applicative

问题 I think I found a flaw in the hackage article for Control.Applicative. As a description of the applicative functor laws, it says: class Functor f => Applicative f where A functor with application, providing operations to embed pure expressions ( pure ), and sequence computations and combine their results ( <*> ). A minimal complete definition must include implementations of these functions satisfying the following laws: identity pure id <*> v = v composition pure (.) <*> u <*> v <*> w = u <*>

问题 here's some xml i'm parsing: <?xml version="1.0" encoding="utf-8"?> <data> <row ows_Document='Weekly Report 10.21.2020' ows_Category='Weekly Report'/> <row ows_Document='Daily Update 10.20.2020' ows_Category='Daily Update'/> <row ows_Document='Weekly Report 10.14.2020' ows_Category='Weekly Report'/> <row ows_Document='Weekly Report 10.07.2020' ows_Category='Weekly Report'/> <row ows_Document='Spanish: Reporte Semanal 07.10.2020' ows_Category='Weekly Report'/> </data> i've been trying to

来源： https://stackoverflow.com/questions/64177058/how-does-sequencea-work-on-lists-of-pairs

来源： https://stackoverflow.com/questions/64177058/how-does-sequencea-work-on-lists-of-pairs

来源： https://stackoverflow.com/questions/59869399/why-does-mutual-yielding-make-arrowapply-and-monads-equivalent-unlike-arrow-and

来源： https://stackoverflow.com/questions/59869399/why-does-mutual-yielding-make-arrowapply-and-monads-equivalent-unlike-arrow-and

来源： https://stackoverflow.com/questions/62949200/why-are-monoidal-and-applicative-laws-telling-us-the-same-thing

问题 The bounty expires tomorrow . Answers to this question are eligible for a +500 reputation bounty. Damián Rafael Lattenero is looking for a canonical answer : Maybe it could be a generalisation of arity in methods requiring multiple typed arguments. Two Main Problems to solve: 1) Type check is lost Using the array argument Single.zip() version I lose the strongly typed arguments. 2) Source argument Cannot be Nullable I cannot send nullable source values as argument of Single.zip() function 3)

问题 It is widely known that Applicative generalises Arrows . In the Idioms are oblivious, arrows are meticulous, monads are promiscuous paper by Sam Lindley, Philip Wadler and Jeremy Yallop it is said that Applicative is equivalent to Static Arrows, that is arrows for which the following isomorphism holds: arr a b :<->: arr () (a -> b) As far as I can understand, it could be illustrated the following way: Note: newtype Identity a = Id { runId :: a } . Klesli Identity is a Static Arrow as it wraps

问题 class Applicative f => Monad f where return :: a -> f a (>>=) :: f a -> (a -> f b) -> f b (<*>) can be derived from pure and (>>=) : fs <*> as = fs >>= (\f -> as >>= (\a -> pure (f a))) For the line fs >>= (\f -> as >>= (\a -> pure (f a))) I am confused by the usage of >>= . I think it takes a functor f a and a function, then return another functor f b . But in this expression, I feel lost. 回答1: Lets start with the type we're implementing: (<*>) :: Monad f => f (a -> b) -> f a -> f b (The