
Well, not 50/50 but not deterministically -- I've caught
"fun1-main1-fun2" sequence,
"main1-fun1-main2" is much more often.
Since you are using -threaded, the Haskell threads are running on top
of the OS threads, so I guess it depends on the OS scheduler.
On 4 September 2012 12:27, Eugene Perederey
I added a few prints to your code
fun m = do putStrLn "fun1" putMVar m 10 putStrLn "fun2"
main = do m <- newEmptyMVar forkIO $ fun m putStrLn "main1" putMVar m 10 putStrLn "main2"
It works 50/50 for me in Mac OS X 10.6.8.
On 4 September 2012 12:16, mukesh tiwari
wrote: Hi Eugene Thank you for reply.
On Wed, Sep 5, 2012 at 12:32 AM, Eugene Perederey
wrote: Why do you think main should block more than once? I see only two possible scenarios: the fun thread puts to mvar first thus blocking main,
So at least in this case I should get thread blocked indefinitely in an MVar operation
or 10 is put into mvar in main, blocking the other thread indefinitely.
or main will execute and fun thread will die. There are 50 - 50 chance for this ( assuming both are equally likely ). I did some modification in my code and Now I am consistently getting "Concurrent: thread blocked indefinitely in an MVar operation"
import Data.List import Control.Concurrent
fun m = do putMVar m 10 return ()
main = do m <- newEmptyMVar forkIO $ fun m putStrLn "I am inside main"
putMVar m 10 return ()
[mukesh.tiwari@ Programming]$ ghc-7.4.1 -threaded -fforce-recomp Concurrent.hs [1 of 1] Compiling Main ( Concurrent.hs, Concurrent.o ) Linking Concurrent ... [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 I am inside main Concurrent: thread blocked indefinitely in an MVar operation
[mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 I am inside main Concurrent: thread blocked indefinitely in an MVar operation
[mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 I am inside main Concurrent: thread blocked indefinitely in an MVar operation
[mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 I am inside main Concurrent: thread blocked indefinitely in an MVar operation
[mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 I am inside main Concurrent: thread blocked indefinitely in an MVar operation
My question is why the outcome in first case is deterministic. Every time the code executing ( at least half the time main thread should be blocked ) .
Regards Mukesh Tiwari
On 4 September 2012 11:54, mukesh tiwari
wrote: Hello All I was going trough Real World Haskell and it says "If we try to put a value into an MVar that is already full, our thread is put to sleep until another thread takes the value out". I wrote a simple code to block main
import Data.List import Control.Concurrent
fun m = do putMVar m 10 return ()
main = do m <- newEmptyMVar forkIO $ fun m putMVar m 10 return ()
What I am expecting that main should be blocked at least couple of times but its behaving more deterministically. [mukesh.tiwari@ Programming]$ ghc-7.4.1 -threaded -fforce-recomp Concurrent.hs [1 of 1] Compiling Main ( Concurrent.hs, Concurrent.o ) Linking Concurrent ... [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$ ./Concurrent +RTS -N2 [mukesh.tiwari@ Programming]$
I am expecting to get thread blocked indefinitely on MVar at least half the time. Could some one please tell me why this deterministic behavior ? Regards Mukesh Tiwari
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-- Best, Eugene Perederey
-- Best, Eugene Perederey
-- Best, Eugene Perederey