[Haskell-beginners] monad question

[David McBride]

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 <mike_k_houghton at 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 <toad3k at 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 <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
>>
>>
>>
>>
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