[Haskell-beginners] MVar blocked or not?

[Adrian Adshead]

Hi everyone,

I have a beginer question on using MVar.

In the simple case I am getting what I expect...


***CODE***
module Main ( main ) where

import Control.Concurrent
import System.IO

main = do
  mv <- newEmptyMVar
  putMVar mv 'a'
  threadDelay (5 * 10^6)
  putMVar mv 'b'
  threadDelay (5 * 10^6)
***/CODE***


Output after the first 5 second delay is :-

Demo.exe: thread blocked indefinitely



It even works as expected when there is a reading thread

***CODE***
module Main ( main ) where

import Control.Concurrent
import Control.Monad
import System.IO

main = do
  mv <- newEmptyMVar
  forkIO (readMV mv)
  putMVar mv 'a'
  threadDelay (5 * 10^6)
  putMVar mv 'b'
  threadDelay (5 * 10^6)

readMV mv = forever $ do
  ch <- takeMVar mv
  putChar ch >> hFlush stdout
***/CODE***

Output including expected delays :-
ab



Then it starts to act unexpectedly

***CODE***
module Main ( main ) where

import Control.Concurrent
import Control.Monad
import System.IO

main = do
  mv <- newEmptyMVar
  forkIO (readMV mv)
  putMVar mv 'a'
  threadDelay (5 * 10^6)
  putMVar mv 'b'
  threadDelay (5 * 10^6)

readMV mv = forever $ do
  ch <- modifyMVar mv modMV
  putChar ch >> hFlush stdout
  threadDelay (9*10^5)

modMV 'a' = return ('c','a')
modMV 'c' = return ('a','c')
modMV x = return (x,x)
***/CODE***

Output is :-
acacac


The problem is that the output stops after 5 seconds
and the program exits after 10 seconds. But why?

Questions:-

1. Why does the output stop? (Shouldn't the reader thread
continue even if the main thread blocks on the second putMVar?)

2. Why does it exit? (Shouldn't it be blocked at the second
putMVar? The readMV thread now only modifies the MVar so it
should never be empty)

3. Is there a race condition using modifyMVar while there
is another thread blocked in a putMVar? (I am assuming the
second putMVar succeeded, but where did the 'b' go?)

Many thanks for any help

Adrian.


      
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