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Today's Topics:
1. Re: monad question (mike h)
----------------------------------------------------------------------
Message: 1
Date: Sat, 14 Oct 2017 19:37:49 +0100
From: mike h <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] monad question
Message-ID: <[email protected]>
Content-Type: text/plain; charset="utf-8"
Thanks David.
> On 13 Oct 2017, at 20:13, David McBride <[email protected]> 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 <[email protected]
> <mailto:[email protected]>> 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 <[email protected]
>> <mailto:[email protected]>> 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 <[email protected]
>> <mailto:[email protected]>> 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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