Re: [Haskell-beginners] Hints on how to remove unsafePerformIO from my function?

Well, I thought about that, but I can't find anything in the network.socket library that would allow me to see if there is any data ready to be read from. But besides that, wouldn't that cause a busy wait as the thread constantly failed to read and yielded over and over again until there was a new packet? I'd rather if this thread chilled out until something worthwhile occurs. On Mon, Jul 19, 2010 at 10:12 AM, Benjamin Edwards wrote:

fork it into a thread and have it run on a loop, yielding if it can't read any data from the udp socket?

On 19 July 2010 16:07, David McBride wrote:

...

I am writing a voip server to take advantage of haskells awesome threading and parsing ability. At first everything was going great, but my thread that fetches udp makes use of unsafePerformIO:

fetchUDPSIP :: TChan B.ByteString -> IO () fetchUDPSIP chan = do sock <- getUDPSocket 5060 let results = (unstrict . repeat . getUDP) sock mapM_ (atomically . writeTChan chan) results where unstrict [] = [] unstrict (x:xs) = unsafePerformIO x:unstrict xs

It fetches it from a socket, and then writes it to a TChan. This results in a stream of bytestrings that another thread can read from. If I don't use unsafePerformIO, then it tries to read all possible packets before returning anything, and so it never writes to the TChan at all. The problem with doing it this way is that unsafePerformIO apparently stops every other thread from doing anything while it is waiting for a packet.

But I can't think of a way to rewrite this function to do what I want. I'm kind of new to this, does anyone have any hints that could help me out?

Here's a simple version without any of the implementation details:

fetchLine = do let results = (unstrict . repeat) getLine mapM_ putStrLn results where unstrict [] = [] unstrict (x:xs) = unsafePerformIO x:unstrict xs

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