在下面的代码中我该怎么做
-
更改
stdoutCharConsumer
,以便在打印输入流中的所有数据后打印换行符 -
实施
mixin
和mixin'
没有进入Pipes.Internal?可能吗?我需要像 生产环境 者的 next
函数 .
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: 现在,在我阅读了基于拉/推的流之后,我认为即使使用Pipes.Internal也是不可能的 . 这是真的吗?
1 回答
Consumers
和Pipes
都不知道上游输入结束 . 你需要pipes-parse
来自pipes-parse
.与
Consumer
相比,Parser
更直接了解Producer
;它们的draw函数(大致类似于await
)在找到输入结束时返回Nothing
.要运行解析器,我们调用
evalStateT
而不是runEffect
:至于
mixin
和mixin'
,我怀疑他们不可能按照预期写作 . 原因是结果Pipe
必须知道上游终止,以便知道何时产生作为参数传递的Producer
的剩余值 .