I am writing a lambda calculus parser in Haskell and I can't find a solution to fix its current problem.
How I parse expressions:
expr :: Parser LamExpr
expr = do terms <- some $ token term
return $ foldl1 LamApp terms
How I parse terms:
term :: Parser LamExpr
term = do symbol "("
e <- expr
symbol ")"
return e
<|> do symbol "\\"
x <- var
symbol "->"
e <- expr
return $ LamAbs x e
<|> do {x <- var; return $ LamVar x}
<|> do {name <- macroName; return $ LamMacro name}
On input "x1 x2) x3" my parser returns
LamApp (LamVar 1) (LamVar 2)
Parsing should fail as it is syntactically incorrect, but it still parses the first application. I think this is because of do terms <- some $ token term which will parse as much as it can due to some.
How can I fix this so that the whole parsing fails instead of one section?
I assume you are using some parsec variant. You just have to add an eof to the end of your parser.
parseInput = do
e <- expr
eof
pure e -- (*)
Or for short using a Control.Applicative combinator:
parseInput = expr <* eof
(*) btw the community is starting to use pure instead of return these days
Related
I'm currently encountering a problem while translating a parser from a CFG-based tool (antlr) to Megaparsec.
The grammar contains lists of expressions (handled with makeExprParser) that are enclosed in brackets (<, >) and separated by ,.
Stuff like <>, <23>, <23,87> etc.
The problem now is that the expressions may themselves contain the > operator (meaning "greater than"), which causes my parser to fail.
<1223>234> should, for example, be parsed into [BinaryExpression ">" (IntExpr 1223) (IntExpr 234)].
I presume that I have to strategically place try somewhere, but the places I tried (to the first argument of sepBy and the first argument of makeExprParser) did unfortunately not work.
Can I use makeExprParser in such a situation or do I have to manually write the expression parser?:
This is the relevant part of my parser:
-- uses megaparsec, text, and parser-combinators
{-# LANGUAGE OverloadedStrings #-}
module Main where
import Control.Monad.Combinators.Expr
import Data.Text
import Data.Void
import System.Environment
import Text.Megaparsec
import Text.Megaparsec.Char
import qualified Text.Megaparsec.Char.Lexer as L
type BinaryOperator = Text
type Name = Text
data Expr
= IntExpr Integer
| BinaryExpression BinaryOperator Expr Expr
deriving (Eq, Show)
type Parser = Parsec Void Text
lexeme :: Parser a -> Parser a
lexeme = L.lexeme sc
symbol :: Text -> Parser Text
symbol = L.symbol sc
sc :: Parser ()
sc = L.space space1 (L.skipLineComment "//") (L.skipBlockCommentNested "/*" "*/")
parseInteger :: Parser Expr
parseInteger = do
number <- some digitChar
_ <- sc
return $ IntExpr $ read number
parseExpr :: Parser Expr
parseExpr = makeExprParser parseInteger [[InfixL (BinaryExpression ">" <$ symbol ">")]]
parseBracketList :: Parser [Expr]
parseBracketList = do
_ <- symbol "<"
exprs <- sepBy parseExpr (symbol ",")
_ <- symbol ">"
return exprs
main :: IO ()
main = do
text : _ <- getArgs
let res = runParser parseBracketList "stdin" (pack text)
case res of
(Right suc) -> do
print suc
(Left err) ->
putStrLn $ errorBundlePretty err
You've (probably) misdiagnosed the problem. Your parser fails on <1233>234> because it's trying to parse > as a left associative operator, like +. In other words, the same way:
1+2+
would fail, because the second + has no right-hand operand, your parser is failing because:
1233>234>
has no digit following the second >. Assuming you don't want your > operator to chain (i.e., 1>2>3 is not a valid Expr), you should first replace InfixL with InfixN (non-associative) in your makeExprParser table. Then, it will parse this example fine.
