write
main :: IO ()
main = evalStateT loop (ProgramState [])
where
loop = do
should_continue <- parseInput "Enter your command."
if should_continue then loop else return ()
If you like, you can import Control.Monad and shorten this to
main :: IO ()
main = evalStateT loop (ProgramState [])
where
loop = do
should_continue <- parseInput "Enter your command."
when should_continue loop
Final note: if you want to capture the final state of your loop, use runStateT instead of evalStateT.

Simplest non-trivial monad transformer example for "dummies", IO+Maybe

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.

How to use State with Pipes?

I have a function with a type Map Int String -> Proxy () a () Void IO b. Right now it awaits, does whatever with the value it got, and then re-calls itself. I'd like to change it to use State (Map Int String) instead of having that passed as an argument, so I can just use forever and don't need to have every branch remember to recurse. I get that I need to use StateT to combine State with another monad, but I don't understand where in that type signature the StateT belongs, or whether or not I need to lift functions like get. What is the correct type for a function that is both a State (Map Int String) and a Proxy () a () Void IO b?
Note: Proxy () a () Void = Consumer a, so I will refer to it as a Consumer for this answer.
The simple way is to put your StateT monad transformer layer outside of the Consumer layer and then run it immediately. Here is an example:
import Control.Monad (forever)
import Control.Monad.Trans.State.Strict
import Pipes
example :: (Show a) => Consumer a IO r
example = flip evalStateT 0 $ forever $ do
-- Inside here we are using `StateT Int (Consumer a IO) r`
a <- lift await
n <- get
lift $ lift $ putStrLn $ "Received value #" ++ show n ++ ": " ++ show a
put (n + 1)
... and this is how it behaves in action:
>>> runEffect $ each ["Test", "ABC"] >-> example
Received value #0: "Test"
Received value #1: "ABC"

Error check within do block in Haskell

i have the following set of actions:
action1 :: IO Bool
action2 :: IO Bool
action3 :: IO Bool
some actions are just composition of another actions
complexAction = do
action1
action2
action3
What i need is the construction that checks result of each action and returns False in a case of false. I can do it manually but i know for sure that haskell does have tools to get rid of that kind of boilerplate.
The simplest way is
complexAction = fmap and (sequence [action1, action2, action3])
But you could also write your own combinator to stop after the first action:
(>>/) :: Monad m => m Bool -> m Bool -> m Bool
a >>/ b = do
yes <- a
if yes then b else return False
You'd want to declare the fixity to make it associative
infixl 1 >>/
Then you can do
complexAction = action1 >>/ action2 >>/ action3
I'd suggest you to use MaybeT monad transformer instead. Using it has many advantages over just returning IO Bool:
Your actions can have different types and return values (not just true/false). If you don't need any results, just use MaybeT IO ().
Later ones can depend on results of preceding ones.
Since MaybeT produces monads that are instances of MonadPlus, you can use all monad plus operations. Namely mzero for a failed action and x mplus y, which will run y iff x fails.
A slight disadvantage is that you have to lift all IO actions to MaybeT IO. This can be solved by writing your actions as MonadIO m => ... -> m a instead of ... -> IO a.
For example:
import Control.Monad
import Control.Monad.IO.Class
import Control.Monad.Trans
import Control.Monad.Trans.Maybe
-- Lift print and putStrLn
print' :: (MonadIO m, Show a) => a -> m ()
print' = liftIO . print
putStrLn' :: (MonadIO m) => String -> m ()
putStrLn' = liftIO . putStrLn
-- Add something to an argument
plus1, plus3 :: Int -> MaybeT IO Int
plus1 n = print' "+1" >> return (n + 1)
plus3 n = print' "+3" >> return (n + 3)
-- Ignore an argument and fail
justFail :: Int -> MaybeT IO a
justFail _ = mzero
-- This action just succeeds with () or fails.
complexAction :: MaybeT IO ()
complexAction = do
i <- plus1 0
justFail i -- or comment this line out <----------------<
j <- plus3 i
print' j
-- You could use this to convert your actions to MaybeT IO:
boolIOToMaybeT :: IO Bool -> MaybeT IO ()
boolIOToMaybeT x = do
r <- lift x
if r then return () else mzero
-- Or you could have even more general version that works with other
-- transformers as well:
boolIOToMaybeT' :: (MonadIO m, MonadPlus m) => IO Bool -> m ()
boolIOToMaybeT' x = do
r <- liftIO x
if r then return () else mzero
main :: IO ()
main = runMaybeT complexAction >>= print'
As Petr says, for anything but a narrow and contained case, you're almost certainly better off wiring your code for proper error handling from the outset. I know I've often regretted not doing this, condemning myself to some very tedious refactoring.
If I may, I'd like to recommend Gabriel Gonzalez's errors package, which imposes a little more coherence on Haskell's various error-handling mechanisms than has been traditional. It allows you to plumb Eithers through your code, and Either is a good type for capturing errors. (By contrast, Maybe will lose information on the error side.) Once you've installed the package, you can write things like this:
module Errors where
import Control.Error
import Data.Traversable (traverse)
data OK = OK Int deriving (Show)
action1, action2, action3 :: IO (Either String OK)
action1 = putStrLn "Running action 1" >> return (Right $ OK 1)
action2 = putStrLn "Running action 2" >> return (Right $ OK 2)
action3 = putStrLn "Running action 3" >> return (Left "Oops on 3")
runStoppingAtFirstError :: [IO (Either String OK)] -> IO (Either String [OK])
runStoppingAtFirstError = runEitherT . traverse EitherT
...with output like
*Errors> runStoppingAtFirstError [action1, action2]
Running action 1
Running action 2
Right [OK 1,OK 2]
*Errors> runStoppingAtFirstError [action1, action3, action2]
Running action 1
Running action 3
Left "Oops on 3"
(But note that the computation here stops at the first error and doesn't soldier on until the bitter end -- which might not be what you had wanted. The errors package is certainly wide-ranging enough that many other variations are possible.)

Composing IO Monads using do

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"

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