[Haskell-beginners] monad question

[David McBride]

Functions are Monads.

:i Monad
class Applicative m => Monad (m :: * -> *) where
  (>>=) :: m a -> (a -> m b) -> m b
  (>>) :: m a -> m b -> m b
  return :: a -> m a
...
instance Monad (Either e) -- Defined in ‘Data.Either’
instance Monad [] -- Defined in ‘GHC.Base’
...
instance Monad ((->) r) -- Defined in ‘GHC.Base’

That last instance means if I have a function whose first argument is type
r, that is a monad.  And if you fill in the types of the various monad
functions you would get something like this

(>>=) :: ((->) r) a -> (a -> ((-> r) b) -> ((-> r) b)
(>>=) :: (r -> a) -> (a -> (r -> b)) -> (r -> b) -- simplified
return :: a -> (r -> a)

So in the same way that (IO String) is a Monad and can use do notation, (a
-> String) is also a Monad, and can also use do notation.  Hopefully that
made sense.

On Fri, Oct 13, 2017 at 2:15 PM, mike h <mike_k_houghton at yahoo.co.uk> wrote:

>
> I have
>
> cap :: String -> String
> cap = toUpper
>
> rev :: String -> String
> rev = reverse
>
> then I make
>
> tupled :: String -> (String, String)
> tupled = do
>     r <- rev
>     c <- cap
>     return (r, c)
>
> and to be honest, yes it’s been a long day at work, and this is coding at
> home rather than coding (java) at work but
> I’m not sure how tupled  works!!!
> My first shot was supplying a param s like this
>
> tupled :: String -> (String, String)
> tupled s = do
>     r <- rev s
>     c <- cap s
>     return (r, c)
>
> which doesn’t compile. But how does the first version work? How does the
> string to be processed get into the rev and cap functions??
>
> Thanks
>
> Mike
>
>
>
>
> _______________________________________________
> Beginners mailing list
> Beginners at haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.haskell.org/pipermail/beginners/attachments/20171013/80941fa7/attachment.html>