Unfortunately, with or without this change your parser will still fail on the simpler test case:
<1233>
because the > is interpreted as an operator within a continuing expression.
In other words, the problem isn't that your parser can't handle expressions with > characters, it's that it's overly aggressive in treating > characters as part of an expression, preventing them from being recognized as the closing angle bracket.
To fix this, you need to figure out exactly what you're parsing. Specifically, you need to resolve the ambiguity in your parser by precisely characterizing the situations where > can be part of a continuing expression and where it can't.
One rule that will probably work is to only consider a > as an operator if it is followed by a valid "term" (i.e., a parseInteger). You can do this with lookAhead. The parser:
symbol ">" <* lookAhead term
will parse a > operator only if it is followed by a valid term. If it fails to find a term, it will consume some input (at least the > symbol itself), so you must surround it with a try:
try (symbol ">" <* lookAhead term)
With the above two fixes applied to parseExpr:
parseExpr :: Parser Expr
parseExpr = makeExprParser term
[[InfixN (BinaryExpression ">" <$ try (symbol ">" <* lookAhead term))]]
where term = parseInteger
you'll get the following parses:
λ> parseTest parseBracketList "<23>"
[IntExpr 23]
λ> parseTest parseBracketList "<23,87>"
[IntExpr 23,IntExpr 87]
λ> parseTest parseBracketList "<23,87>18>"
[IntExpr 23,BinaryExpression ">" (IntExpr 87) (IntExpr 18)]
However, the following will fail:
λ> parseTest parseBracketList "<23,87>18"
1:10:
|
1 | <23,87>18
| ^
unexpected end of input
expecting ',', '>', or digit
λ>
because the fact that the > is followed by 18 means that it is a valid operator, and it is parse failure that the valid expression 87>18 is followed by neither a comma nor a closing > angle bracket.
If you need to parse something like <23,87>18, you have bigger problems. Consider the following two test cases:
<1,2>3,4,5,6,7,...,100000000000,100000000001>
<1,2>3,4,5,6,7,...,100000000000,100000000001
It's a challenge to write an efficient parser that will parse the first one as a list of 10000000000 expressions but the second one as a list of two expression:
[IntExpr 1, IntExpr 2]
followed by some "extra" text. Hopefully, the underlying "language" you're trying to parse isn't so hopelessly broken that this will be an issue.
Parsec provides an operator to choose between two parsers:
(<|>)
:: Text.Parsec.Prim.ParsecT s u m a
-> Text.Parsec.Prim.ParsecT s u m a
-> Text.Parsec.Prim.ParsecT s u m a
Is there a similar function to chain two parsers? I didn't find one with the same signature using Hoogle.
As an example, let's say I want to parse any word optionally followed by a single digit. My first idea was to use >> but it doesn't seem to work.
parser = many1 letter >> optional (fmap pure digit)
I used fmap pure in order to convert the digit to an actual string and thus match the parsed type of many1 letter. I don't know if it is useful.
Try this:
parser = (++) <$> many1 letter <*> option "" (fmap pure digit)
This is equivalent to:
parser = pure (++) <*> many1 letter <*> option "" (fmap pure digit)
option [] (fmap pure digit) return empty string if the parser digit have failed and a string from one digital char otherwise.
You can also use do-notation for more readable code:
parser = do
s1 <- many1 letter
s2 <- option "" (fmap pure digit)
return (s1 ++ s2)
I have two parsers parser1 :: Parser a and parser2 :: Parser a.
I would like now to parse a list of as interspersing them with parser2
The desired signature is something like
interspersedParser :: Parser b -> Parser a -> Parser [a]
For example, if Parser a parses the 'a' character and Parser b parser the 'b' character, then the interspersedParser should parse
""
"a"
"aba"
"ababa"
...
I'm using megaparsec. Is there already some combinator which behaves like this, which I'm currently not able to find?
