monads

Scala: how to understand the flatMap method of Try?

由 ☆樱花仙子☆ 提交于 2019-12-05 16:16:46
问题 The flatMap method of the Success is implemented like this: def flatMap[U](f: T => Try[U]): Try[U] = try f(value) catch { case NonFatal(e) => Failure(e) } I kinda understand what this method is doing, it helps us to avoid writing a lot of catching code. But in what sense is it similar to the regular flatMap? A regular flatMap takes a sequence of sequences, and put all the elements into one big "flat" sequence. But the flatMap method of Try is not really flattening anything. So, how to

Scotty Using MongoDB

由 随声附和 提交于 2019-12-05 15:25:10
I'm relatively new to Haskell, and this is my first time working with monad transformers. I'd really appreciate some help. runQuery :: Pipe -> Query -> ActionM (Either Failure [Document]) runQuery pipe query = access pipe master "nutrition" (find query >>= rest) main = do pipe <- runIOE $ connect $ host "127.0.0.1" scotty 3000 $ do post "/" $ do b <- body let user :: Either String User = eitherDecode b case user of Left err -> text . pack $ "Could not decode the user:" ++ err ++ ":\n" ++ (show b) Right u -> do let query::Query = (select ["i_name" =: ["$in" =: map (unpack . name) (foods u)]]

Direct translation of Haskell monad into Scala

由 风流意气都作罢 提交于 2019-12-05 14:39:58
Trying to learn how to program monads in Scala, got some troubles Given the quick code sample import Control.Monad newtype LJ a = LJ { session :: a } instance Monad LJ where return s = LJ s (>>=) m f = f ( session m ) instance Functor LJ where fmap f m = LJ . f $ session m type SimpleLJ = LJ String auth :: String -> String -> SimpleLJ auth = undefined readFeed :: String -> SimpleLJ readFeed = undefined closeFeed :: String -> SimpleLJ closeFeed = undefined proceed = auth "123" "456" >>= readFeed >>= closeFeed how do I write the same thing in Scala (not scalaz)? As far as I learned, it's enough

Make an arbitrary class in Scala as a monad instance

由 非 Y 不嫁゛ 提交于 2019-12-05 14:38:18
问题 In order to make anything operable in monad context, if using Haskell - I just add implementation of class Monad for given type anywhere. So I don't touch a source of the data type definition at all. Like (something artificial) data Z a = MyZLeft a | MyZRight a swap (MyZLeft x) = MyZRight x swap (MyZRight x) = MyZLeft x instance Monad Z where return a = MyZRight a (>>=) x f = case x of MyZLeft s -> swap (f s) MyZRight s -> swap (f s) so I'm not touching definition of Z, but make it as a monad

scala Either.RightProjection confusion (for comprehension de-sugaring)

由 北战南征 提交于 2019-12-05 14:27:56
问题 I can use an = in a scala for-comprehension (as specified in section 6.19 of the SLS) as follows: Option Suppose I have some function String => Option[Int] : scala> def intOpt(s: String) = try { Some(s.toInt) } catch { case _ => None } intOpt: (s: String)Option[Int] Then I can use it thus scala> for { | str <- Option("1") | i <- intOpt(str) | val j = i + 10 //Note use of = in generator | } | yield j res18: Option[Int] = Some(11) It was my understanding that this was essentially equivalent to:

Applicative is to monad what X is to comonad

由 |▌冷眼眸甩不掉的悲伤 提交于 2019-12-05 14:23:55
问题 Can we solve this equation for X ? Applicative is to monad what X is to comonad 回答1: After giving it some thought, I think this is actually a backward question. One might think that ComonadApply is to Comonad what Applicative is to Monad , but that is not the case. But to see this, let us use PureScript's typeclass hierarchy: class Functor f where fmap :: (a -> b) -> f a -> f b class Functor f => Apply f where apply :: f (a -> b) -> f a -> f b -- (<*>) class Apply f => Applicative f where

Is this a monad?

由 一曲冷凌霜 提交于 2019-12-05 13:35:01
问题 I'm trying to understand the concept of monads and I want to know if this code is an implementation of this concept (in JavaScript). I have function M which return new object that have set method which create wrapper method var foo = M().set('getX', function() { return this.x; }).set('setX', function(x) { this.x = x; }).set('addX', function(x) { this.x += x; }); And then I can chain method of foo foo.setX(10).addX(20).addX(30).getX() will return 60 and the same if I have object with methods

Applicative instance for MaybeT m assumes Monad m

由 时间秒杀一切 提交于 2019-12-05 13:15:44
问题 I've been using the Haxl monad (described here: http://www.reddit.com/r/haskell/comments/1le4y5/the_haxl_project_at_facebook_slides_from_my_talk), which has the interesting feature that <*> for its Applicative instance isn't the same as ap from Control.Monad. This is a key feature that allows it to do concurrent computations without blocking. For example, if hf and ha are long computations, then let hf :: Haxl (a -> b) = ... ha :: Haxl a = ... in do f <- hf a <- ha return (f a) will do them

Implementing an MFunctor instance for RVarT

由 跟風遠走 提交于 2019-12-05 13:02:01
Is it possible to implement an MFunctor instance for RVarT ? So far I've come up with the following: {-# LANGUAGE RankNTypes #-} import Data.RVar -- from rvar import Control.Monad.Trans.Class (lift) -- from transformers hoistRVarT :: Monad m => (forall t. n t -> m t) -> RVarT n a -> RVarT m a hoistRVarT f rv = sampleRVarTWith (lift . f) rv However this can not be used as a definition of hoist for MFunctor , due to the Monad constraint on m incurred by lift . The problem is, that I couldn't find another way to lift the the resulting monad into RVarT without lift . But I think conceptually it

Scala equivalent of Haskell's do-notation (yet again)

由 丶灬走出姿态 提交于 2019-12-05 12:46:49
问题 I know that Haskell's do x <- [1, 2, 3] y <- [7, 8, 9] let z = (x + y) return z can be expressed in Scala as for { x <- List(1, 2, 3) y <- List(7, 8, 9) z = x + y } yield z But, especially with monads, Haskell often has statements inside the do block that don't correspond to either <- or = . For example, here's some code from Pandoc that uses Parsec to parse something from a string. -- | Parse contents of 'str' using 'parser' and return result. parseFromString :: GenParser tok st a -> [tok] -
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