At Wed, 29 Jun 2011 21:13:47 +1000, John Ky wrote:
Hi Haskell Cafe,
I've written an echo server using just sockets:
...
When I send text to it, it will echo it back immediately after my newline.
I then modified it to user IterIO:
import Control.Concurrent import Control.Exception import Control.Monad import Control.Monad.Trans import Data.IterIO import Data.IterIO.Inum import Network import System.IO import System.IO.Error (isEOFError) import qualified Data.ByteString.Lazy as L
iterHandle' :: (MonadIO m) => Handle -> IO (Iter L.ByteString m (), Onum L.ByteString m a) iterHandle' = iterHandle
main = withSocketsDo $ do sListen <- listenOn (PortNumber 8000) putStrLn "Listening on Port 8000" forkIO $ forever $ do (sSession, hostname, port) <- accept sListen putStrLn ("Connected to " ++ hostname ++ ":" ++ show port) forkIO $ do (iter, enum) <- iterHandle' sSession enum |$ iter return () putStrLn "Press <CTRL-D> to quit." exitOnCtrlD
exitOnCtrlD = try getLine >>= either (\e -> unless (isEOFError e) $ ioError e) (const exitOnCtrlD)
It works, however it doesn't send anything back to me until end of file.
I fixed that problem with my sockets version by flushing after each line, but I don't know if IterIO will let me flush on every newline.
The buffering is actually happening in the Handle code. One way to avoid this is to change your code to call hSetBuffering as follows: hSetBuffering sSession NoBuffering (iter, enum) <- iterHandle' sSession This is mentioned in the documentation of handleI, but arguably should also be there in the docs for iterHandle. I think I'll add a mention there. If you use Sockets and call iterStream, that should also avoid the problem. David