How to combine two parsers in Haskell ReadP anonymously? - parsing

The Problem
I want to chain two ReadP parsers anonymously.
Example
Input characters are in {'a', 'o', '+', ' '} where ' ' is a space.
I want to parse this input according to the following rules:
the first o or a is not preceeded by a space/plus
every other o is preceeded by a space
every other a is preceeded by a plus
The BNF
In case this is not clear, I came up with the following BNF:
{-
<Input> :: = <O> <Expr> | <A> <Expr>
<Expr> ::= <Space> <O> <Expr> | <Plus> <A> <Expr> | <EOL>
<O> ::= "o"
<A> ::= "a"
<Space> ::= " "
<Plus> ::= "+"
-}
Specific questions
The idea is, that the rules "o preceeded by space" and "a preceeded by plus"
(if not at the beginning) should not be part of the o or a parser, nor do
I want to create an explicit/named parser, because that is a part of the
definition of an expression (Expr).
How do I chain the oParser with the spaceParser?
Composition (does not work): spaceParser . oParser
(+++) is the symmetric choice: oParser (+++) spaceParser. And related to that: Is (+++) equivalent to <|> from Control.Applicative?
ReadP.choice [oParser, spaceParser] creates a new parser, which obviously does not work for a string, but I did that previously and the behaviour of choice surprised me. How does choice work?
The Code
--------------------------------------------------------------------------------
module Parser (main) where
--------------------------------------------------------------------------------
-- modules
import Text.ParserCombinators.ReadP as ReadP
import Data.Char as D
import Control.Applicative
--------------------------------------------------------------------------------
-- this parses one o
oParser :: ReadP Char
oParser = satisfy (== 'o')
-- this parses one a
aParser :: ReadP Char
aParser = satisfy (== 'a') --(\v -> v == 'a')
-- this parses one plus
pParser :: ReadP Char
pParser = satisfy (== '+')
spaceParser :: ReadP Char
spaceParser = satisfy D.isSpace
parseExpr :: ReadP String
parseExpr = -- ReadP.choice [oParser, spaceParser]
main :: IO ()
main = print [ x | (x, "") <- ReadP.readP_to_S parseExpr "o +a o+a+a o o"]
Thank you a lot for reading all of that :) (And: Which haskell parsing libraries would you recommend?)

Related

Using makeExprParser with ambiguity

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.

