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 ((+) <$> (+3)) and that has to be resolved before looking at <*>.
On Mon, Aug 22, 2016 at 9:07 AM, Olumide <50295@web.de mailto:50295@web.de> wrote:
Hi List,
I'm struggling to relate the definition of a function as a function
instance Applicative ((->) r) where pure x = (\_ -> x) f <*> g = \x -> f x (g x)
with the following expression
ghci> :t (+) <$> (+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 _______________________________________________ Beginners mailing list Beginners@haskell.org mailto:Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
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