In parsec there is a sepBy parser which does that. The same parser seems to be available in megaparsec as well: https://hackage.haskell.org/package/megaparsec-4.4.0/docs/Text-Megaparsec-Combinator.html
Sure, you can use sepBy, but isn't this just:
interspersedParser sepP thingP = (:) <$> thingP <*> many (sepP *> thingP)
EDIT: Oh, this requires at least one thing to be there. You also wanted empty, so just stick a <|> pure [] on the end.
In fact, this is basically how sepBy1 (a variant of sepBy that requires at least one) is implemented:
-- | #sepBy p sep# parses /zero/ or more occurrences of #p#, separated
-- by #sep#. Returns a list of values returned by #p#.
--
-- > commaSep p = p `sepBy` comma
sepBy :: Alternative m => m a -> m sep -> m [a]
sepBy p sep = sepBy1 p sep <|> pure []
{-# INLINE sepBy #-}
-- | #sepBy1 p sep# parses /one/ or more occurrences of #p#, separated
-- by #sep#. Returns a list of values returned by #p#.
sepBy1 :: Alternative m => m a -> m sep -> m [a]
sepBy1 p sep = (:) <$> p <*> many (sep *> p)
{-# INLINE sepBy1 #-}
I'm attempting to write a parser in Haskell using Parsec. Currently I have a program that can parse
test x [1,2,3] end
The code that does this is given as follows
testParser = do {
reserved "test";
v <- identifier;
symbol "[";
l <- sepBy natural commaSep;
symbol "]";
p <- pParser;
return $ Test v (List l) p
} <?> "end"
where commaSep is defined as
commaSep = skipMany1 (space <|> char ',')
Now is there some way for me to parse a similar statement, specifically:
test x [1...3] end
Being new to Haskell, and Parsec for that matter, I'm sure there's some nice concise way of doing this that I'm just not aware of. Any help would be appreciated.
Thanks again.
I'll be using some functions from Control.Applicative like (*>). These functions are useful if you want to avoid the monadic interface of Parsec and prefer the applicative interface, because the parsers become easier to read that way in my opinion.
If you aren't familiar with the basic applicative functions, leave a comment and I'll explain them. You can look them up on Hoogle if you are unsure.
As I've understood your problem, you want a parser for some data structure like this:
data Test = Test String Numbers
data Numbers = List [Int] | Range Int Int
A parser that can parse such a data structure would look like this (I've not compiled the code, but it should work):
-- parses "test <identifier> [<numbers>] end"
testParser :: Parser Test
testParser =
Test <$> reserved "test" *> identifier
<*> symbol "[" *> numbersParser <* symbol "]"
<* reserved "end"
<?> "test"
numbersParser :: Parser Numbers
numbersParser = try listParser <|> rangeParser
-- parses "<natural>, <natural>, <natural>" etc
listParser :: Parser Numbers
listParser =
List <$> sepBy natural (symbol ",")
<?> "list"
-- parses "<natural> ... <natural>"
rangeParser :: Parser Numbers
rangeParser =
Range <$> natural <* symbol "..."
<*> natural
<?> "range"
I'm relatively new to Haskell with main programming background coming from OO languages. I am trying to write an interpreter with a parser for a simple programming language. So far I have the interpreter at a state which I am reasonably happy with, but am struggling slightly with the parser.
Here is the piece of code which I am having problems with
data IntExp
= IVar Var
| ICon Int
| Add IntExp IntExp
deriving (Read, Show)
whitespace = many1 (char ' ')
parseICon :: Parser IntExp
parseICon =
do x <- many (digit)
return (ICon (read x :: Int))
parseIVar :: Parser IntExp
parseIVar =
do x <- many (letter)
prime <- string "'" <|> string ""
return (IVar (x ++ prime))
parseIntExp :: Parser IntExp
parseIntExp =
do x <- try(parseICon)<|>try(parseIVar)<|>parseAdd
return x
parseAdd :: Parser IntExp
parseAdd =
do x <- parseIntExp
whitespace
string "+"
whitespace
y <- parseIntExp
return (Add x y)
runP :: Show a => Parser a -> String -> IO ()
runP p input
= case parse p "" input of
Left err ->
do putStr "parse error at "
print err
Right x -> print x
The language is slightly more complex, but this is enough to show my problem.
