Re: [Haskell-beginners] 'Simple' function

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:

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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

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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

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