Hello, all. I'm experimenting with concurrent haskell using GHC 6.4. I wrote these fuctions as described in "Tackling the Awkward Squad": par_io :: IO a -> IO a -> IO a par_io t1 t2 = do c <- newEmptyMVar :: IO (MVar a) id1 <- forkIO $ wrapper c t1 id2 <- forkIO $ wrapper c t2 res <- takeMVar c killThread id1 killThread id2 return res where wrapper :: MVar a -> IO a -> IO () wrapper mvar io = do res <- io putMVar mvar res timeout :: Int -> IO a -> IO (Maybe a) timeout n t = do res <- par_io thr timer return res where thr = do res <- t return $ Just res timer = do threadDelay n return Nothing Now, then I try something like timeout 1000000 (print $ cycle "test") it behaves correctly: repeatingly prints "test" for one second and then returns Nothing. But then I try this: timeout 1000000 (print $ 2^2^2^2^2^2) it hangs for about eight seconds. Currently I'm playing with theorem-proving using resolution. So I need some technique to break a computation, if takes too long. It seems that using "timeout" there does nothing. I waited for several minutes for invalid theorem and timeout didn't expire. "Control.Concurrent" page of "Haskell Hierarchical Libraries" says, that a thread may be pre-empted whenever it allocates some memory, and that tight loops which do no allocation tend to lock out other threads. But resolution function, i wrote, hogs a *lot* of memory and (print $ 2^2^2^2^2^2) does either. So i'm curious about this behavior of timeout. So what is the correct way of running such computations in parallel? Maybe I'm missing something?