问题
I am trying to use channels/STM to implement message passing in Haskell. Maybe this is a terrible idea, and there is a better way to implement/use message passing in Haskell. If this is the case, do let me know; however, my quest has opened some basic questions on concurrent Haskell.
I have heard great things about STM, and in particular the implementation in Haskell. Since it supports reading to and writing from, and has some safety benefits, I figured one would start there. This brings up my biggest question: does
msg <- atomically $ readTChan chan
where chan is a TChan Int, cause a wait that waits for the channel to have a value on it?
Consider the following program:
p chan = do
atomically $ writeTChan chan 1
atomically $ writeTChan chan 2
q chan = do
msg1 <- atomically $ readTChan chan
msg2 <- atomically $ readTChan chan
-- for testing purposes
putStrLn $ show msg1
putStrLn $ show msg2
main = do
chan <- atomically $ newTChan
p chan
q chan
Compile this with ghc --make -threaded, and then run the program, and indeed you get 1 followed by 2 printed to console. Now, suppose we do
main = do
chan <- atomically $ newTChan
forkIO $ p chan
forkIO $ q chan
instead. Now, if we use - threaded, it will either print nothing, 1, or 1 followed by 2 to the terminal; however, if you don't compile with -threaded it always prints 1 followed by 2. Question 2: what is the difference between -threaded and not? I imagine that they aren't really running as concurrent things, and they are just run one after the other. This is consistent with what follows.
Now, in my thinking if I had p and q running concurrently; i.e. I forkIO'd them, they should be able to run in the opposite order. Supposing
main = do
chan <- atomically newTChan
forkIO $ q chan
forkIO $ p chan
Now, if I compile this without -threaded, I never get anything printed to console. If I compile with -threaded, I sometimes do. Although, it is very rare to get 1 followed by 2 -- usually just 1 or nothing. I tried this with Control.Concurrent.Chan as well, and got consistent results.
Second big question: how do channels and fork play together, and what is going on in the above program?
At any rate, it seems that I can't so naively simulate message passing with STM. Perhaps Cloud Haskell is an option that solves these problems -- I really don't know. Any information on how to get message passing going short of serialize ~~> write to socket ~~> read from socket ~~> deserialize would be hugely appreciated.
回答1:
No your idea is right - this is kindof what TChans are for - you just missed a minor point of forkIO:
The problem is that your main thread will not wait for the termination of the threads created with forkIO (see here for reference)
so if I use the hint given in the reference:
import Control.Concurrent
import Control.Concurrent.STM
p :: Num a => TChan a -> IO ()
p chan = do
atomically $ writeTChan chan 1
atomically $ writeTChan chan 2
q chan = do
msg1 <- atomically $ readTChan chan
msg2 <- atomically $ readTChan chan
-- for testing purposes
putStrLn $ show msg1
putStrLn $ show msg2
main :: IO ()
main = do
children <- newMVar []
chan <- atomically $ newTChan
_ <- forkChild children $ p chan
_ <- forkChild children $ q chan
waitForChildren children
return ()
waitForChildren :: MVar [MVar ()] -> IO ()
waitForChildren children = do
cs <- takeMVar children
case cs of
[] -> return ()
m:ms -> do
putMVar children ms
takeMVar m
waitForChildren children
forkChild :: MVar [MVar ()] -> IO () -> IO ThreadId
forkChild children io = do
mvar <- newEmptyMVar
childs <- takeMVar children
putMVar children (mvar:childs)
forkFinally io (\_ -> putMVar mvar ())
it works as expected:
d:/Temp $ ghc --make -threaded tchan.hs
[1 of 1] Compiling Main ( tchan.hs, tchan.o )
Linking tchan.exe ...
d:/Temp $ ./tchan.exe
1
2
d:/Temp $
and of course it will continue to work if you switch the calls to p and q too
来源：https://stackoverflow.com/questions/30340159/haskell-channels-stm-threaded-message-passing