Parsing a series of lambda calculus terms

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

Correctly parsing nested data using megaparsec

I am trying to get more familiar with megaparsec, and I am running into some issues with presedences. By 'nested data' in the title I refer to the fact that I am trying to parse types, which in turn could contain other types. If someone could explain why this does not behave as I would expect, please don't hesitate to tell me.
I am trying to parse types similar to those found in Haskell. Types are either base types Int, Bool, Float or type variables a (any lowercase word).
We can also construct algebraic data types from type constructors (Uppercase words) such as Maybe and type parameters (any other type). Examples are Maybe a and Either (Maybe Int) Bool. Functions associate to the right and are constructed with ->, such as Maybe a -> Either Int (b -> c). N-ary tuples are a sequence of types separated by , and enclosed in ( and ), such as (Int, Bool, a). A type can be wrapped in parenthesis to raise its precedence level (Maybe a). A unit type () is also defined.
I am using this ADT to describe this.
newtype Ident = Ident String
newtype UIdent = UIdent String
data Type a
= TLam a (Type a) (Type a)
| TVar a Ident
| TNil a
| TAdt a UIdent [Type a]
| TTup a [Type a]
| TBool a
| TInt a
| TFloat a
I have tried to write a megaparsec parser to parse such types, but I get unexpected results. I attach the relevant code below after which I will try to describe what I experience.
{-# LANGUAGE OverloadedStrings #-}
module Parser where
import AbsTinyCamiot
import Text.Megaparsec
import Text.Megaparsec.Char
import qualified Text.Megaparsec.Char.Lexer as Lexer
import Text.Megaparsec.Debug
import Control.Applicative hiding (many, some, Const)
import Control.Monad.Combinators.Expr
import Control.Monad.Identity
import Data.Void
import Data.Text (Text, unpack)
type Parser a = ParsecT Void Text Identity a
-- parse types
pBaseType :: Parser (Type ())
pBaseType = choice [
TInt () <$ label "parse int" (pSymbol "Int"),
TBool () <$ label "parse bool" (pSymbol "Bool"),
TFloat () <$ label "parse float" (pSymbol "Float"),
TNil () <$ label "parse void" (pSymbol "()"),
TVar () <$> label "parse type variable" pIdent]
pAdt :: Parser (Type ())
pAdt = label "parse ADT" $ do
con <- pUIdent
variables <- many $ try $ many spaceChar >> pType
return $ TAdt () con variables
pType :: Parser (Type ())
pType = label "parse a type" $
makeExprParser
(choice [ try pFunctionType
, try $ parens pType
, try pTupleType
, try pBaseType
, try pAdt
])
[]--[[InfixR (TLam () <$ pSymbol "->")]]
pTupleType :: Parser (Type ())
pTupleType = label "parse a tuple type" $ do
pSymbol "("
fst <- pType
rest <- some (pSymbol "," >> pType)
pSymbol ")"
return $ TTup () (fst : rest)
pFunctionType :: Parser (Type ())
pFunctionType = label "parse a function type" $ do
domain <- pType
some spaceChar
pSymbol "->"
some spaceChar
codomain <- pType
return $ TLam () domain codomain
parens :: Parser a -> Parser a
parens p = label "parse a type wrapped in parentheses" $ do
pSymbol "("
a <- p
pSymbol ")"
return a
pUIdent :: Parser UIdent
pUIdent = label "parse a UIdent" $ do
a <- upperChar
rest <- many $ choice [letterChar, digitChar, char '_']
return $ UIdent (a:rest)
pIdent :: Parser Ident
pIdent = label "parse an Ident" $ do
a <- lowerChar
rest <- many $ choice [letterChar, digitChar, char '_']
return $ Ident (a:rest)
pSymbol :: Text -> Parser Text
pSymbol = Lexer.symbol pSpace
pSpace :: Parser ()
pSpace = Lexer.space
(void spaceChar)
(Lexer.skipLineComment "--")
(Lexer.skipBlockComment "{-" "-}")
This might be overwhelming, so let me explain some key points. I understand that I have a lot of different constructions that could match on an opening parenthesis, so I've wrapped those parsers in try, such that if they fail I can try the next parser that might consume an opening parenthesis. Perhaps I am using try too much? Does it affect performance to potentially backtrack so much?
I've also tried to make an expression parser by defining some terms and an operator table. You can see now that I've commented out the operator (function arrow), however. As the code looks right now I loop infinitely when I try to parse a function type. I think this might be due to the fact that when I try to parse a function type (invoked from pType) I immediately try to parse a type representing the domain of the function, which again call pType. How would I do this correctly?
If I decide to use the operator table instead, and not use my custom parser for function types, I parse things using wrong precedences. E.g Maybe a -> b gets parsed as Maybe (a -> b), while I would want it to be parsed as (Maybe a) -> b. Is there a way where I could use the operator table and still have type constructors bind more tightly than the function arrow?
