I'm trying to parse a string using megaparsec.
Part of it is a repetition of strings separated by a separator and I'm using sepBy for this.
Consider for example
sepBy (char 'a') (char 's')
This parses correctly "", "a", "asa", ...
The problem appears if I need to continue parsing with another parser which starts with my separator, as in
(,) <$> sepBy (char 'a') (char 's') <*> string "something"
If I try to parse the string "asasomething" with this parser I'd expect to get ("aa", "something"). Instead I get an error because I don't have an a after the second s.
I tried also with sepEndBy but the result is the same
I solved it as follows.
The implementation of sepBy used by megapersec is
sepBy :: MonadPlus m => m a -> m sep -> m [a]
sepBy p sep = do
r <- C.optional p
case r of
Nothing -> return []
Just x -> (x:) <$> many (sep >> p)
I modified it to
sepBy :: Parser a -> Parser sep -> Parser [a]
sepBy p sep = do
r <- optional p
case r of
Nothing -> return []
Just x -> (x:) <$> many (try $ sep >> p)
to specialise it to Parsec add a try to avoid eager parsing
Related
I'm practicing writing parsers. I'm using Tsodings JSON Parser video as reference. I'm trying to add to it by being able to parse arithmetic of arbitrary length and I have come up with the following AST.
data HVal
= HInteger Integer -- No Support For Floats
| HBool Bool
| HNull
| HString String
| HChar Char
| HList [HVal]
| HObj [(String, HVal)]
deriving (Show, Eq, Read)
data Op -- There's only one operator for the sake of brevity at the moment.
= Add
deriving (Show, Read)
newtype Parser a = Parser {
runParser :: String -> Maybe (String, a)
}
The following functions is my attempt of implementing the operator parser.
ops :: [Char]
ops = ['+']
isOp :: Char -> Bool
isOp c = elem c ops
spanP :: (Char -> Bool) -> Parser String
spanP f = Parser $ \input -> let (token, rest) = span f input
in Just (rest, token)
opLiteral :: Parser String
opLiteral = spanP isOp
sOp :: String -> Op
sOp "+" = Add
sOp _ = undefined
parseOp :: Parser Op
parseOp = sOp <$> (charP '"' *> opLiteral <* charP '"')
The logic above is similar to how strings are parsed therefore my assumption was that the only difference was looking specifically for an operator rather than anything that's not a number between quotation marks. It does seemingly begin to parse correctly but it then gives me the following error:
λ > runParser parseOp "\"+\""
Just ("+\"",*** Exception: Prelude.undefined
CallStack (from HasCallStack):
error, called at libraries/base/GHC/Err.hs:80:14 in base:GHC.Err
undefined, called at /DIRECTORY/parser.hs:110:11 in main:Main
I'm confused as to where the error is occurring. I'm assuming it's to do with sOp mainly due to how the other functions work as intended as the rest of parseOp being a translation of the parseString function:
stringLiteral :: Parser String
stringLiteral = spanP (/= '"')
parseString :: Parser HVal
parseString = HString <$> (charP '"' *> stringLiteral <* charP '"')
The only reason why I have sOp however is that if it was replaced with say Op, I would get the error that the following doesn't exist Op :: String -> Op. When I say this my inclination was that the string coming from the parsed expression would be passed into this function wherein I could return the appropriate operator. This however is incorrect and I'm not sure how to proceed.
charP and Applicative Instance
charP :: Char -> Parser Char
charP x = Parser $ f
where f (y:ys)
| y == x = Just (ys, x)
| otherwise = Nothing
f [] = Nothing
instance Applicative Parser where
pure x = Parser $ \input -> Just (input, x)
(Parser p) <*> (Parser q) = Parser $ \input -> do
(input', f) <- p input
(input', a) <- q input
Just (input', f a)
The implementation of (<*>) is the culprit. You did not use input' in the next call to q, but used input instead. As a result you pass the string to the next parser without "eating" characters. You can fix this with:
instance Applicative Parser where
pure x = Parser $ \input -> Just (input, x)
(Parser p) <*> (Parser q) = Parser $ \input -> do
(input', f) <- p input
(input'', a) <- q input'
Just (input'', f a)
With the updated instance for Applicative, we get:
*Main> runParser parseOp "\"+\""
Just ("",Add)
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 ?
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 ..... (..... ..... ......)
I am writing a simple parser/interpreter for lambda calculus in Haskell.
I am using the syntax "/(x.x)" for a lambda function, "x" for a name and "f y" for application, such that "/(x.x) y" is the application of identity to y and should return y.
My strategy is to have a parser for names, lambda functions and applications. The name and function parsers work, but not the application parser.
When I try to parse "/(x.x) y" it fails. It should try to parse the source as a name, as a lambda function, and after not being able to use both (because it would not parse all the source) try to parse it as an application and then succeed.
