And just as a note, you can't really ever get the value inside the IO monad out. IO is not pure / non-deterministic, since it depends on something outside the program, and there's no way to "make it pure", as it were. You have to do all your operations on that String within the context of an IO -aldiyen

On Jun 10, 2015, at 12:47, Steven Williams wrote:

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1

Here is return's type signature:

return :: Monad m => a -> m a

What you are doing with the do notation can also be expressed as ioStr

...

...

= (\str -> return str).

do notation and bind both require you to have a value that has the same monad as before.

Steven Williams My PGP Key: http://pgp.mit.edu/pks/lookup?op=get&search=0xCACA6C74669A54 FA

...

On 10/06/15 12:35, Mike Houghton wrote: Hi,

I’ve been tryimg to write a function with signature

asString :: IO String -> String

Does someone please have the patience to explain to me what the compiler error messages really mean for these two attempts and exactly what I’m doing (!!!) If I *do not* give this function any type signature then it works i.e..

asString ioStr = do str <- ioStr return $ str

and the compiler tells me its signature is

asString :: forall (m :: * -> *) b. Monad m => m b -> m b

which, at this stage of my Haskell progress, is just pure Voodoo. Why isn’t it’s signature asString :: IO String -> String ?

Another naive attempt is asString ioStr = str where str <- ioStr

and then compiler says parse error on input ‘<-’

Many Thanks

Mike

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners -----BEGIN PGP SIGNATURE----- Version: GnuPG v2

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Thanks for all the replies! It’s become a little clearer. However… (again this is naive begginer stuff.. ) if the signature is asString :: IO String -> String why is this not a pure function? The IO string has already been supplied - maybe via keyboard input - and so for the same IO String the function will always return the same value. Surely this behaviour is different to a monadic function that reads the keyboard and its output (rather than the input) could be different. ie if I give asString an input of IO “myString” then it will always return “myString” every time I invoke it with IO “myString” Many thanks Mike

On 10 Jun 2015, at 18:20, Imants Cekusins wrote:

Mike, if you are trying to run a "hello world" program in ghci, here are 2 working functions.

-- #1 : all it does is prompts for input and sends the value back to IO

module Text where

ioStr :: IO() ioStr = do putStrLn "enter anything" str <- getLine putStrLn str

-- #2 this program prepends the string you pass to it as an arg with "Hello"

str2str:: String -> String str2str s = "Hello " ++ s

-- how to run: -- #1 : ioStr -- #2 : str2str "some text"

hope this helps

On 10 June 2015 at 19:08, aldiyen wrote:

...

And just as a note, you can't really ever get the value inside the IO monad out. IO is not pure / non-deterministic, since it depends on something outside the program, and there's no way to "make it pure", as it were. You have to do all your operations on that String within the context of an IO

-aldiyen

...

On Jun 10, 2015, at 12:47, Steven Williams wrote:

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1

Here is return's type signature:

return :: Monad m => a -> m a

What you are doing with the do notation can also be expressed as ioStr

...

...

= (\str -> return str).

do notation and bind both require you to have a value that has the same monad as before.

Steven Williams My PGP Key: http://pgp.mit.edu/pks/lookup?op=get&search=0xCACA6C74669A54 FA

...

On 10/06/15 12:35, Mike Houghton wrote: Hi,

I’ve been tryimg to write a function with signature

asString :: IO String -> String

Does someone please have the patience to explain to me what the compiler error messages really mean for these two attempts and exactly what I’m doing (!!!) If I *do not* give this function any type signature then it works i.e..

asString ioStr = do str <- ioStr return $ str

and the compiler tells me its signature is

asString :: forall (m :: * -> *) b. Monad m => m b -> m b

which, at this stage of my Haskell progress, is just pure Voodoo. Why isn’t it’s signature asString :: IO String -> String ?

Another naive attempt is asString ioStr = str where str <- ioStr

and then compiler says parse error on input ‘<-’

Many Thanks

Mike

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners -----BEGIN PGP SIGNATURE----- Version: GnuPG v2

iQIcBAEBAgAGBQJVeGo5AAoJEMrKbHRmmlT6N8UP/i/tAhDtyHiG3sgH3e5xAqyt JAsyX2JaBQVjERRVaJQy1+Pg9hNdGBCrVljxY0BH5B8np956bnuIEyZKtSc2i2Jc HM0lBesyzCYqw29QxAyFFno07iXQllocZaHUIgC4AoNYO5zNGSPYcNaB4O5SYoKl 83Cjz97BHgAHkvHpsLDLOpizOkP+CsXwi8s/KRKoidLkbQpmv9SpqiFvmm9u+UK1 emZF/4veFE4Ay3AvIsxMpn7M5hVoKgat1xyGX02IrenvkOL69IIYc+4OvzK49Lxg e8jrAehJDMh+U7zN+qVCY1ZyJbJF+uGawFC+XoswOdAra+Q23te77RKkligkmN7s ACut72hwTejZN/sIaORqZXuy+HUY1LjlJnlz0RCdG1CLkr3EaKG5ZCX3E2N8RnxL 1CKtEdtFJGDeBcIBh5my/7IC22loTpVhBhPU2DPo+iOP2sRsUs0nllbqbjGfGpuE m37dR/tfq9FKwqYS5RUuAcZ8fWuPdojmO2WvI4thHBGJhsRK4gqhAI4MnKLHBEoL xfyHSaoFif/jC7peF/+ZPjKSsIpCJU+R/tDUBM9u22o3IVeTs1sWGZXM7J32tlGc K/MTF/F3phcxwSCqb99WBHhXOIkKSgp47gx1INgDZFug/CgjUI1Sl4jvZ5j/45D5 +RlHcYv+qp4J8nI59pFW =Vunc -----END PGP SIGNATURE----- _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

---

Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

---

Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

On Wed, Jun 10, 2015 at 11:15 AM Marcin Mrotek wrote:

That's why, this is not a pure function because every time you call it, it may return a different string.

