[Haskell-beginners] Functions as Applicatives

[Lee Duhem]

On Tue, Aug 23, 2016 at 10:30 PM, Olumide <50295 at web.de> wrote:
> On 23/08/2016 12:39, Tony Morris wrote:
>>
>> All functions in Haskell always take one argument.
>
>
> I know that. All functions accept one argument and return a value _or_
> another function. Is f the latter type?

In the ((->) r) case, yes.

lee

>
> - Olumide
>
>
>>
>>
>> On 23/08/16 21:28, Olumide wrote:
>>>
>>> I must be missing something. I thought f accepts just one argument.
>>>
>>> - Olumide
>>>
>>> On 23/08/2016 00:54, Theodore Lief Gannon wrote:
>>>>
>>>> Yes, (g x) is the second argument to f. Consider the type signature:
>>>>
>>>> (<*>) :: Applicative f => f (a -> b) -> f a -> f b
>>>>
>>>> In this case, the type of f is ((->) r). Specialized to that type:
>>>>
>>>> (<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)
>>>> f <*> g = \x -> f x (g x)
>>>>
>>>> Breaking down the pieces...
>>>> f :: r -> a -> b
>>>> g :: r -> a
>>>> x :: r
>>>> (g x) :: a
>>>> (f x (g x)) :: b
>>>>
>>>> The example is made a bit confusing by tossing in an fmap. As far as the
>>>> definition above is concerned, 'f' in the example is ((+) <$> (+3)) and
>>>> that has to be resolved before looking at <*>.
>>>>
>>>>
>>>> On Mon, Aug 22, 2016 at 9:07 AM, Olumide <50295 at web.de
>>>> <mailto:50295 at web.de>> wrote:
>>>>
>>>>     Hi List,
>>>>
>>>>     I'm struggling to relate the definition of a function as a function
>>>>
>>>>     instance Applicative ((->) r) where
>>>>         pure x = (\_ -> x)
>>>>         f <*> g = \x -> f x (g x)
>>>>
>>>>     with the following expression
>>>>
>>>>     ghci> :t (+) <$> (+3) <*> (*100)
>>>>     (+) <$> (+3) <*> (*100) :: (Num a) => a -> a
>>>>     ghci> (+) <$> (+3) <*> (*100) $ 5
>>>>     508
>>>>
>>>>     From chapter 11 of LYH http://goo.gl/7kl2TM .
>>>>
>>>>     I understand the explanation in the book: "we're making a function
>>>>     that will use + on the results of (+3) and (*100) and return that.
>>>>     To demonstrate on a real example, when we did (+) <$> (+3) <*>
>>>>     (*100) $ 5, the 5 first got applied to (+3) and (*100), resulting in
>>>>     8 and 500. Then, + gets called with 8 and 500, resulting in 508."
>>>>
>>>>     The problem is that I can't relate that explanation with the
>>>>     definition of a function as an applicative; especially f <*> g = \x
>>>>     -> f x (g x) . Is (g x) the second argument to f?
>>>>
>>>>     Regards,
>>>>
>>>>     - Olumide
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>>>>
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