[Haskell-beginners] Applicative: how <*> really works

[Jeffrey Brown]

If you post there, you could put the link on this thread; I'd be interested.

On Fri, Aug 4, 2017 at 3:04 PM, Yassine <yassine912 at gmail.com> wrote:

> Ok, thanks for your answer
>
> 2017-08-04 20:04 GMT+02:00 David McBride <toad3k at gmail.com>:
> > This is a bit complicated for this list.  You might have a bit more
> > luck posting this to stackoverflow.com.
> >
> > On Thu, Aug 3, 2017 at 3:19 PM, Yassine <yassine912 at gmail.com> wrote:
> >> Hi,
> >>
> >> I have a question about functor applicate.
> >>
> >> I know that:
> >> pure (+1) <*> Just 2
> >>
> >>
> >> produce: Just 3
> >> because pure (+1) produce Just (+1) and then Just (+1) <*> Just 2
> >> produce Just (2+1)
> >>
> >>
> >> but in more complex case like:
> >> newtype Parser a = P (String -> [(a,String)])
> >>
> >> parse :: Parser a -> String -> [(a,String)]
> >> parse (P p) inp = p inp
> >>
> >>
> >> item :: Parser Char
> >> item = P (\inp -> case inp of
> >>  []     -> []
> >> (x:xs) -> [(x,xs)])
> >>
> >> instance Functor Parser where
> >> fmap g p = P (\inp -> case p inp of
> >> []              -> []
> >>  [(v, out)]      -> [(g v, out)])
> >>
> >> instance Applicative Parser where
> >> pure v = P (\inp -> [(v, inp)])
> >> pg <*> px = P (\inp -> case parse pg inp of
> >> []              -> []
> >> [(g, out)]      -> parse (fmap g px) out)
> >>
> >>
> >> When I do:
> >> parse (pure (\x y -> (x,y)) <*> item <*> item) "abc"
> >>
> >> The answer is:
> >> [(('a','b'),"c")]
> >>
> >> But I don't understand what exactly happens.
> >> First:
> >> pure (\x y -> (x,y)) => P (\inp -> [(\x y -> (x,y), inp)])
> >>
> >> Then:
> >> P (\inp -> [(\x y -> (x,y), inp)]) <*> item => ???
> >>
> >> Can someone explain what's happens step by step please.
> >>
> >> Thank you.
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