Lastly, as I am learning megaparsec as I go, if anyone sees any misunderstandings or things that are wierd/unexpected, please tell me. I've read most of this tutorial to get my this far.
Please let me know of any edits I can make to increase the quality of my question!
Your code does not handle precedences at all, and also as a result of this it uses looping left-recursion.
To give an example of left-recursion in your code, pFunctionType calls pType as the first action, which calls pFunctionType as the first action. This is clearly a loop.
For precedences, I recommend to look at tutorials on "recursive descent operator parsing", a quick Google search reveals that there are several of them. Nevertheless I can summarize here the key points. I write some code.
{-# language OverloadedStrings #-}
import Control.Monad.Identity
import Data.Text (Text)
import Data.Void
import Text.Megaparsec
import Text.Megaparsec.Char
import qualified Text.Megaparsec.Char.Lexer as Lexer
type Parser a = ParsecT Void Text Identity a
newtype Ident = Ident String deriving Show
newtype UIdent = UIdent String deriving Show
data Type
= TVar Ident
| TFun Type Type -- instead of "TLam"
| TAdt UIdent [Type]
| TTup [Type]
| TUnit -- instead of "TNil"
| TBool
| TInt
| TFloat
deriving Show
pSymbol :: Text -> Parser Text
pSymbol = Lexer.symbol pSpace
pChar :: Char -> Parser ()
pChar c = void (char c <* pSpace)
pSpace :: Parser ()
pSpace = Lexer.space
(void spaceChar)
(Lexer.skipLineComment "--")
(Lexer.skipBlockComment "{-" "-}")
keywords :: [String]
keywords = ["Bool", "Int", "Float"]
pUIdent :: Parser UIdent
pUIdent = try $ do
a <- upperChar
rest <- many $ choice [letterChar, digitChar, char '_']
pSpace
let x = a:rest
if elem x keywords
then fail "expected an ADT name"
else pure $ UIdent x
pIdent :: Parser Ident
pIdent = try $ do
a <- lowerChar
rest <- many $ choice [letterChar, digitChar, char '_']
pSpace
return $ Ident (a:rest)
Let's stop here.
I changed the names of constructors in Type to conform to how they are called in Haskell. I also removed the parameter on Type, to have less noise in my example, but you can add it back of course.
Note the changed pUIdent and the addition of keywords. In general, if you want to parse identifiers, you have to disambiguate them from keywords. In this case, Int could parse both as Int and as an upper case identifier, so we have to specify that Int is not an identifier.
Continuing:
pClosed :: Parser Type
pClosed =
(TInt <$ pSymbol "Int")
<|> (TBool <$ pSymbol "Bool")
<|> (TFloat <$ pSymbol "Float")
<|> (TVar <$> pIdent)
<|> (do pChar '('
ts <- sepBy1 pFun (pChar ',') <* pChar ')'
case ts of
[] -> pure TUnit
[t] -> pure t
_ -> pure (TTup ts))
pApp :: Parser Type
pApp = (TAdt <$> pUIdent <*> many pClosed)
<|> pClosed
pFun :: Parser Type
pFun = foldr1 TFun <$> sepBy1 pApp (pSymbol "->")
pExpr :: Parser Type
pExpr = pSpace *> pFun <* eof
We have to group operators according to binding strength. For each strength, we need to have a separate parsing function which parses all operators of that strength. In this case we have pFun, pApp and pClosed in increasing order of binding strength. pExpr is just a wrapper which handles top-level expressions, and takes care of leading whitespace and matches the end of the input.
When writing an operator parser, the first thing we should pin down is the group of closed expressions. Closed expressions are delimited by a keyword or symbol both on the left and the right. This is conceptually "infinite" binding strength, since text before and after such expressions don't change their parsing at all.
Keywords and variables are clearly closed, since they consist of a single token. We also have three more closed cases: the unit type, tuples and parenthesized expressions. Since all of these start with a (, I factor this out. After that, we have one or more types separated by , and we have to branch on the number of parsed types.
The rule in precedence parsing is that when parsing an operator expression of given strength, we always call the next stronger expression parser when reading the expressions between operator symbols.
, is the weakest operator, so we call the function for the second weakest operator, pFun.
pFun in turn calls pApp, which reads ADT applications, or falls back to pClosed. In pFun you can also see the handling of right associativity, as we use foldr1 TFun to combine expressions. In a left-associative infix operator, we would instead use foldl1.
Note that parser functions always parse all stronger expressions as well. So pFun falls back on pApp when there is no -> (because sepBy1 accepts the case with no separators), and pApp falls back on pClosed when there's no ADT application.

