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
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
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