Alberto G. Corona wrote:
however, It happens that fails in my windows box with ghc 6.10.1 , single core
here is the code and the results:
-----------begin code: module Main where
import Control.Concurrent.STM
import Control.Concurrent import System.IO.Unsafe import GHC.Conc
mtxs= unsafePerformIO $ mapM newTMVarIO $ take 5 $ repeat 0
proc i= atomically $ do unsafeIOToSTM $ putStr $ "init of process "++ show i++"\n"
As Sterling points out, unsafeIOToSTM is really unsafe. A fundamental restriction is that the IO action used with unsafeIOToSTM may not block. However, putStr may block. It's actually possible to do the logging without blocking:
cut here >>> module Main where
import Control.Concurrent import Control.Concurrent.STM import GHC.Conc import Control.Concurrent.MVar import Data.IORef import Control.Monad data Logger = Logger (IORef [String]) (MVar ()) newLogger :: IO Logger newLogger = do ref <- newIORef [] wake <- newEmptyMVar return $ Logger ref wake logLogger :: Logger -> String -> IO () logLogger (Logger ref wake) msg = do atomicModifyIORef ref $ \msgs -> (msg:msgs, ()) tryPutMVar wake () return () dumpLogger :: Logger -> IO () dumpLogger (Logger ref wake) = forever $ do takeMVar wake msgs <- atomicModifyIORef ref $ \msgs -> ([], msgs) putStr $ unlines . reverse $ msgs proc log mtxs i = do let logSTM = unsafeIOToSTM . log xs' <- atomically $ do logSTM $ "init of process " ++ show i xs <- mapM takeTMVar mtxs mapM (\(mtx, x) -> putTMVar mtx (x+1)) $ zip mtxs xs xs' <- mapM readTMVar mtxs logSTM $ "End of processs " ++ show i ++ " result = " ++ show xs' return xs' log $ "Final result of process " ++ show i ++ " = " ++ show xs' main = do log <- newLogger forkIO $ dumpLogger log mtxs <- replicateM 5 $ newTMVarIO 0 mapM (forkIO . proc (logLogger log) mtxs) [1..5] threadDelay 1000000 <<< And that gives reasonable results, for example: init of process 1 End of processs 1 result= [1,1,1,1,1] Final result of process 1 = [1,1,1,1,1] init of process 2 End of processs 2 result= [2,2,2,2,2] Final result of process 2 = [2,2,2,2,2] init of process 3 init of process 4 End of processs 4 result= [3,3,3,3,3] Final result of process 4 = [3,3,3,3,3] init of process 5 End of processs 5 result= [4,4,4,4,4] Final result of process 5 = [4,4,4,4,4] End of processs 3 result= [3,3,3,3,3] init of process 3 End of processs 3 result= [5,5,5,5,5] Final result of process 3 = [5,5,5,5,5] HTH, Bertram