[Haskell-beginners] 'Simple' function

[Steven Williams]

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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

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 at haskell.org 
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> 
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