This is a common source of confusion. A value of type IO a for some a is not an impure function because it is not a function. Its *evaluation* is completely pure and referentially transparent: every time you evaluate `getLine`, you get the same IO String value. The only observable difference is under execution, but *we don't expect execution to be pure*: we only expect evaluation to be pure.

impossible: asString :: IO String -> String here is an article which in addition to this thread clarified it for me: https://wiki.haskell.org/All_About_Monads search for: 4.3 No way out The IO monad is a familiar example of a one-way monad in Haskell. Because you can't escape from the IO monad, it is impossible to write a function that does a computation in the IO monad but whose result type does not include the IO type constructor. This means that any function whose result type does not contain the IO type constructor is guaranteed not to use the IO monad. ... There is no way to get rid of the IO type constructor in the signature of any function that uses it, so the IO type constructor acts as a kind of tag that identifies all functions that do I/O. Furthermore, such functions are only useful within the IO monad. So a one-way monad effectively creates an isolated computational domain in which the rules of a pure functional language can be relaxed. Functional computations can move into the domain, but dangerous side-effects and non-referentially-transparent functions cannot escape from it.

To be complete, my message actually assumes a function of that type *with the behavior you want*, which would be that of unsafePerformIO. Of course a trivial *pure* function with that type is for instance: asString _ = "Hi" But it does not have the behavior you want, it just ignores its argument. -- Yannis On 10/06/2015 20:22, Yannis Juglaret wrote:

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA256

Assuming

asString :: IO String -> String

we have

getLine :: IO String

asString getLine :: String

Yet

asString getLine

could be "Hello" the first time you use it, then "Hi" the second time you use it. Same argument, different result, so this is not a pure function.

- -- Yannis

On 10/06/2015 19:50, Mike Houghton wrote:

...

Thanks for all the replies! It’s become a little clearer. However… (again this is naive begginer stuff.. ) if the signature is

asString :: IO String -> String

why is this not a pure function? The IO string has already been supplied - maybe via keyboard input - and so for the same IO String the function will always return the same value. Surely this behaviour is different to a monadic function that reads the keyboard and its output (rather than the input) could be different. ie if I give asString an input of IO “myString” then it will always return “myString” every time I invoke it with IO “myString”

Many thanks

Mike

...

On 10 Jun 2015, at 18:20, Imants Cekusins wrote:

Mike, if you are trying to run a "hello world" program in ghci, here are 2 working functions.

-- #1 : all it does is prompts for input and sends the value back to IO

module Text where

ioStr :: IO() ioStr = do putStrLn "enter anything" str <- getLine putStrLn str

-- #2 this program prepends the string you pass to it as an arg with "Hello"

str2str:: String -> String str2str s = "Hello " ++ s

-- how to run: -- #1 : ioStr -- #2 : str2str "some text"

hope this helps

On 10 June 2015 at 19:08, aldiyen wrote:

...

And just as a note, you can't really ever get the value inside the IO monad out. IO is not pure / non-deterministic, since it depends on something outside the program, and there's no way to "make it pure", as it were. You have to do all your operations on that String within the context of an IO

-aldiyen

...

On Jun 10, 2015, at 12:47, Steven Williams wrote:

Here is return's type signature:

return :: Monad m => a -> m a

What you are doing with the do notation can also be expressed as ioStr

...

...

...

...

>> = (\str -> return str).

do notation and bind both require you to have a value that has the same monad as before.

Steven Williams My PGP Key: http://pgp.mit.edu/pks/lookup?op=get&search=0xCACA6C74669A54 FA

...

...

...

...

> On 10/06/15 12:35, Mike Houghton wrote: Hi, > > I’ve been tryimg to write a function with signature > > asString :: IO String -> String > > > Does someone please have the patience to explain to me > what the compiler error messages really mean for these > two attempts and exactly what I’m doing (!!!) If I *do > not* give this function any type signature then it works > i.e.. > > asString ioStr = do str <- ioStr return $ str > > and the compiler tells me its signature is > > asString :: forall (m :: * -> *) b. Monad m => m b -> m > b > > which, at this stage of my Haskell progress, is just pure > Voodoo. Why isn’t it’s signature asString :: IO String > -> String ? > > > Another naive attempt is asString ioStr = str where str > <- ioStr > > and then compiler says parse error on input ‘<-’ > > > Many Thanks > > Mike > > _______________________________________________ Beginners > mailing list Beginners@haskell.org > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

...

>

---

...

...

...

...

Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

- -- Yannis JUGLARET

-- Yannis JUGLARET