
If you are using do notation, you can't. If you aren't you can write
tupled s = (rev s, cap s)
Your old tupled is equivalent to this
tupled = rev >>= \s -> cap >>= \c -> return (s, c)
which is quite different.
On Fri, Oct 13, 2017 at 3:05 PM, mike h
That certainly helps me David, thanks. How then would you write
tupled :: String -> (String, String)
with the parameter written explicitly? i.e.
tupled s = do …
or does the question not make sense in light of your earlier reply?
Thanks
Mike
On 13 Oct 2017, at 19:35, David McBride
wrote: 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
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
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