Hi ..
Thanks ..
But I keep get error when I run the program in order to see the result . It
says 'var' is not in the scope.
Mozhgan
Date: Tue, 16 Dec 2008 05:54:40 -0700
From: [email protected]
To: [email protected]
Subject: Re: [Haskell-cafe] Re: MVar and Par ..
CC: [email protected]; [email protected]
Okay, that's a bit clearer. Control.Parallel is not what you want; that is for
parallelizing pure code (and is very nice inside its little domain). But you
want concurrent code: multiple threads that change state at the same time.
That is in Control.Concurrent.
In particular, the "shared variable" is Data.IORef.
Simple example from which you should be able to extrapolate.
import Control.Concurrent
import Control.Concurrent.MVar
import Data.IORef
threadBody = do
x <- readIORef var
writeIORef var (x+1)
main = do
var <- newIORef 0
-- spawn two threads
forkIO threadBody
forkIO threadBody
-- look at the value of the variable
x <- readIORef var
print x
You can scatter threadDelay throughout this code to mess with the scheduler and
show how a failure could happen.
Conversion to MVars should be pretty straightforward.
Luke
2008/12/16 Mozhgan Kabiri <[email protected]>
Hi ,
Yeah , first I want to get unexpected answer ! And then I want to prove that
when I use MVar I will get the write answer.I want to show that when we have a
shared variable, we should use MVar.Besides we use for example two processors
to do different tasks.But the point is these processors have to share a
variable and I want to show it with MVar.But I don't know how to write a simple
code to show it while undrestand it.
Mozhgan
> To: [email protected]
> From: [email protected]
> Date: Tue, 16 Dec 2008 12:34:23 +0000
> Subject: [Haskell-cafe] Re: MVar and Par ..
>
> Mozhgan Kabiri wrote:
> > Hi .. Hope you are doing well . I've just joined this group.
>
> Hi.
>
> > Recently, I am struggling to do some simple experiment with haskell
> > language about parallelism and wrong answers that we can get while using
> > a shared variable .
>
> Your goal is still unclear.
> Are you trying to create an example which shows unexpected answers?
> Are you trying to create an example which shows the expected answer?
>
> > I tried to write a simple program, for example calculationg 'n=n+1' few
> > times.And then I tried to do it in parallel by using 'par' and 'pseq' .
> > The aim was to get the wrong answer because we have to share a variable
> > here,and without using 'MVar' function we will get the wrong answer for
> > the calculation .
>
> MVar is a mutable storage cell. One can use forkIO to create IO threads
> which
> race to change MVar, and if this is done badly then you can get unexpected
> answers.
>
> One cannot use the pure "par" and "pseq" to launch IO threads, so you cannot
> use
> "par" and "pseq" to create race conditions, so one can only get the expected
> answer.
>
> > I don't kno
w how to write it in parallel in order to get a wrong answer
> > when we don't use MVar,because we have a shared variable here. I read
> > about MVars as well,but also I don't know how to combine MVar and Par
> > together to get the program to work.
>
> I do not immediately see how MVar and "par" can be sensibly combined at all.
>
> > I wrote this :
> >
> > module Main where f :: Int -> Int -> Int f i n = g 1 i n where g x i n |
> > x <= i = g (x+1) i (n+1) | otherwise = n main :: IO () main = do
> > putStrLn "starting..." let r = f 10 5 putStrLn (show r) putStrLn "finished"
> > I want to make to work in parallel by using 'Par'.And also use MVar for
> > this simple example to work.
> > All of the example about MVar are a little bit complicated and I
> > couldn't figure it that how can I write one,the same !
> >
> >
; Can any one help me with this ? I want a simple example that I can feel
> > the need of MVar when I run my program in parallel and while I am using
> > a shared variable.
> >
> > Regards; Mozhgan
>
> I have run out of time, so I leave this part of your question to others.
>
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