I’m new to Haskell. This is my parser:

data Parser a = MkParser (String -> Maybe a)

This parses any string, gives the first character:

-- anyChar
anyChar :: Parser Char
anyChar = MkParser sf
    where
        sf "" = Nothing
        sf (c:cs) = Just c

It works. Now, I am learning from a tutorial. It says I can convert an answer of a parser and run it through a function, like this (creating a new parser):

-- convert a parsers answer based on a function
convert :: (a -> b) -> Parser a -> Parser b
convert f (MkParser p1) = MkParser sf
    where
        sf inp = case p1 inp of
            Nothing -> Nothing
            Just x -> Just (f x)

This looks like it works! I realized here that I can access the input string in my parser functions. (Even though the definition of convert doesn’t take an input string. I want to recreate anyChar so that it also uses the input string (Just so I can understand the syntax and what is going on). (I came from using Python and I’m a rookie at Haskell)

This is what I tried

-- apparently we can use "inp" to signifiy the input string
-- let's recreate anyChar to use input
anyCharInp :: Parser Char
anyCharInp = MkParser sf
    where
        case inp of
            (c:cs) -> Just c
            _ -> Nothing

But it’s giving an indentation error. Any ideas?