[Haskell-beginners] IO vars

[Corentin Dupont]

Hi Mihai,
maybe the term "thread" in my mail is not correct.
What I mean is that a value gets stored by f and discovered by g.

*f,g :: IO ()
f = withFile "toto" WriteMode (flip hPutStr "42")
g = withFile "toto" ReadMode hGetLine >>= (\s -> putStrLn $ "Answer:" ++ s)
main = f >> g*

Is it possible to do the same without files (the types must remain IO())?


On Wed, Aug 29, 2012 at 11:04 AM, Mihai Maruseac
<mihai.maruseac at gmail.com>wrote:

> On Wed, Aug 29, 2012 at 10:58 AM, Corentin Dupont
> <corentin.dupont at gmail.com> wrote:
> > Hi all,
> > there is something very basic that it seems escaped me.
> > For example with the following program f and g have type IO () and I can
> > thread a value between the two using a file.
> > Can I do the exact same (not changing the types of f and g) without a
> file?
> >
> > f,g :: IO ()
> > f = withFile "toto" WriteMode (flip hPutStr "toto")
> > g = withFile "toto" ReadMode hGetLine >>= putStrLn
> > main = f >> g
>
> Of course:
>
> f,g :: IO ()
> f = putStr "Answer: "
> g = print 42
>
> Main> f >> g
> Answer: 42
>
> The () is threaded by >>, not the file content
>
> --
> MM
>
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