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
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.
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