arrows

来源： https://stackoverflow.com/questions/63191868/array-aggregation-across-a-link-table-in-opaleye

问题 I tried writing down joinArr :: ??? a => a r (a r b) -> a r b . I came up with a solution which uses app , therefore narrowing the a down to ArrowApply 's: joinArr :: ArrowApply a => a r (a r b) -> a r b joinArr g = g &&& Control.Category.id >>> app Is it possible to have this function written down for arrows? My guess is no. Control.Monad.join could have been a good stand-in for >>= in the definition of the Monad type class: m >>= k = join $ k <$> m . Having joinArr :: Arrow a => a r (a r b)

问题 Description of loop from Control.Arrow : The loop operator expresses computations in which an output value is fed back as input, although the computation occurs only once. It underlies the rec value recursion construct in arrow notation. Its source code, and its instantiation for (->) : class Arrow a => ArrowLoop a where loop :: a (b,d) (c,d) -> a b c instance ArrowLoop (->) where loop f b = let (c,d) = f (b,d) in c This immediately reminds me of fix , the fixpoint combinator: fix :: (a -> a)

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 I'm trying to understand arrow notation, in particularly how it works with Monads. With Monads I can define the following: f = (*2) g = Just 5 >>= (return . f) and g is Just 10 How do I do the above but using arrow notation? 回答1: Changing your Monad thinking to Arrow thinking The first step to translating into Arrow is to move from thinking about m b on its own to thinking about a -> m b . With a monad, you'd write use x = do ..... .... doThis = do .... ... thing = doThis >>= use whereas an

问题 I'm trying to understand arrow notation, in particularly how it works with Monads. With Monads I can define the following: f = (*2) g = Just 5 >>= (return . f) and g is Just 10 How do I do the above but using arrow notation? 回答1: Changing your Monad thinking to Arrow thinking The first step to translating into Arrow is to move from thinking about m b on its own to thinking about a -> m b . With a monad, you'd write use x = do ..... .... doThis = do .... ... thing = doThis >>= use whereas an

问题 I'm trying to grok & work with Arrows, and am having some difficulty. I have a context where I need an Arrow [a] [b] , and I want to write an Arrow a b and map/sequence it inside the arrow, a la mapM . Specifically, the arrow is a Hakyll Compiler , but I don't think that matters much for the answer. Given an arrow myInnerArrow :: Arrow a => a b c How can I lift this into an arrow myOuterArrow :: Arrow a => a [b] [c] ? I have scoured the base library, particularly in Data.List and Control

问题 I'm looking to replace all title elements with h1 , h2 , ... , h6 elements depending on how many ancestors are section elements. Example input/output: Input.xml <document> <section> <title>Title A</title> <section> <title>Title B</title> </section> <section> <title>Title C</title> <section> <title>Title D</title> </section> </section> </section> </document> Output.xml <document> <section> <h1>Title A</h1> <section> <h2>Title B</h2> </section> <section> <h2>Title C</h2> <section> <h3>Title D<

问题 I'm looking to replace all title elements with h1 , h2 , ... , h6 elements depending on how many ancestors are section elements. Example input/output: Input.xml <document> <section> <title>Title A</title> <section> <title>Title B</title> </section> <section> <title>Title C</title> <section> <title>Title D</title> </section> </section> </section> </document> Output.xml <document> <section> <h1>Title A</h1> <section> <h2>Title B</h2> </section> <section> <h2>Title C</h2> <section> <h3>Title D<

问题 Apparently, every Arrow is a Strong profunctor. Indeed ^>> and >>^ correspond to lmap and rmap . And first' and second' are just the same as first and second . Similarly every ArrowChoice is also Choice. What profunctors lack compared to arrows is the ability to compose them. If we add composition, will we get an arrow? In other words, if a (strong) profunctor is also a category, is it already an arrow? If not, what's missing? 回答1: What profunctors lack compared to arrows is the ability to