How can I issue multiple calls to SDL.pollEvent :: IO Event until the output is SDL.NoEvent and collect all the results into a list?
In imperative terms something like this:
events = []
event = SDL.pollEvent
while ( event != SDL.NoEvent ) {
events.add( event )
event = SDL.pollEvent
}
James Cook was so kind to extend monad-loops with this function:
unfoldWhileM :: Monad m => (a -> Bool) -> m a -> m [a]
used with SDL:
events <- unfoldWhileM (/= SDL.NoEvent) SDL.pollEvent
You could use something like:
takeWhileM :: (a -> Bool) -> IO a -> IO [a]
takeWhileM p act = do
x <- act
if p x
then do
xs <- takeWhileM p act
return (x : xs)
else
return []
Instead of:
do
xs <- takeWhileM p act
return (x : xs)
you can also use:
liftM (x:) (takeWhileM p act) yielding:
takeWhileM :: (a -> Bool) -> IO a -> IO [a]
takeWhileM p act = do
x <- act
if p x
then liftM (x:) (takeWhileM p act)
else return []
Then you can use: takeWhileM (/=SDL.NoEvent) SDL.pollEvent
You can use monadic lists:
import Control.Monad.ListT (ListT)
import Control.Monad.Trans.Class (lift) -- transformers, not mtl
import Data.List.Class (takeWhile, repeat, toList)
import Prelude hiding (takeWhile, repeat)
getEvents :: IO [Event]
getEvents =
toList . takeWhile (/= NoEvent) $ do
repeat ()
lift pollEvent :: ListT IO Event
ListT from the "List" package on hackage.
Using these stubs for Event and pollEvent
data Event = NoEvent | SomeEvent
deriving (Show,Eq)
instance Random Event where
randomIO = randomRIO (0,1) >>= return . ([NoEvent,SomeEvent] !!)
pollEvent :: IO Event
pollEvent = randomIO
and a combinator, borrowed and adapted from an earlier answer, that stops evaluating the first time the predicate fails
spanM :: (Monad m) => (a -> Bool) -> m a -> m [a]
spanM p a = do
x <- a
if p x then do xs <- spanM p a
return (x:xs)
else return [x]
allows this ghci session, for example:
*Main> spanM (/= NoEvent) pollEvent
[SomeEvent,SomeEvent,NoEvent]
i eventually stumbled over this code snippet in an actual SDL game from hackage
getEvents :: IO Event -> [Event] -> IO [Event]
getEvents pEvent es = do
e <- pEvent
let hasEvent = e /= NoEvent
if hasEvent
then getEvents pEvent (e:es)
else return (reverse es)
thanks for your answers btw!
Related
How to call IO () function inside another IO () function? I want to print to standard output then call function to do the same.
For example,
p :: String -> IO ()
p [x] = putStr x
p xs = q xs
q :: String -> IO ()
q (x:xs) = putStr x ++ p xs
Your first problem is with typing
p [x] = putStr x
{- putStr :: String -> IO ()
x :: Char, not a String
-}
and
q (x:xs) = putStr x ++ p xs
{- (++) :: [a] -> [a] -> [a]
putStr x :: IO (), not a list of anything.
-}
Let's look at q first, since it follows from p. You're breaking it down into characters, so you should use putChar rather than putStr
Also we're looking at sequencing actions, so we should either use (>>) or (>>=) depending on whether or not you need the result. In this case the result is a value of the unit type (()) which is a useless result and safe to ignore.
q :: String -> IO ()
q (x:xs) = putChar x >> p xs
{- or using `do` notation:
q (x:xs) = do
putChar x
p xs
-}
p can be changed likewise to use putChar rather than putStr
p :: String -> IO ()
p [x] = putChar x
p xs = q xs
though be aware that you haven't matched an empty list on either p or q.
About this time you should notice that substituting putChar for putStr just so you can break strings down to Chars is kind of backward thinking. p = putStr and you're done. However, if you're committed to this backward thinking:
import Control.Monad (foldM_, mapM_)
p = foldM_ (\_ x -> putChar x) ()
-- or
p = foldM_ ((putChar .) . flip const) ()
-- or
p = mapM_ putChar
I'm trying to write a function with StateT only to learn more about it.
