I have this monadic object.
data Parser a = Parser (String -> Maybe (a, String))
instance Functor Parser where
-- fmap :: (a -> b) -> Parser a -> Parser b
fmap f (Parser pa) = Parser $ \input -> case pa input of
Nothing -> Nothing
Just (a, rest) -> Just (f a, rest)
instance Applicative Parser where
pure = return
(<*>) = ap
instance Monad Parser where
--return :: a -> Parser a
return a = Parser $ \input -> Just (a, input)
--(>>=) :: Parser a -> (a -> Parser b) -> Parser b
(Parser pa) >>= f = Parser $ \input -> case pa input of
Nothing -> Nothing
Just (a,rest) -> parse (f a) rest
And I have this definition of an item
which I am told "reads in a character" but I don't really see any reading going on.
item :: Parser Char
item = Parser $ \ input -> case input of "" -> Nothing
(h:t) -> Just (h, t)
But ok, fine, maybe I should just relax about how literal to take the word "read" and jibe with it. Moving on, I have
failParse :: Parser a
failParse = Parser $ \ input -> Nothing
sat :: (Char -> Bool) -> Parser Char
sat p = do c <- item
if p c
then return c
else failParse
And this is where I get pretty confused. What is getting stored in the variable c
? Since item
is a Parser
with parameter Char
, my first guess is that c
is storing such an object. But after a second of thought I know that's not now the do
notation works, you don't get the monad, you get the contents of the monad. Great, but then that tells me c
is then the function
\ input -> case input of "" -> Nothing
(h:t) -> Just (h, t)
But clearly that's wrong since the next line of the definition of sat
treats c
like a character. Not only is that not what I expect, but it's about three levels of structure down from what I expected! It's not the function, it's not the Maybe
object, and it's not the tuple, but it's the left coordinate of the Just
tuple buried inside the function! How is that little character working all that way outside? What is instructing the <-
to extract this part of the monad?
As comment mentioned,
<-
just be do notation syntax sugar and equivalent to:Okay, let see what is
c
? consider the definition of(>>=)
or more readable way:
And Now, matches it with above expression
item >>= (\c->if p c then return c else failParse)
give:and
and
item
has type:so, we can now replace
a
in(>>=)
by Char, givesand now
\c->if p c then return c else failParse
also have type:(Char -> Parser b)
and so
c
is a Char, and the whole expression can be extended to: