Haskell Language Communicating between Threads with `MVar`


Example

It is very easy to pass information between threads using the MVar a type and its accompanying functions in Control.Concurrent:

  • newEmptyMVar :: IO (MVar a) -- creates a new MVar a
  • newMVar :: a -> IO (MVar a) -- creates a new MVar with the given value
  • takeMVar :: MVar a -> IO a -- retrieves the value from the given MVar, or blocks until one is available
  • putMVar :: MVar a -> a -> IO () -- puts the given value in the MVar, or blocks until it's empty

Let's sum the numbers from 1 to 100 million in a thread and wait on the result:

import Control.Concurrent
main = do
  m <- newEmptyMVar
  forkIO $ putMVar m $ sum [1..10000000]
  print =<< takeMVar m  -- takeMVar will block 'til m is non-empty!

A more complex demonstration might be to take user input and sum in the background while waiting for more input:

main2 = loop
  where 
    loop = do
        m <- newEmptyMVar
        n <- getLine
        putStrLn "Calculating. Please wait"
        -- In another thread, parse the user input and sum
        forkIO $ putMVar m $ sum [1..(read n :: Int)]
        -- In another thread, wait 'til the sum's complete then print it
        forkIO $ print =<< takeMVar m
        loop

As stated earlier, if you call takeMVar and the MVar is empty, it blocks until another thread puts something into the MVar, which could result in a Dining Philosophers Problem. The same thing happens with putMVar: if it's full, it'll block 'til it's empty!

Take the following function:

concurrent ma mb = do
  a <- takeMVar ma
  b <- takeMVar mb
  putMVar ma a
  putMVar mb b

We run the the two functions with some MVars

concurrent ma mb     -- new thread 1 
concurrent mb ma     -- new thread 2

What could happen is that:

  1. Thread 1 reads ma and blocks ma
  2. Thread 2 reads mb and thus blocks mb

Now Thread 1 cannot read mb as Thread 2 has blocked it, and Thread 2 cannot read ma as Thread 1 has blocked it. A classic deadlock!