In f, I'd like to access to the Int in the last type argument of StateT [Int] IO Int:
f :: StateT [Int] IO Int
f = state $ \xs -> update (error "I want a") xs
update :: Int -> [Int] -> (Int, [Int])
update x [] = (x, [])
update x (y:ys) = (x+y, ys)
Here's how I'd like to call it:
let x = return 55 :: StateT [Int] IO Int
Referencing runStateT:
*Main> :t runStateT
runStateT :: StateT s m a -> s -> m (a, s)
I'd expect to run it:
runStateT (f x) [1,2,3]
to get the following from GHCI, i.e. the IO (Int, [Int]) gets printed:
(56, [2,3])
since the inner a, i.e. 55, + 1, i.e. from [1,2,3], returns (56, [2,3]).
How can I write the above function, getting access to the a?
Ok, here's what say you want:
>>> let x = return 55 :: StateT [Int] IO Int
>>> runStateT (f x) [1,2,3]
(56, [2,3])
So let's work backwards from that.
From the use of f, we can infer its type -
f :: StateT [Int] IO Int -> StateT [Int] IO Int
Note the difference from your given type for f in the question - namely f is a function between values of type StateT [Int] IO Int, not a value of that type.
To define f, we need (>>=) :: Monad m => m a -> (a -> m b) -> m b. This will allow us to take our input of type StateT [Int] IO Int and run some computation on the Int the input computes.
f x = x >>= \i -> state (splitAt 1) >>= \[j] -> return (i + j)
or, using do-notation:
f x = do
i <- x
[j] <- state (splitAt 1)
return (i + j)
Which gives us exactly the result we want.
While this works, it's highly non-idiomatic. Rather than passing monadic values in as inputs to functions and binding them inside the function, it's far more common to define functions that take regular values and return monadic ones, using the bind operator (>>=) outside.
So it'd be far more normal to define
shiftAdd :: Int -> StateT [Int] IO Int
shiftAdd i = do
[j] <- state (splitAt 1)
return (i + j)
So now we can run not only
>>> runStateT (shiftAdd 55) [1,2,3]
(56,[2,3])
but also
>>> runStateT (shiftAdd 55 >>= shiftAdd >>= shiftAdd)
(61,[])
It's still not as idiomatic as it could be as:
I made it unnecessarily partial by using splitAt (it'll throw an exception if the state list is empty)
it's unnecessarily specific (doesn't use IO at all, but we can't use it with other base monads)
Fixing that up gives us:
shiftAdd' :: (Monad m, Num a) => a -> StateT [a] m a
shiftAdd' i = state $ \js -> case js of
[] -> (i, [])
j : js -> (i + j, js)
Which works just fine:
>>> runStateT (return 55 >>= shiftAdd') [1,2,3]
(56,[2,3])
>>> runStateT (return 55 >>= shiftAdd' >>= shiftAdd' >>= shiftAdd') [1,2,3]
(61,[])
>>> runStateT (return 55 >>= shiftAdd' >>= shiftAdd' >>= shiftAdd') []
(55,[])
I can write the following:
f :: [Int] -> [Int]
f x = 0:(map (+1) x)
g :: [Int] -> [Int]
g x = map (*2) x
a = f b
b = g a
main = print $ take 5 a
And things work perfectly fine (ideone).
However, lets say I want g to do something more complex than multiply by 2, like ask the user for a number and add that, like so:
g2 :: [Int] -> IO [Int]
g2 = mapM (\x -> getLine >>= (return . (+x) . read))
How do I then, well, tie the knot?
Clarification:
Basically I want the list of Ints from f to be the input of g2 and the list of Ints from g2 to be the input of f.
The effectful generalization of lists is ListT:
import Control.Monad
import Pipes
f :: ListT IO Int -> ListT IO Int
f x = return 0 `mplus` fmap (+ 1) x
g2 :: ListT IO Int -> ListT IO Int
g2 x = do
n <- x
n' <- lift (fmap read getLine)
return (n' + n)
a = f b
b = g2 a
main = runListT $ do
n <- a
lift (print n)
mzero
You can also implement take like functionality with a little extra code:
import qualified Pipes.Prelude as Pipes
take' :: Monad m => Int -> ListT m a -> ListT m a
take' n l = Select (enumerate l >-> Pipes.take n)
main = runListT $ do
n <- take' 5 a
lift (print n)
mzero
Example session:
>>> main
0
1<Enter>
2
2<Enter>
3<Enter>
7
4<Enter>
5<Enter>
6<Enter>
18
7<Enter>
8<Enter>
9<Enter>
10<Enter>
38
You can learn more about ListT by reading the pipes tutorial, specifically the section on ListT.
