Re: [Haskell-beginners] how does hgearman-worker work?

6 Apr 2017


      Thank you very much, David.
...
If you want to run it from within StateT GearmanClient IO, you must
use liftIO.
The execution of the worker implementation below shows the ThreadId but the worker doesn't grab any job from gearmand as expected. GRAB_JOB, wich sends gmLoop (https://github.com/jperson/hgearman-client/blob/master/Network/Gearman/Worke...), appears in gearmand logs but the worker close the connection before gearmand sends GEARMAN_COMMAND_JOB_ASSIGN replay. It looks like the worker does not execute gmWait.


{-# LANGUAGE LambdaCase #-}

import qualified Control.Monad.State as S
import qualified Data.ByteString.Char8 as B
import qualified Network.Gearman.Client as C
import qualified Network.Gearman.Worker as W
import Network.Gearman.Internal (Function, Port)
import Network.Socket (HostName)
import GHC.Conc.Sync (ThreadId)

main :: IO ()
main = do
  work >>= \ case
    Nothing -> putStrLn "nothing"
    Just t -> putStrLn $ show t
  return ()

work :: IO (Maybe ThreadId)
work = do
  connect >>= \case
    Left e -> error $ B.unpack e
    Right gc -> do
      (res, _) <- flip S.runStateT gc $ do
          g <- W.registerWorker ((B.pack "foo")::Function)  (\_ -> B.pack "bar")
          t <- S.liftIO $ W.runWorker gc (return g)
          return $ Just t
      return res
  where
    connect = C.connectGearman (B.pack "worker-id-123") ("localhost"::HostName)  (4730::Port)
...
This is just a guess based on what I know about gearman and that
particular api choice.  He may have intended you to use runWorker
outside of the setup phase.  He certainly doesn't prevent it.
someprocedure' :: IO ()
someprocedure' = do
  gs <- connectGearman >>= \case
    Left e -> return []
    Right gc -> do
      (res, _) <- flip runStateT gc $ do
        g <- registerWorker undefined undefined
        g2 <- registerWorker undefined undefined
        return $ [g,g2]
      return res
mapM_ (\g -> runWorker g (return ())) gs
I'm not sure it could work in this way because runWorker :: GearmanClient -> Gearman () -> IO ThreadId and connectGearman result is of type IO (Either GearmanError GearmanClient)

Best regards,
Alexei
...
On 06 April 2017 at 19:54 David McBride  wrote:
There are a couple problems.  One is that runWorker has a type of IO
ThreadId.  I have no idea why he would write it that way in his API.
If you want to run it from within StateT GearmanClient IO, you must
use liftIO.
liftIO :: (MonadIO m) => IO a -> StateT s IO
instance MonadIO (StateT s IO) where
  liftIO :: IO a -> StateT s IO a
liftIO $ runWorker gc whatever.
When you are working in monadic code, you connect monadic components
based on their types.  If you are a procedure
someprocedure :: IO ???
Then every statement you used must some form of ???.  runWorker
returns (IO ThreadId), return () returns (IO ()), return res returns
IO (whatever type res is).  I'm not sure what you intend to do with
the threadId, save it or ignore it, but you might try something like
this.
someprocedure' :: IO (Maybe ThreadId)
someprocedure' = do
  connectGearman >>= \case
    Left e -> return Nothing
    Right gc -> do
      (res, _) <- flip runStateT gc $ do
        g <- registerWorker undefined undefined
        t <- liftIO $ runWorker gc undefined
        return $ Just t
      return res
This is just a guess based on what I know about gearman and that
particular api choice.  He may have intended you to use runWorker
outside of the setup phase.  He certainly doesn't prevent it.
someprocedure' :: IO ()
someprocedure' = do
  gs <- connectGearman >>= \case
    Left e -> return []
    Right gc -> do
      (res, _) <- flip runStateT gc $ do
        g <- registerWorker undefined undefined
        g2 <- registerWorker undefined undefined
        return $ [g,g2]
      return res
mapM_ (\g -> runWorker g (return ())) gs
On Thu, Apr 6, 2017 at 11:37 AM,   wrote:
...
A while ago I asked similar question about hgearman client. With help I got in the List (https://mail.haskell.org/pipermail/beginners/2017-March/017435.html) and I implemented a gearman client in Haskell. (here the implementation http://stackoverflow.com/questions/42774191/how-does-hgearman-client-work)
Unfortunately I need again some help be implementation of gearman worker.
I post here only the snippet with the badly implemented code in hope to find again some help. (Complete implementation: http://stackoverflow.com/questions/43155857/how-does-hgearman-worker-work)
Right gc -> do
  (res, _) <- flip S.runStateT gc $ do
    g <- (W.registerWorker name func)
    t <- W.runWorker gc (return ())
    return t >> return  ()
return res
This throws exception:
Couldn't match expected type `S.StateT
                                Network.Gearman.Internal.GearmanClient IO a0'
                with actual type `IO GHC.Conc.Sync.ThreadId'
    In a stmt of a 'do' block: t <- W.runWorker gc (return ())
    In the second argument of `($)', namely
      `do { g <- (W.registerWorker name func);
            t <- W.runWorker gc (return ());
            return t >> return () }
What do I wrong with W.runWorker gc (return ())?
runWorker :: GearmanClient -> Gearman () -> IO ThreadId
https://hackage.haskell.org/package/hgearman-0.1.0.2/docs/Network-Gearman-Wo...
Best regards,
Alexei
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Re: [Haskell-beginners] how does hgearman-worker work?

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