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Basic Haskell IO Monad FilePath join


Basic Haskell IO Monad FilePath join

By : Surag
Date : November 14 2020, 04:48 PM
wish helps you Exploring the basics of Haskell IO Monads, I would like to create a handle to a temporary file. The temporary directory FilePath is within an IO context, and I need to map a function into this context to join the path with a filename. , I think the answer to your question is to use the flip function, e.g.
code :
createTempFile filename = fmap (flip joinDrive filename) getTemporaryDirectory
createTempFile filename = fmap (`joinDrive` filename) getTemporaryDirectory


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Haskell with IO monad error, from IO to basic type

Haskell with IO monad error, from IO to basic type


By : Rob
Date : March 29 2020, 07:55 AM
like below fixes the issue liftM fsum vals has type IO Float. You're giving it a name with let res = ..., but later trying to use it as a Float. You should bind it instead, producing a Float as desired, using <-:
code :
res <- liftM fsum vals
let res = fsum vals
haskell join multi-level monad

haskell join multi-level monad


By : dpace
Date : March 29 2020, 07:55 AM
may help you . If you note that m (n _) is a monad transformer then you can define this operation. This is certainly the case when we notice that IO (Maybe a) is the same as MaybeT IO a: we then just use MaybeT's join. We can do this in no small part due to the fact that Maybe and IO "layer" especially nicely.
On the other hand, not all "composed" monads are monad transformers. In particular, IO [a] isn't really one. The true ListT transformer we'd like to have and to join looks like
code :
newtype ListT m a = ListT { runListT :: m (Maybe (a, ListT m a)) }
Define bind without join for the list monad in Haskell

Define bind without join for the list monad in Haskell


By : abhanah
Date : March 29 2020, 07:55 AM
will be helpful for those in need Sure. The actual definition in GHC/Base.hs is in terms of the equivalent list comprehension:
code :
instance Monad []  where
    xs >>= f             = [y | x <- xs, y <- f x]
(>>=) :: [a] -> (a -> [b]) -> [b]
[] >>= f = ???
(x:xs) >>= f = ???
[] >>= f = []
(x:xs) >>= f = f x ++ (xs >>= f)
Haskell: Join on State Monad

Haskell: Join on State Monad


By : Rahul Sangareddy
Date : March 29 2020, 07:55 AM
may help you . How to formally calculate/interpret the following expression? , If you specialise the type of join for State s you get:
Haskell: How to prove(test) whether a self-defined Monad instance follow Monad laws?

Haskell: How to prove(test) whether a self-defined Monad instance follow Monad laws?


By : Rishav Garg
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , FWIW, here's a set of QuickCheck properties I recently wrote that tests the Monad laws for a Maybe implementation derived from its F-Algebra:
code :
testProperty "Monad left identity law" $ do
  a :: String <- arbitrary
  k :: String -> MaybeFix Integer <- (fromMaybe .) <$> arbitrary

  let left = return a >>= k
  let right = k a

  return $ left == right
,
testProperty "Monad right identity law" $ do
  m :: MaybeFix Integer <- fromMaybe <$> arbitrary

  let left = m >>= return
  let right = m

  return $ left == right
,
testProperty "Monad associativity law" $ do
  m :: MaybeFix String <- fromMaybe <$> arbitrary
  k :: String -> MaybeFix Integer <- (fromMaybe .) <$> arbitrary
  h :: Integer -> MaybeFix Ordering <- (fromMaybe .) <$> arbitrary

  let left = m >>= (\x -> k x >>= h)
  let right = (m >>= k) >>= h

  return $ left == right
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