Haskell - intersperse a parser with another one

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 #-}

Parse string to list in Haskell

I would like to parse string to list in Haskell, but I have no idea how to write it proper.
String structure is:
A [B [], C [D [], E []]]
and it represents structure
A-+-B
|
`-C-+-D
|
`-E
And ideas?
Thanks!
Your best bet when parsing in Haskell is almost always Parsec. Here's an example.
import Text.ParserCombinators.Parsec
data Tree a = Tree [Tree a] | Leaf a deriving Show
parseLeaf :: Parser (Tree Char)
parseLeaf = noneOf "[]" >>= return.Leaf
parseNode :: Parser (Tree Char)
parseNode = do
char '['
a <- parseTree
char ']'
return (Tree a)
parseTree = many1 (parseLeaf <|> parseNode)
Testing this out:
> parseTest parseTree "a[aa]"
[Leaf 'a',Tree [Leaf 'a',Leaf 'a']]
> parseTest parseTree "a[aa]aaa"
[Leaf 'a',Tree [Leaf 'a',Leaf 'a'],Leaf 'a',Leaf 'a',Leaf 'a']
> parseTest parseTree "a[aa[aaaaaa]]aaa"
[Leaf 'a',Tree [Leaf 'a',Leaf 'a',Tree [Leaf 'a',Leaf 'a',Leaf 'a',Leaf 'a',Leaf 'a',Leaf 'a']],Leaf 'a',Leaf 'a',Leaf 'a']
Here's how it works. Parsers in Parsec are monadic, and so support the usual do-notation. (You can write parsers applicatively as well, you can look up how to do that elsewhere.)
We start with a simple data structure
data Tree a = Tree [Tree a] | Leaf a deriving Show
This is not quite the same as what you wanted, but as I wasn't entirely sure what your semantics were, I've used this example. You should be able to adapt it for your purposes.
We then need a parser for each possible part of the tree. A leaf is fairly simple, it is just anything that isn't a bracket:
parseLeaf :: Parser (Tree Char)
parseLeaf = noneOf "[]" >>= return.Leaf
Note that this could have been written out in longhand like
parseLeaf = do
a <- noneOf "[]"
return (Leaf a)
Then to parse a branching part of the tree, we need to parse the opening and closing brackets. In between the brackets we can have a whole tree again.
parseNode :: Parser (Tree Char)
parseNode = do
char '['
a <- parseTree
char ']'
return (Tree a)
So what is parseTree? It is just many of either of the parsers we have written. The <|> operator allows the parser to choose either parser, whichever parses correctly first. So we have
parseTree = many1 (parseLeaf <|> parseNode)
You should be able to adapt this to your purposes. It looks like your structure might be somewhat more like this:
data Structure a = Node (Structure a) a a | Leaf a
By following the same principles, working out what the parser is needed for each possibility and then combining them, you should be parsing in no time.
UPDATE
Here is a very quick and dirty version of parsing the data structure you asked about. It does not support spaces or commas, but should help demonstrate the basic principle.
data Tree = Tree Char [Tree] deriving Show
parseTree :: Parser Tree
parseTree = do
character <- noneOf "[]"
subtree <- parseSubTree
return $ Tree character subtree
parseSubTree :: Parser [Tree]
parseSubTree = do
char '['
trees <- many parseTree
char ']'
return trees
And here is a version with the commas and whitespace added in a fairly simple way. There are a lot of useful combinators in the parsec library that could simplify and improve this, you should investigate them yourself. Note also the applicative style used for the symbol shortcut parser definition. Many people prefer applicative style for parsers and it can be a lot more succinct, so that is worth finding out about as well.
data Tree = Tree Char [Tree] deriving Show
symbol :: String -> Parser String
symbol s = string s <* spaces
parseTree :: Parser Tree
parseTree = do
character <- noneOf "[]"
spaces
subtree <- parseSubTree
return $ Tree character subtree
parseSubTree :: Parser [Tree]
parseSubTree = do
symbol "["
trees <- sepBy parseTree (symbol ",")
symbol "]"
return trees
Here it is working:
> parseTest parseTree "A [ A [ B [ ] , C [ ], D [ ] ] ] "
Tree 'A' [Tree 'A' [Tree 'B' [],Tree 'C' [],Tree 'D' []]]

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