I would like to generate random sequences from a Markov chain. To generate the Markov chain I use the following code.
module Main where
import qualified Control.Monad.Random as R
import qualified Data.List as L
import qualified Data.Map as M
type TransitionMap = M.Map (String, String) Int
type MarkovChain = M.Map String [(String, Int)]
addTransition :: (String, String) -> TransitionMap -> TransitionMap
addTransition k = M.insertWith (+) k 1
fromTransitionMap :: TransitionMap -> MarkovChain
fromTransitionMap m =
M.fromList [(k, frequencies k) | k <- ks]
where ks = L.nub $ map fst $ M.keys m
frequencies a = map reduce $ filter (outboundFor a) $ M.toList m
outboundFor a k = fst (fst k) == a
reduce e = (snd (fst e), snd e)
After collecting the statistics and generating a Markov Chain object I would like to generate random sequences. I could imagine this method could look something like that (pseudo-code)
generateSequence mc s
| s == "." = s
| otherwise = s ++ " " ++ generateSequence mc s'
where s' = drawRandomlyFrom $ R.fromList $ mc ! s
I would greatly appreciate if someone could explain to me, how I should implement this function.
Edit
If anyone's interested it wasn't as difficult as I thought.
module Main where
import qualified Control.Monad.Random as R
import qualified Data.List as L
import qualified Data.Map as M
type TransitionMap = M.Map (String, String) Rational
type MarkovChain = M.Map String [(String, Rational)]
addTransition :: TransitionMap -> (String, String) -> TransitionMap
addTransition m k = M.insertWith (+) k 1 m
fromTransitionMap :: TransitionMap -> MarkovChain
fromTransitionMap m =
M.fromList [(k, frequencies k) | k <- ks]
where ks = L.nub $ map fst $ M.keys m
frequencies a = map reduce $ filter (outboundFor a) $ M.toList m
outboundFor a k = fst (fst k) == a
reduce e = (snd (fst e), snd e)
generateSequence :: (R.MonadRandom m) => MarkovChain -> String -> m String
generateSequence m s
| not (null s) && last s == '.' = return s
| otherwise = do
s' <- R.fromList $ m M.! s
ss <- generateSequence m s'
return $ if null s then ss else s ++ " " ++ ss
fromSample :: [String] -> MarkovChain
fromSample ss = fromTransitionMap $ foldl addTransition M.empty $ concatMap pairs ss
where pairs s = let ws = words s in zipWith (,) ("":ws) ws
sample :: [String]
sample = [ "I am a monster."
, "I am a rock star."
, "I want to go to Hawaii."
, "I want to eat a hamburger."
, "I have a really big headache."
, "Haskell is a fun language."
, "Go eat a big hamburger."
, "Markov chains are fun to use."
]
main = do
s <- generateSequence (fromSample sample) ""
print s
The only tiny annoyance is the fake "" starting node.
Not sure if this is what you're looking for. This compiles though:
generateSequence :: (R.MonadRandom m) => MarkovChain -> String -> m String
generateSequence mc s | s == "." = return s
| otherwise = do
s' <- R.fromList $ rationalize (mc M.! s)
s'' <- generateSequence mc s'
return $ s ++ " " ++ s''
rationalize :: [(String,Int)] -> [(String,Rational)]
rationalize = map (\(x,i) -> (x, toRational i))
All random number generation needs to happen in either the Random monad or the IO monad. For your purpose, it's probably easiest to understand how to do that in the IO monad, using evalRandIO. In the example below, getRandom is the function we want to use. Now getRandom operates in the Random monad, but we can use evalRandIO to lift it to the IO monad, like this:
main :: IO ()
main = do
x <- evalRandIO getRandom :: IO Double
putStrLn $ "Your random number is " ++ show x
Note: The reason we have to add the type signature to the line that binds x is because in this particular example there are no other hints to tell the compiler what type we want x to be. However, if we used x in some way that makes it clear that we want it to be a Double (e.g., multiplying by another Double), then the type signature wouldn't be necessary.
