使用TVar时如何等待M_完成？

Question

I'm writing a function, where I process a list using forM_ , and append the result to a TVar list:

import Control.Concurrent.STM
import Control.Concurrent.STM.TVar
import Control.Concurrent (forkIO)
import Control.Monad (forM_)

insert :: a -> [a] -> [a]
insert = (:) -- stub

my_func_c :: (a -> a) -> [a] -> IO [a]
my_func_c my_function arr = do

    res_var <- atomically $ newTVar ([]::[a])

    forkIO $ forM_ arr $ \x -> atomically $ do
        let y = id $! my_function x
        modifyTVar res_var (insert y)

    atomically $ readTVar res_var

The result is always empty, if I compile it with -threaded . How is it possible to wait for the threads to finish? I can not use MVar or Async . I have to solve this problem using TVar , or other data structures which are based on TVar

Answer 1

The idiomatic solution would be to use TMVar s:

my_func_c :: (a -> a) -> [a] -> IO [a]
my_func_c my_function arr = do
    res_var <- atomically $ newTVar []
    finished <- atomically $ newEmptyTMVar

    forkIO $ do
        forM_ arr $ \x -> atomically $ do
            let y = id $! my_function x
            modifyTVar res_var (insert y)
        atomically $ putTMVar finished ()

    atomically $ takeTMVar finished >> readTVar res_var

but if you are really only allowed to use TVar s, you can simulate a TMVar () with a TVar Bool :

my_func_c :: (a -> a) -> [a] -> IO [a]
my_func_c my_function arr = do
    res_var <- atomically $ newTVar []
    finished <- atomically $ newTVar False

    forkIO $ do
        forM_ arr $ \x -> atomically $ do
            let y = id $! my_function x
            modifyTVar res_var (insert y)
        atomically $ writeTVar finished True

    atomically $ waitTVar finished >> readTVar res_var

waitTVar :: TVar Bool -> STM ()
waitTVar tv = do
    finished <- readTVar tv
    unless finished retry

Under the hood, TMVar a is just a TVar (Maybe a) (so TMVar () ~ TVar (Maybe ()) ~ TVar Bool ) with takeTMVar doing a readTVar and a retry , so the above two solutions are operationally completely equivalent.