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@gmail.com> wrote:
On Wed, Aug 29, 2012 at 10:58 AM, Corentin Dupont
<corentin.dupont@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