Combining parsers in Haskell

问题

I'm given the following parsers

newtype Parser a = Parser { parse :: String -> Maybe (a,String) }

instance Functor Parser where
   fmap f p = Parser $ \s -> (\(a,c) -> (f a, c)) <$> parse p s

instance Applicative Parser where
   pure a = Parser $ \s -> Just (a,s)
   f <*> a = Parser $ \s ->
     case parse f s of
       Just (g,s') -> parse (fmap g a) s'
       Nothing -> Nothing

instance Alternative Parser where
   empty = Parser $ \s -> Nothing
   l <|> r = Parser $ \s -> parse l s <|> parse r s

 ensure :: (a -> Bool) -> Parser a -> Parser a
 ensure p parser = Parser $ \s ->
    case parse parser s of
      Nothing -> Nothing
      Just (a,s') -> if p a then Just (a,s') else Nothing

 lookahead :: Parser (Maybe Char)
 lookahead = Parser f
   where f [] = Just (Nothing,[])
         f (c:s) = Just (Just c,c:s)

 satisfy :: (Char -> Bool) -> Parser Char
 satisfy p = Parser f
   where f [] = Nothing
         f (x:xs) = if p x then Just (x,xs) else Nothing

 eof :: Parser ()
 eof = Parser $ \s -> if null s then Just ((),[]) else Nothing

 eof' :: Parser ()
 eof' = ???

I need to write a new parser eof' that does exactly what eof does but is built only using the given parsers and the Functor/Applicative/Alternative instances above. I'm stuck on this as I don't have experience in combining parsers. Can anyone help me out ?

回答1:

To understand it easier, we can write it in an equational pseudocode, while we substitute and simplify the definitions, using Monad Comprehensions for clarity and succinctness.

Monad Comprehensions are just like List Comprehensions, only working for any MonadPlus type, not just []; while corresponding closely to do notation, e.g. [ (f a, s') | (a, s') <- parse p s ] === do { (a, s') <- parse p s ; return (f a, s') }.

This gets us:

newtype Parser a = Parser { parse :: String -> Maybe (a,String) }

instance Functor Parser where
   parse (fmap f p)  s  =  [ (f a, s') | (a, s') <- parse p s ]

instance Applicative Parser where
   parse (pure a)    s  =  pure (a, s)
   parse (pf <*> pa) s  =  [ (g a, s'') | (g, s')  <- parse pf s 
                                        , (a, s'') <- parse pa s' ]

instance Alternative Parser where
   parse empty s      =  empty
   parse (l <|> r) s  =  parse l s <|> parse r s

ensure :: (a -> Bool) -> Parser a -> Parser a
parse (ensure pred p) s  =  [ (a, s') | (a, s') <- parse p s, pred a ]

lookahead :: Parser (Maybe Char)
parse lookahead []       =  pure (Nothing, [])
parse lookahead s@(c:_)  =  pure (Just c,  s )

satisfy :: (Char -> Bool) -> Parser Char
parse (satisfy p) []      =  mzero
parse (satisfy p) (x:xs)  =  [ (x, xs) | p x ]

eof :: Parser ()
parse eof s  =  [ ((), []) | null s ]  

eof' :: Parser ()
eof'  =  ???

By the way thanks to the use of Monad Comprehensions and the more abstract pure, empty and mzero instead of their concrete representations in terms of the Maybe type, this same (pseudo-)code will work with a different type, like [] in place of Maybe, viz. newtype Parser a = Parser { parse :: String -> [(a,String)] }.

So we have

ensure    :: (a -> Bool) -> Parser a           -> Parser a
lookahead ::                Parser (Maybe Char)

_{(satisfy is no good for us here .... why?)}

Using that, we can have

ensure  .......  ......                        :: Parser (Maybe Char)

_{(... what does ensure id (pure False) do? ...)}

but we'll have a useless Nothing result in case the input string was in fact empty, whereas the eof parser given to use produces the () as its result in such case (and otherwise it produces nothing).

No fear, we also have

fmap :: (  a      ->   b )                     -> Parser a        ->  Parser b

which can transform the Nothing into () for us. We'll need a function that will always do this for us,

alwaysUnit nothing  =  ()

which we can use now to arrive at the solution:

eof'  =  fmap ..... (..... ..... ......)

来源：https://stackoverflow.com/questions/60221086/combining-parsers-in-haskell