On Wed, Apr 14, 2010 at 2:13 PM, Gregory Collins <[email protected]>wrote:
> Jesper Louis Andersen <[email protected]> writes: > > > This post describes some odd behaviour I have seen in GHC 6.12.1 when > writing > > Combinatorrent. The post is literate Haskell so you can run it. The > executive > > summary: A space leak occurs when a new process is spawned from inside > another > > process - and I can't figure out why. I am asking for help on > haskell-cafe. > > > > ...[snip]... > >> > >> import Control.Monad.State > > Does the problem go away if you use "Control.Monad.State.Strict"? > Nope :) That was the first thing I tried here. I tried playing with optimization level too. Next I tried making two versions that were as similar as possible and then comparing the core with ghc-core. I can't see a difference between a version that uses 1MB and a version that uses 160MB (on my system 160MB is the worst I can get it to blow up). The two versions I compared: Low memory: \begin{code} > startp4 :: IO ThreadId > startp4 = spawn () () (return ()) > startp3 :: IO ThreadId > startp3 = spawn () () (forever $ > do liftIO startp4 > liftIO $ putStrLn "Delaying" > liftIO $ threadDelay (3 * 1000000)) > main1 = do > putStrLn "Main thread starting" > startp3 > threadDelay (1 * 1000000) > > main = main1 \end{code} Too much memory: \begin{code} > startp4 :: IO ThreadId > startp4 = spawn () () (forever $ return ()) > startp3 :: IO ThreadId > startp3 = spawn () () (forever $ > do liftIO startp4 > liftIO $ putStrLn "Delaying" > liftIO $ threadDelay (3 * 1000000)) > main1 = do > putStrLn "Main thread starting" > startp3 > threadDelay (1 * 1000000) > > main = main1 \end{code} The difference is whether or not the threads must keep returning () or if they returns it once. I'm not sure what to make of it. My conclusion is that keeping the thread alive via forever is the problem, but when I test this hypothesis with a threadDelay the space leak goes away: \begin{code} > startp4 :: IO ThreadId > startp4 = spawn () () (liftIO $ threadDelay (100 * 1000000)) > startp3 :: IO ThreadId > startp3 = spawn () () (forever $ > do liftIO startp4 > liftIO $ putStrLn "Delaying" > liftIO $ threadDelay (3 * 1000000)) > main1 = do > putStrLn "Main thread starting" > startp3 > threadDelay (1 * 1000000) > > main = main1 \end{code} It will be interesting to hear what fixes this! Jason
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