Using your MarkovChain type, for a current state you can trivially get the available transitions in the form [(nextState,probability)]. (I'm using the word "probability" loosely, it doesn't need to be a true probability; any numeric weight is fine). This is what fromList in Control.Monad.Random is designed for. Again, it operates in the Random monad, but we can use evalRandIO to lift it to the IO monad. Suppose transitions is your list of transitions, having the type [(nextState,probability)]. Then, in the IO monad you can call:
nextState <- evalRandIO $ fromList transitions
You might instead want to create your own function that operates in the Random monad, like this:
getRandomTransition :: RandomGen g => MarkovChain -> String -> Rand g String
getRandomTransition currState chain = do
let transitions = lookup currState chain
fromList transitions
Then you can call this function in the IO monad using evalRandIO, e.g.
nextState <- evalRandIO $ getRandomTransition chain
I have a simple list that I would like to iterate over "yield"ing between each element and printing that element to the output. I am trying to use the ContT monad to do this but running into issues. Here's what I have so far:
data K a = Nil | K (a,() -> K a)
listIterator :: (Monad m) => [r] -> m (K r)
listIterator [] = return Nil
listIterator (x:xs) = return (ContT (\k -> K (x,k))) >> listIterator xs
runIterator :: IO ()
runIterator = do
a <- listIterator ([1,2,3] :: [Int])
let loop Nil = liftIO $ print "nil"
loop (K (curr,newI)) =
do
liftIO $ print curr
loop (newI ())
loop a
The expected output is:
1
2
3
nil
What I get is:
nil
Any help is appreciated!
listIterator (x:xs) = return (ContT (\k -> K (x,k))) >> listIterator xs
does not do what you expect, equational reasoning
listIterator (x:xs)
= return (ContT (\k -> K (x,k))) >> listIterator xs
= (return (ContT (\k -> K (x,k)))) >>= \_ -> listIterator xs
= (\_ -> listIterator xs) (ContT (\k -> K (x,k)))
= listIterator xs
I'm not sure exactly why you want to use an iterator. Haskell is already lazy, so iteration patterns like this are mostly used only when you have resource management issues that need to interact well with a demand driven usage pattern. And, you don't need the continuation monad at all:
Instead of writing the K constructor to take a tuple it is more idiomatic to
data K a = Nil | K a (() -> K a)
intuitively, the type for the listIterator does not use its monadic structure: it just constructs a value, so
listIterator ::[r] -> K r
listIterator [] = Nil
listIterator (x:xs) = K x (\_ -> listIterator xs)
now life is trivial
runIterator :: IO ()
runIterator = do
let a = listIterator ([1,2,3] :: [Int])
loop Nil = liftIO $ print "nil"
loop (K curr newI) =
do
liftIO $ print curr
loop (newI ())
loop a
which would probably be best to write without the use of do notation.
This may not be the answer you were looking for, but if you are interested in this style of programming, you should look into pipes and similar libraries. (conduit is the rising star in the "real world", but pipes provides a simpler tool for teaching which is why I use it here.)
$ cabal update && cabal install pipes
Pipes are like iterators, except they come in three flavors: those that can acquire input (Consumers), those that produce output (Producers), and those that do both (Pipes). If you connect pipes such that the input and output ends are all satisfied, then it is called a "Pipeline", and it is a self-contained unit that can be run without any additional input.
Pipe provides a monad instance for convenience in creating pipes. The >+> operator connects two pipes together.
import Control.Pipe
import Control.Monad.Trans.Class
import Control.Monad.IO.Class
-- annoyingly, Pipe does not provide a MonadIO instance
instance (MonadIO m) => MonadIO (Pipe a b m) where
liftIO = lift . liftIO
listIterator :: Monad m => [a] -> Producer (Maybe a) m ()
listIterator (x:xs) = yield (Just x) >> listIterator xs
listIterator [] = yield Nothing
printer :: (MonadIO m, Show a) => Consumer (Maybe a) m ()
printer = do
mx <- await
case mx of
Just x -> liftIO (print x) >> printer
Nothing -> liftIO (putStrLn "nil")
main = runPipe $ listIterator [1, 2, 3] >+> printer
The source for Control.Pipe is delightfully simple, especially if you have been reading Gabriel's recent blog posts about Free monads, particularly Why free monads matter and Purify code using free monads.