Re: [Haskell-beginners] IO vars

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 wrote:

On Wed, Aug 29, 2012 at 10:58 AM, Corentin Dupont 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