在reddit上有一个归档线程,上面写着本质上管道/管道不能是箭头,b/c箭头必须是同步的。该线程在此处链接https://www.reddit.com/r/haskell/comments/rq1q5/conduitssinks_and_refactoring_arrows/
我不明白“同步”是从哪里来的,因为它不是箭头定义的一部分。另外,我在github https://github.com/cmahon/interactive-brokers上偶然发现了这个项目,它明确地将管道视为箭头。为了方便起见,我在这里粘贴了实例定义。这里我漏掉了什么?
-- The code in this module was provided by Gabriel Gonzalez
{-# LANGUAGE RankNTypes #-}
module Pipes.Edge where
import Control.Arrow
import Control.Category (Category((.), id))
import Control.Monad ((>=>))
import Control.Monad.Trans.State.Strict (get, put)
import Pipes
import Pipes.Core (request, respond, (\>\), (/>/), push, (>~>))
import Pipes.Internal (unsafeHoist)
import Pipes.Lift (evalStateP)
import Prelude hiding ((.), id)
newtype Edge m r a b = Edge { unEdge :: a -> Pipe a b m r }
instance (Monad m) => Category (Edge m r) where
id = Edge push
(Edge p2) . (Edge p1) = Edge (p1 >~> p2)
instance (Monad m) => Arrow (Edge m r) where
arr f = Edge (push />/ respond . f)
first (Edge p) = Edge $ \(b, d) ->
evalStateP d $ (up \>\ unsafeHoist lift . p />/ dn) b
where
up () = do
(b, d) <- request ()
lift $ put d
return b
dn c = do
d <- lift get
respond (c, d)
instance (Monad m) => ArrowChoice (Edge m r) where
left (Edge k) = Edge (bef >=> (up \>\ (k />/ dn)))
where
bef x = case x of
Left b -> return b
Right d -> do
_ <- respond (Right d)
x2 <- request ()
bef x2
up () = do
x <- request ()
bef x
dn c = respond (Left c)
runEdge :: (Monad m) => Edge m r a b -> Pipe a b m r
runEdge e = await >>= unEdge e发布于 2017-02-24 04:46:12
考虑一下这个管道:yield '*' :: Pipe x Char IO ()。我们可以将它包装在newtype PipeArrow a b = PipeArrow { getPipeArrow :: Pipe a b IO () }这样的新型适配器中,并尝试在那里定义Arrow实例。
购买如何编写一个在yield '*'上工作的first :: PipeArrow b c -> PipeArrow (b, d) (c, d)?管道从不等待来自上游的值。我们将不得不凭空产生一个d,来伴随'*'。
管道满足箭头(和ArrowChoice)的大多数法律,但是first不能以合法的方式实现。
您发布的代码没有为Pipe定义Arrow实例,而是为一个从上游获取值并返回Pipe的函数定义。
https://stackoverflow.com/questions/42418842
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