applicative

What exactly does “effectful” mean

由 爷，独闯天下 提交于 2019-11-26 19:25:47
问题 Time and again I read the term effectful , but I am still unable to give a clear definition of what it means. I assume the correct context is effectful computations , but I've also seen the term effectful values ) I used to think that effectful means having side effects . But in Haskell there are no side-effects (except to some extent IO). Still there are effectful computations all over the place. Then I read that monads are used to create effectful computations. I can somewhat understand

Monoidal Functor is Applicative but where is the Monoid typeclass in the definition of Applicative?

由 喜夏-厌秋 提交于 2019-11-26 18:20:18
问题 Applicative is a Monoidal Functor : mappend :: f -> f -> f $ :: (a -> b) -> a -> b <*> :: f(a -> b) -> f a -> f b But I don't see any reference about Monoid in the definition of the Applicative typeclass, could you tell me why ? Definition : class Functor f => Applicative (f :: * -> *) where pure :: a -> f a (<*>) :: f (a -> b) -> f a -> f b GHC.Base.liftA2 :: (a -> b -> c) -> f a -> f b -> f c (*>) :: f a -> f b -> f b (<*) :: f a -> f b -> f a {-# MINIMAL pure, ((<*>) | liftA2) #-} No

How to handle side effect with Applicative?

由 一曲冷凌霜 提交于 2019-11-26 17:18:48
问题 I see everywhere that Applicative can handle side effects, but all the simple examples I've seen are just combining stuff together like: > (,,) <$> [1,2] <*> ["a", "b", "c"] <*> ["foo", "bar"] [(1,"a","foo"),(1,"a","bar"),(1,"b","foo"),(1,"b","bar"), (1,"c","foo"),(1,"c","bar"),(2,"a","foo"),(2,"a","bar"), (2,"b","foo"),(2,"b","bar"),(2,"c","foo"),(2,"c","bar")] Which is cool but I can't see how that links to side effects. My understanding is that Applicative is a weak monad and so you can

What advantage does Monad give us over an Applicative?

由 情到浓时终转凉″ 提交于 2019-11-26 12:58:51
问题 I\'ve read this article, but didn\'t understand last section. The author says that Monad gives us context sensitivity, but it\'s possible to achieve the same result using only an Applicative instance: let maybeAge = (\\futureYear birthYear -> if futureYear < birthYear then yearDiff birthYear futureYear else yearDiff futureYear birthYear) <$> (readMay futureYearString) <*> (readMay birthYearString) It\'s uglier for sure without do-syntax, but beside that I don\'t see why we need Monad. Can

Difference between Monad and Applicative in Haskell

由 我与影子孤独终老i 提交于 2019-11-26 10:16:51
问题 I just read the following from typeclassopedia about the difference between Monad and Applicative . I can understand that there is no join in Applicative . But the following description looks vague to me and I couldn\'t figure out what exactly is meant by \"the result\" of a monadic computation/action. So, if I put a value into Maybe , which makes a monad, what is the result of this \"computation\"? Let’s look more closely at the type of (>>=). The basic intuition is that it combines two

When and why should one use Applicative Functors in Scala

由 此生再无相见时 提交于 2019-11-26 10:10:02
问题 I know that Monad can be expressed in Scala as follows: trait Monad[F[_]] { def flatMap[A, B](f: A => F[B]): F[A] => F[B] } I see why it is useful. For example, given two functions: getUserById(userId: Int): Option[User] = ... getPhone(user: User): Option[Phone] = ... I can easily write function getPhoneByUserId(userId: Int) since Option is a monad: def getPhoneByUserId(userId: Int): Option[Phone] = getUserById(userId).flatMap(user => getPhone(user)) ... Now I see Applicative Functor in Scala

Distinction between typeclasses MonadPlus, Alternative, and Monoid?

由 。_饼干妹妹 提交于 2019-11-26 10:09:04
问题 The standard-library Haskell typeclasses MonadPlus , Alternative , and Monoid each provide two methods with essentially the same semantics: An empty value: mzero , empty , or mempty . An operator a -> a -> a that joins values in the typeclass together: mplus , <|> , or mappend . All three specify these laws to which instances should adhere: mempty `mappend` x = x x `mappend` mempty = x Thus, it seems the three typeclasses are all providing the same methods. ( Alternative also provides some

Applicatives compose, monads don&#39;t

由 岁酱吖の 提交于 2019-11-26 09:47:43
Applicatives compose, monads don't. What does the above statement mean? And when is one preferable to other? If we compare the types (<*>) :: Applicative a => a (s -> t) -> a s -> a t (>>=) :: Monad m => m s -> (s -> m t) -> m t we get a clue to what separates the two concepts. That (s -> m t) in the type of (>>=) shows that a value in s can determine the behaviour of a computation in m t . Monads allow interference between the value and computation layers. The (<*>) operator allows no such interference: the function and argument computations don't depend on values. This really bites. Compare

Applicatives compose, monads don&#39;t

由 ♀尐吖头ヾ 提交于 2019-11-26 03:26:14
问题 Applicatives compose, monads don\'t. What does the above statement mean? And when is one preferable to other? 回答1: If we compare the types (<*>) :: Applicative a => a (s -> t) -> a s -> a t (>>=) :: Monad m => m s -> (s -> m t) -> m t we get a clue to what separates the two concepts. That (s -> m t) in the type of (>>=) shows that a value in s can determine the behaviour of a computation in m t . Monads allow interference between the value and computation layers. The (<*>) operator allows no

Good examples of Not a Functor/Functor/Applicative/Monad?

由 筅森魡賤 提交于 2019-11-26 00:31:18
问题 While explaining to someone what a type class X is I struggle to find good examples of data structures which are exactly X. So, I request examples for: A type constructor which is not a Functor. A type constructor which is a Functor, but not Applicative. A type constructor which is an Applicative, but is not a Monad. A type constructor which is a Monad. I think there are plenty examples of Monad everywhere, but a good example of Monad with some relation to previous examples could complete the
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