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 mailto:mike_k_houghton@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

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Thanks David.

On 13 Oct 2017, at 20:13, David McBride wrote:

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 mailto:mike_k_houghton@yahoo.co.uk> wrote: 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 mailto:toad3k@gmail.com> 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 mailto:mike_k_houghton@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

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