Ok so I’m trying to learn Haskell.

This is my Parser.

import Data.Char
data Parser a = MkParser (String -> Maybe (String, a))

This is a parser which parses a string once, and depending on what it parses, it runs another parser.

-- second parse depends on the first
(<<<=) :: Parser a -> (a -> Parser b) -> Parser b
-- a parse, gives Nothing | Just (string, ans depending on a)
(MkParser pa) <<<= k = MkParser sf
    where
        sf inp = case pa inp of
            Nothing -> Nothing
            Just (cs, c) -> unParser (k c) cs

unParser :: Parser a -> String -> Maybe (String, a)
-- applies a parser manually
unParser (MkParser pa) inp = pa inp

Now I use it here (isDigit just checks if the character is a digit, and isAlpha checks if it’s a letter). I also use this basic parser (parses if character is what I am looking for)

char :: Char -> Parser Char
char wanted = MkParser sf
  where
    sf (c:cs) | c == wanted = Just (cs, c)
    sf _ = Nothing

Now when I do this:

t = satisfy isAlpha <<<= x -> satisfy isDigit <<<= y -> char x

it works!

But the moment I put brackets around this

t = satisfy isAlpha <<<= (x -> satisfy isDigit) <<<= y -> char x

it gives an error saying variable x is not in scope… Why?

I initially just placed the brackets because it is easier to read. But turns out my code only works when I remove the brackets.

I tried working with it and realized this works too:

t = satisfy isAlpha <<<= (x -> satisfy isDigit <<<= y -> char x)

Question 1) But why? If it does (x -> satisfy isDigit <<<= y -> char x) first, it wouldn’t know what x is, so how does it work?

Question 2) What’s the evaluation order if I don’t use brackets?

Question 3) And why does it not work when I do (x -> satisfy isDigit)?

All questions are related, and I’m sure I am just misunderstanding the evaluation order. I used to program in OOP languages and I feel like I’m misunderstanding the brackets and evaluation order in Haskell?