io-monad

问题 I'm just learning Haskell and IO monads. I'm wondering why wouldn't this force the program to output "hi" as well as "bye": second a b = b main = print ((second $! ((print "hi") >>= (\r -> return ()))) "bye") As far as I understand, the $! operator would force the first argument of second to be evaluated, and the >>= operator would need to run print "hi" in order to get a value off of it and pass it to \r -> return () , which would print "hi" to the screen. What's wrong with my reasoning? And

问题 I'm just learning Haskell and IO monads. I'm wondering why wouldn't this force the program to output "hi" as well as "bye": second a b = b main = print ((second $! ((print "hi") >>= (\r -> return ()))) "bye") As far as I understand, the $! operator would force the first argument of second to be evaluated, and the >>= operator would need to run print "hi" in order to get a value off of it and pass it to \r -> return () , which would print "hi" to the screen. What's wrong with my reasoning? And

来源： https://stackoverflow.com/questions/63712633/what-is-zio-error-channel-and-how-to-get-a-feeling-about-what-to-put-in-it

问题 I have the following code fastShuffle :: [a] -> IO [a] fastShuffle a = <some code> prop_fastShuffle_correct :: [Int] -> Property prop_fastShuffle_correct s = monadicIO ( do sh <- run (fastShuffle s) return ( True ==> ( insertionSort sh == insertionSort s && if length s > 10 then s /= sh else True ) ) ) And .. it is working. I can't understand how what looks to be a pure function ( prop_fastShuffle_correct ) can call a non pure function that has side effects ( fastShuffle ). Hope that someone

问题 I have a Logger type of kind * -> * which can take any type and log the value in a file. I am trying to implement this in a monadic way so that I log and keep working the same. My code looks like import Control.Applicative import Control.Monad import System.IO import Control.Monad.IO.Class instance Functor Logger where fmap = liftM instance Applicative Logger where pure = return (<*>) = ap newtype Logger a = Logger a deriving (Show) instance Monad (Logger) where return = Logger Logger logStr

问题 I have some code that currently uses a ST monad for evaluation. I like not putting IO everywhere because the runST method produces a pure result, and indicates that such result is safe to call (versus unsafePerformIO ). However, as some of my code has gotten longer, I do want to put debugging print statements in. Is there any class that provides a dual-personality monad [or typeclass machinery], one which can be a ST or an IO (depending on its type or a "isDebug" flag)? I recall SPJ

问题 Mutable vectors in Haskell have three element-level mutators: read :: PrimMonad m => MVector (PrimState m) a -> Int -> m a write :: PrimMonad m => MVector (PrimState m) a -> Int -> a -> m () swap :: PrimMonad m => MVector (PrimState m) a -> Int -> Int -> m () Now I can use these fine -- import Data.Vector import Data.Vector.Mutable import Control.Monad.ST import Control.Monad.Primitive incrAt :: Vector Double -> Int -> Vector Double incrAt vec i = runST $ do mvec <- thaw vec oldval <- read

问题 Mutable vectors in Haskell have three element-level mutators: read :: PrimMonad m => MVector (PrimState m) a -> Int -> m a write :: PrimMonad m => MVector (PrimState m) a -> Int -> a -> m () swap :: PrimMonad m => MVector (PrimState m) a -> Int -> Int -> m () Now I can use these fine -- import Data.Vector import Data.Vector.Mutable import Control.Monad.ST import Control.Monad.Primitive incrAt :: Vector Double -> Int -> Vector Double incrAt vec i = runST $ do mvec <- thaw vec oldval <- read

问题 Hello i was wondering how can you unwrap a value at a later time in the IO monad? If a<-expression binds the result to a then can't i use (<-expression) as a parameter for a given method eg: method (<-expression) where method method accepts the result of the evaluation? Code let inh=openFile "myfile" WriteMode let outh=openFile "out.txt" WriteMode hPutStrLn (<-outh) ((<-inh)>>=getLine) I have not entered the Monad chapter just basic <- and do blocks but i suppose it has to do with monads.

问题 I'm trying to understand why a function I have written with a do-block can't be rewritten to fmap a similar lambda expression over a list. I have the following: -- This works test1 x = do let m = T.pack $ show x T.putStrLn m test1 1 Produces 1 But -- This fails fmap (\x -> do let m = T.pack $ show x T.putStrLn m ) [1..10] -- And this also fails fmap (\x -> do T.putStrLn $ T.pack $ show x ) [1..10] With error: <interactive>:1:1: error: • No instance for (Show (IO ())) arising from a use of