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
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 multilevel 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
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
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 selfdefined 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 FAlgebra: 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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