[Haskell-beginners] Functions as Applicatives
Tony Morris
tonymorris at gmail.com
Tue Aug 23 11:39:43 UTC 2016
All functions in Haskell always take one argument.
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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