point-free + IO Monad

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Quickcheck2 - writing a modifier...

Ovidiu D

31 Mar 2013 31 Mar '13

2:55 a.m.

I have the following code: f1 :: String -> String f2 :: IO String -> IO () f2 a = do b <- a putStr $ f1 b How can I write the function f2 in a point-free style? I have tried this: f2 = return.f1 >>= putStr ...but it doesn't work. Thanks!

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Franco

31 Mar 31 Mar

4:08 a.m.

Ovidiu D writes:

...

I have the following code:

f1 :: String -> String

f2 :: IO String -> IO () f2 a = do b <- a putStr $ f1 b

How can I write the function f2 in a point-free style?

maybe f2 = (=<<) (putStrLn . f1) I still prefer the non point-free but clearer f2 = a >>= (putStrLn . f1)

Ovidiu D

5:27 a.m.

That's good enough. Thanks! On Sun, Mar 31, 2013 at 11:08 AM, Franco wrote:

...

Ovidiu D writes:

...
I have the following code:

f1 :: String -> String

f2 :: IO String -> IO () f2 a = do b <- a putStr $ f1 b

How can I write the function f2 in a point-free style?

maybe

f2 = (=<<) (putStrLn . f1)

I still prefer the non point-free but clearer

f2 = a >>= (putStrLn . f1)

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Tony Morris

5:59 a.m.

You can drop the parens if you like. f2 a = putStrLn . f1 =<< a -- my preference f2 a = a >>= putStrLn . f1 On 31/03/2013 7:29 PM, "Ovidiu D" wrote:

...

That's good enough. Thanks!

On Sun, Mar 31, 2013 at 11:08 AM, Franco wrote:

...
Ovidiu D writes:

...
I have the following code:

f1 :: String -> String

f2 :: IO String -> IO () f2 a = do b <- a putStr $ f1 b

How can I write the function f2 in a point-free style?

maybe

f2 = (=<<) (putStrLn . f1)

I still prefer the non point-free but clearer

f2 = a >>= (putStrLn . f1)

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

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Franco
Ovidiu D
Tony Morris