如何找到Pipe的結尾

Question

在下面的代碼中我該怎么做

• 更改stdoutCharConsumer以便在打印輸入流中的所有數據后打印換行符

• 實施mixin和mixin'

沒有進入Pipes.Internal？ 可能嗎？ 我需要像生產者的next功能一樣。

我使用Pipes 4.1.0

#!/usr/bin/env runhaskell
{-# OPTIONS_GHC -Wall #-}

import Pipes

digits, characters :: Monad m => Producer Char m ()
digits = each "0123456789"
characters = each "abcdefghijklmnopqrstuvwxyz"

interleave :: Monad m => Producer a m () -> Producer a m () -> Producer a m ()
interleave a b = do
  n <- lift $ next a
  case n of
    Left () -> b
    Right (x, a') -> do
      yield x
      interleave b a'

stdoutCharConsumer :: Consumer Char IO ()
stdoutCharConsumer = await >>= liftIO . putChar >> stdoutCharConsumer

-- first element of the mixin should go first
mixin :: Monad m => Producer b m () -> Pipe a b m ()
mixin = undefined

-- first element of the pipe should go first
mixin' :: Monad m => Producer b m () -> Pipe a b m ()
mixin' = undefined

main :: IO ()
main = do

    -- this prints "a0b1c2d3e4f5g6h7i8j9klmnopqrstuvwxyz"
    runEffect $ interleave characters digits >-> stdoutCharConsumer
    putStrLn ""

    -- this prints "0a1b2c3d4e5f6g7h8i9jklmnopqrstuvwxyz"
    runEffect $ interleave digits characters >-> stdoutCharConsumer
    putStrLn ""

    -- should print "0a1b2c3d4e5f6g7h8i9jklmnopqrstuvwxyz"
    runEffect $ characters >-> mixin digits >-> stdoutCharConsumer
    putStrLn ""

    -- should print "a1b2c3d4e5f6g7h8i9jklmnopqrstuvwxyz"
    runEffect $ digits >-> mixin characters >-> stdoutCharConsumer
    putStrLn ""

    -- should print "a1b2c3d4e5f6g7h8i9jklmnopqrstuvwxyz"
    runEffect $ characters >-> mixin' digits >-> stdoutCharConsumer
    putStrLn ""

    -- should print "0a1b2c3d4e5f6g7h8i9jklmnopqrstuvwxyz"
    runEffect $ digits >-> mixin' characters >-> stdoutCharConsumer
    putStrLn ""

UPD：現在我讀到基於拉/推的流后，我認為即使使用Pipes.Internal也是不可能的。 這是真的嗎？

Answer 1

Consumers和Pipes都不知道上游輸入結束。 你需要Parser來自pipes-parse 。

與Consumer相比， Parser對Producer更直接的了解; 它們的draw函數（大致類似於await ）在找到輸入結束時返回Nothing 。

import qualified Pipes.Parse as P

stdoutCharParser :: P.Parser Char IO ()
stdoutCharParser = P.draw >>= \ma ->
    case ma of 
        Nothing -> liftIO (putStrLn "\n")
        Just c -> liftIO (putChar c) >> stdoutCharParser

要運行解析器，我們調用evalStateT而不是runEffect ：

P.evalStateT stdoutCharParser (interleave characters digits) 

至於mixin和mixin' ，我懷疑他們不可能按照預期寫作。 原因是Pipe必須知道上游終止，以便知道何時產生作為參數傳遞的Producer的剩余值。