monads

问题 Operation >> description is the following: Sequentially compose two actions, discarding any value produced by the first, like sequencing operators (such as the semicolon) in imperative languages. Here is the example which confuses me: > ([1] ++ [2]) >> ([2] ++ [3]) [2,3,2,3] I'm expecting the list [2,3] which would be result of right part of expression. How can result of [2,3,2,3] be explained? 回答1: (>>) is by default defined as a >> b = a >>= (\_ -> b) so the value being ignored is an a in a

问题 I have been trying to write mfix down using Control.Arrow.loop. I came up with different definitions and would like to see which one is mfix 's actual workalike. So, the solution I reckon to be the right one is the following: mfix' :: MonadFix m => (a -> m a) -> m a mfix' k = let f ~(_, d) = sequenceA (d, k d) in (flip runKleisli () . loop . Kleisli) f As one can see, the loop . Kleisli 's argument works for Applicative instances. I find it to be a good sign as we mostly have our knot-tying

问题 I have been trying to write mfix down using Control.Arrow.loop. I came up with different definitions and would like to see which one is mfix 's actual workalike. So, the solution I reckon to be the right one is the following: mfix' :: MonadFix m => (a -> m a) -> m a mfix' k = let f ~(_, d) = sequenceA (d, k d) in (flip runKleisli () . loop . Kleisli) f As one can see, the loop . Kleisli 's argument works for Applicative instances. I find it to be a good sign as we mostly have our knot-tying

问题 From the Haskell wiki: Monads can be viewed as a standard programming interface to various data or control structures, which is captured by the Monad class. All common monads are members of it: class Monad m where (>>=) :: m a -> (a -> m b) -> m b (>>) :: m a -> m b -> m b return :: a -> m a fail :: String -> m a In addition to implementing the class functions, all instances of Monad should obey the following equations, or Monad Laws: return a >>= k = k a m >>= return = m m >>= (\x -> k x >>=

问题 From the Haskell wiki: Monads can be viewed as a standard programming interface to various data or control structures, which is captured by the Monad class. All common monads are members of it: class Monad m where (>>=) :: m a -> (a -> m b) -> m b (>>) :: m a -> m b -> m b return :: a -> m a fail :: String -> m a In addition to implementing the class functions, all instances of Monad should obey the following equations, or Monad Laws: return a >>= k = k a m >>= return = m m >>= (\x -> k x >>=

问题 So I want to understand the practical cases where monads in JavaScript are helpful. I read a bunch on articles on Monads in JavaScript and understand that jQuery is one example of its use. But besides the "chaining" pattern, what other issues can be effectively solved using Monads in front-end engineering? Ref: http://importantshock.wordpress.com/2009/01/18/jquery-is-a-monad/ http://igstan.ro/posts/2011-05-02-understanding-monads-with-javascript.html 回答1: Well I think the first article is

问题 I tried to validate the construction of a Record with Applicatives and the Either Monad . It works fine. But I can't see all Error Messages. Only the first is visible because the Right Path of the Either Monad ignores them. Here is my code: import Data.Either (either) import Text.Printf (printf) data Record = Record { fieldA :: String , fieldB :: String , fieldC :: String } deriving (Show, Eq) type Err = String setField :: String -> String -> Either Err String setField field value | length

问题 I am trying to write a new monad that only can contain a Num. When it fails, it returns 0 much like the Maybe monad returns Nothing when it fails. Here is what I have so far: data (Num a) => IDnum a = IDnum a instance Monad IDnum where return x = IDnum x IDnum x >>= f = f x fail :: (Num a) => String -> IDnum a fail _ = return 0 Haskell is complaining that there is No instance for (Num a) arising from a use of `IDnum' It suggests that I add a add (Num a) to the context of the type signature

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