Re: [Haskell-beginners] IO vars

Of course, not. What you do with a file is a sequence of side effects done by f, then g. If you want to reuse the value of type IO a returned by f, your g function would need to have type g:: a->IO () so that you combine the actions: f >>= g On 29 August 2012 02:21, Corentin Dupont wrote:

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

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