So in the type IntExp ICon is a constant and IVar is a variable, but now onto the problem. This for example runs successfully
runP parseAdd "5 + 5"
which gives (Add (ICon 5) (ICon 5)), which is the expected result. The problem arises when using IVars rather than ICons eg
runP parseAdd "n + m"
This causes the program to error out saying there was an unexpected "n" where a digit was expected. This leads me to believe that parseIntExp isn't working as I intended. My intention was that it will try to parse an ICon, if that fails then try to parse an IVar and so on.
So I either think the problem exists in parseIntExp, or that I am missing something in parseIVar and parseICon.
I hope I've given enough info about my problem and I was clear enough.
Thanks for any help you can give me!
Your problem is actually in parseICon:
parseICon =
do x <- many (digit)
return (ICon (read x :: Int))
The many combinator matches zero or more occurrences, so it's succeeding on "m" by matching zero digits, then probably dying when read fails.
And while I'm at it, since you're new to Haskell, here's some unsolicited advice:
Don't use spurious parentheses. many (digit) should just be many digit. Parentheses here just group things, they're not necessary for function application.
You don't need to do ICon (read x :: Int). The data constructor ICon can only take an Int, so the compiler can figure out what you meant on its own.
You don't need try around the first two options in parseIntExp as it stands--there's no input that would result in either one consuming some input before failing. They'll either fail immediately (which doesn't need try) or they'll succeed after matching a single character.
It's usually a better idea to tokenize first before parsing. Dealing with whitespace at the same time as syntax is a headache.
It's common in Haskell to use the ($) operator to avoid parentheses. It's just function application, but with very low precedence, so that something like many1 (char ' ') can be written many1 $ char ' '.
Also, doing this sort of thing is redundant and unnecessary:
parseICon :: Parser IntExp
parseICon =
do x <- many digit
return (ICon (read x))
When all you're doing is applying a regular function to the result of a parser, you can just use fmap:
parseICon :: Parser IntExp
parseICon = fmap (ICon . read) (many digit)
They're the exact same thing. You can make things look even nicer if you import the Control.Applicative module, which gives you an operator version of fmap, called (<$>), as well as another operator (<*>) that lets you do the same thing with functions of multiple arguments. There's also operators (<*) and (*>) that discard the right or left values, respectively, which in this case lets you parse something while discarding the result, e.g., whitespace and such.
Here's a lightly modified version of your code with some of the above suggestions applied and some other minor stylistic tweaks:
whitespace = many1 $ char ' '
parseICon :: Parser IntExp
parseICon = ICon . read <$> many1 digit
parseIVar :: Parser IntExp
parseIVar = IVar <$> parseVarName
parseVarName :: Parser String
parseVarName = (++) <$> many1 letter <*> parsePrime
parsePrime :: Parser String
parsePrime = option "" $ string "'"
parseIntExp :: Parser IntExp
parseIntExp = parseICon <|> parseIVar <|> parseAdd
parsePlusWithSpaces :: Parser ()
parsePlusWithSpaces = whitespace *> string "+" *> whitespace *> pure ()
parseAdd :: Parser IntExp
parseAdd = Add <$> parseIntExp <* parsePlusWithSpaces <*> parseIntExp
I'm also new to Haskell, just wondering:
will parseIntExp ever make it to parseAdd?
It seems like ICon or IVar will always get parsed before reaching 'parseAdd'.
e.g. runP parseIntExp "3 + m"
would try parseICon, and succeed, giving
(ICon 3) instead of (Add (ICon 3) (IVar m))
Sorry if I'm being stupid here, I'm just unsure.