My code is:
module Main where
import Text.ParserCombinators.Parsec
data Expression =
Var String
| Lambda Expression Expression
| Application Expression Expression
deriving Eq
instance Show Expression where
show (Var v) = v
show (Lambda v e) = "/(" ++ show v ++ "." ++ show e ++ ")"
show (Application e1 e2) = "" ++ show e1 ++ " " ++ show e2 ++ ""
-- Parser
parseSource :: String -> Expression
parseSource source =
case parse (parseExpression <* eof) "" source of
Right e -> e
_ -> error "Parsing error"
parseExpression :: Parser Expression
parseExpression =
parseVar <|> parseLambda <|> parseApplication
parseVar :: Parser Expression
parseVar = Var <$> many1 letter
parseLambda :: Parser Expression
parseLambda = do
_ <- char '/'
_ <- char '('
arg <- parseVar
_ <- char '.'
body <- parseExpression
_ <- char ')'
return $ Lambda arg body
parseApplication :: Parser Expression
parseApplication = do
e1 <- parseExpression
_ <- spaces
e2 <- parseExpression
return $ Application e1 e2
Given a data type data CI = CI Int Int, representing a complex number, I want to build a parser for CI that can convert "a" to CI a 0 and "(a,b)" to CI a b. For example, I want a function parseCI such runParser parseCI "(1,2)" returns the value [(CI 1 2, "")] (ideally, but something similar is fine). I also want to make CI an instance of read.
I would like to do this using functions and definitions from the code below (basically, without anything advanced, like Parsec), but I'm not sure where to start. Some starting code to set me on the right track and/or a hint would be helpful. I'm not looking for a full answer, as I'd like to figure that out myself.
module Parser where
import Control.Applicative
import Control.Monad
newtype Parser a = Parser { runParser :: String -> [(a,String)] }
satisfy :: (Char -> Bool) -> Parser Char
satisfy f = Parser $ \s -> case s of
[] -> []
a:as -> [(a,as) | f a]
char :: Char -> Parser Char
char = satisfy . (==)
string :: String -> Parser String
string str = Parser $ \s -> [(t,u) | let (t,u) = splitAt (length str) s, str == t]
instance Functor Parser where
fmap f p = Parser $ \s ->
[ (f a,t)
| (a,t) <- runParser p s
]
instance Applicative Parser where
pure a = Parser $ \s -> [(a,s)]
af <*> aa = Parser $ \s ->
[ (f a,u)
| (f,t) <- runParser af s
, (a,u) <- runParser aa t
]
instance Alternative Parser where
empty = Parser $ \s -> []
p1 <|> p2 = Parser $ (++) <$> runParser p1 <*> runParser p2`
instance Monad Parser where
return = pure
ma >>= f = Parser $ \s ->
[ (b,u)
| (a,t) <- runParser ma s
, (b,u) <- runParser (f a) t
]
instance MonadPlus Parser where
mzero = empty
mplus = (<|>)
You've probably already seen it, but in case you haven't: Monadic Parsing in Haskell sets up parsing like this.
Since you have two different ways of parsing CI, you might want to approach this as two problems: make one parser parseCI1 that parses "a" to CI a 0 and make another parser parseCI2 that parses "(a,b)" to CI a b. Then, you can combine these into one with
parseCI = parseCI1 <|> parseCI2
For both of these subparsers, you will need some way of parsing integers: parseInt :: Parser Int. When making parseInt, you will likely want to use some combination of satisfy, isDigit, read, and possibly some (depending on how you go about solving this).
Making CI an instance of read is a bit more straightforward once you have parseCI done:
instance Read CI where
readsPrec _ = runParser parseCI
I'm trying to write a Parser for S Expressions from Prof. Yorgey's 2013 homework.
newtype Parser a = Parser { runParser :: String -> Maybe (a, String) }
Given the following definitions, presented in the homework:
type Ident = String
-- An "atom" is either an integer value or an identifier.
data Atom = N Integer | I Ident
deriving Show
-- An S-expression is either an atom, or a list of S-expressions.
data SExpr = A Atom
| Comb [SExpr]
deriving Show
I wrote a parser for Parser Atom and Parser SExpr for A Atom.
parseAtom :: Parser Atom
parseAtom = alt n i
where n = (\_ z -> N z) <$> spaces <*> posInt
i = (\ _ z -> I z) <$> spaces <*> ident
parseAAtom :: Parser SExpr
parseAAtom = fmap (\x -> A x) parseAtom
Then, I attempted to write a parser to handle a Parser SExpr for the Comb ... case:
parseComb :: Parser SExpr
parseComb = (\_ _ x _ _ _ -> x) <$> (zeroOrMore spaces) <*> (char '(') <*>
(alt parseAAtom parseComb) <*> (zeroOrMore spaces)
<*> (char ')') <*> (zeroOrMore spaces)
Assuming that parseComb was right, I could simply make usage of oneOrMore for Parser [SExpr].
parseCombElements :: Parser [SExpr]
parseCombElements = oneOrMore parseComb
So, my two last functions compile, but running runParser parseComb "( foo )" never terminates.
What's wrong with my parseComb definition? Please don't give me the whole answer, but rather a hint - for my own learning.
I am very suspicious of zeroOrMore spaces, because spaces is usually a parser which itself parses zero or more spaces. Which means that it can parse the empty string if there aren't any spaces at that point. In particular, the spaces parser always succeeds.
But when you apply zeroOrMore to a parser that always succeeds, the combined parser will never stop - because zeroOrMore only stops trying again once its parser argument fails.
As an aside, Applicative expressions like (\_ _ x _ _ _ -> x) <$> ... <*> ... <*> ...... which only use a single of the subparsers can usually be written more succinctly with the *> and <* combinators:
... *> ... *> x_parser_here <* ... <* ...