Broken beyond repair: Control.Concurrent.SampleVar

[Chris Kuklewicz]

and I am dumb, sorry.  Yes, you are right.  In fact the code I had posted on the
wiki at http://haskell.org/haskellwiki/SafeConcurrent#SampleVar is

> writeSampleVar :: SampleVar a -> a -> IO ()
> writeSampleVar svar value = do
>   withMVar (lockedStore svar) $ \ store -> do
>     _ <- tryTakeMVar store
>     putMVar store value

which is functionally identical to yours.  I should not write so hastily on my
way out the door...


Felipe Lessa wrote:
> On Sun, Apr 12, 2009 at 05:12:15PM +0100, Chris Kuklewicz wrote:
>> Felipe Lessa wrote:
>>>> writeSampleVar :: SampleVar a -> a -> IO ()
>>>> writeSampleVar s x = block $ withMVar (svLock s) $ const $ do
>>>>   tryTakeMVar (svData s)
>>>>   putMVar (svData s) x
>> That is obvious, but 'block $' is not atomic.
>>
>> Threads 1 and 2 get past the tryTakeMVar.
> 
> That's why there is a "withMVar (svLock s)", the "const $ do"
> part is executed serially. It is impossible to have two threads
> past the "tryTakeMVar".
> 
>> Then thread 1 succeeds with putMVar and thread 2 blocks.
>>
>> Thread 2 is not supposed to block, so this is a failure to agree with the
>> specification.
> 
> Well, it doesn't block. In the worst case I can imagine a thread
> could be killed after the 'tryTakeMVar', leaving the 'SampleVar'
> empty, but I guess the 'block' would prevent that.
> 
> --
> Felipe.