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

[Yassine]

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.
>> _______________________________________________
>> Beginners mailing list
>> Beginners at haskell.org
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> _______________________________________________
> Beginners mailing list
> Beginners at haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners