Re: [Haskell-cafe] Timeouts that don't cause data growth.

23 Mar 2010


      Actually this isn't good enough either as I'm potentially leaving the
"action" thread in a state where it never times out...  I guess I have to do
all thread killing in the main thread.

On Tue, Mar 23, 2010 at 12:23 PM, David Leimbach  wrote:
...
Is this just a problem of spawning too many forkIO resources that never
produce a result?
I was thinking of trying something like the following in System.Timeout's
place:
...
module Main where
...
import Control.Concurrent.MVar
import Control.Concurrent
import Data.Maybe
...
timeout :: Int -> IO a -> IO (Maybe a)
timeout time action = do
  someMVar <- newEmptyMVar   -- MVar is a Maybe
  timeoutThread <- forkIO $ nothingIzer time someMVar
  forkIO $ actionRunner action someMVar timeoutThread
  takeMVar someMVar >>= return
    where
      nothingIzer time mvar = threadDelay time >> putMVar mvar Nothing
      actionRunner action mvar timeoutThread = do
                        res <- action
                        killThread timeoutThread
                       putMVar mvar $ Just res
...
main :: IO ()
main = do
 res <- timeout (5 * 10 ^ 6) (getLine >>= putStrLn)
 case res of
    Nothing -> putStrLn "Timeout"
    Just x -> putStrLn "Success"
On Tue, Mar 23, 2010 at 11:31 AM, Roel van Dijk wrote:
...
I tried a few things. First I added another timeout to main, so the
program kills itself after a few seconds.
doit :: IO (Maybe ())
doit = timeout 12000000 $ {- yield >> -} return ()
main :: IO ()
main = do _ <- timeout 5000000 $ forever doit
         return ()
This program failed to terminate. But when I compiled -with threaded
and added a yield to doit, it worked (kinda). If the timeout in doit
is not too long, like 200 milliseconds, the program has constant space
usage. But when I increased the timeout in doit to 12 seconds I got a
stack overflow.
I'll investigate further when I have more time.
Regards,
Roel