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我正在使用管道生态系统编写流功能，特别是管道并发，它基于操作库，允许我快速制作小程序片段，我通过网络向服务器或标准输入/输出发出命令一个shell命令，然后读回响应。在这种情况下，它是星号，但可以概括为任何类似的东西。

我最初写这个时考虑到管道，但它不起作用。以下代码不起作用的原因是 astPipe 返回 a Pipe _ _ IO a，而来自管道并发的 i 和 o 都返回Consumer/Producer _ IO ()。我想过让astPipeyield Maybe ByteString，然后让输出Consumer消耗Maybe ByteString，但这仍然不能解决退货的Producer问题()。

我觉得我真的很接近解决方案，但我无法完全解决。您应该能够在此文件上运行堆栈以进行复制。

#!/usr/bin/env stack
-- stack --resolver lts-6.20 runghc --package pipes  --package pipes-concurrency  --package operational --package process-streaming

{-# LANGUAGE OverloadedStrings, LambdaCase #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE GADTs #-}
module West.Asterisk where

import System.Process.Streaming as PS
import Control.Monad.Operational as Op

import Pipes as P
import Pipes.Concurrent as PC;

import qualified Data.ByteString.Char8 as B

import Control.Concurrent.Async

import GHC.IO.Exception (ExitCode)

data Version = Version String
data Channels = Channels

data AsteriskInstruction a where
    Login :: AsteriskInstruction (Maybe Version)
    CoreShowChannels :: AsteriskInstruction (Maybe Channels)

type Asterisk a = Program AsteriskInstruction a

runAsterisk :: forall a. Asterisk a -> IO a
runAsterisk m = 
  let

    runAsterisk' :: Producer B.ByteString {- TODO Response -} IO () -> Consumer B.ByteString IO () -> Asterisk a -> IO a
    runAsterisk' i o m' = runEffect $ i >-> astPipe m' >-> o
      where
        astPipe :: Asterisk a -> Pipe B.ByteString B.ByteString IO a
        astPipe k = 
          case Op.view m' of

            Return a -> return a

            Login :>>= k -> do
              yield logincmd
              resp <- await -- :: Response
              let v = undefined resp :: Maybe Version
              astPipe (k v)

            CoreShowChannels :>>= k -> do
              yield coreshowchannelscmd
              resp <- await
              let c = undefined resp :: Maybe Channels
              astPipe (k c)

  in do
    withSpawn unbounded $ \(out1, in1) -> do
        async $ asteriskManager (fromInput in1) (toOutput out1)
        runAsterisk' (fromInput in1) (toOutput out1) m 

asteriskManager :: Producer B.ByteString IO () -> Consumer B.ByteString IO () -> IO ExitCode
asteriskManager prod cons = do
  let ssh = shell "nc someserver 5038"
  execute (piped ssh) (foldOut (withConsumer cons) *> feedProducer prod *> exitCode)


logincmd, coreshowchannelscmd :: B.ByteString
logincmd = "action: login\nusername: username\nsecret: pass\nevents: off\n\n"
coreshowchannelscmd = "action: coreshowchannels\n\n"

错误：

  Blah.hs:38:45:
    Couldn't match type ‘a’ with ‘()’
      ‘a’ is a rigid type variable bound by
          the type signature for runAsterisk :: Asterisk a -> IO a
          at Blah.hs:33:23
    Expected type: Proxy () B.ByteString () B.ByteString IO ()
      Actual type: Pipe B.ByteString B.ByteString IO a
    Relevant bindings include
      astPipe :: Asterisk a -> Pipe B.ByteString B.ByteString IO a
        (bound at Blah.hs:41:9)
      m' :: Asterisk a (bound at Blah.hs:38:22)
      runAsterisk' :: Producer B.ByteString IO ()
                      -> Consumer B.ByteString IO () -> Asterisk a -> IO a
        (bound at Blah.hs:38:5)
      m :: Asterisk a (bound at Blah.hs:34:13)
      runAsterisk :: Asterisk a -> IO a (bound at Blah.hs:34:1)
    In the second argument of ‘(>->)’, namely ‘astPipe m'’
    In the first argument of ‘(>->)’, namely ‘i >-> astPipe m'’
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1 回答 1

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Producer返回的s和Consumers()可以自行停止。Producer在返回类型上具有多态性的 s 和Consumers 永远不会自行停止。

要在您的案例中统一返回类型，请将每个返回类型放在 an 的不同分支中Either，使用fmap.

runAsterisk' :: Producer B.ByteString IO () 
             -> Consumer B.ByteString IO () 
             -> Asterisk a 
             -> IO (Either () a)
runAsterisk' i o m' = runEffect $ fmap Left i >-> fmap Right (astPipe m') >-> fmap Left o

上的模式匹配Either将揭示哪个组件停止了管道。

此外，您可以使用drain将 aConsumer a IO ()转换为永远不会自行停止的使用者：

neverStop :: Consumer a IO () -> Consumer a IO r
neverStop consumer = consumer *> drain

原始消费者停止后收到的所有输入都将被丢弃。

于 2016-10-05T00:01:58.040 回答