Hello Paul,
As far as I can see you're calling the parens accessor function of the
TokenParser record instead of supplying a parser. Here is a working example
grammar I made for my bachelor paper a while ago:
module ExpressionsWithLexer where
import Text.ParserCombinators.Parsec.Expr
import Text.ParserCombinators.Parsec
import qualified Text.ParserCombinators.Parsec.Token as P
import Text.ParserCombinators.Parsec.Language
{-
Leftassociative: +, -, *, /
Rightassociative: ^
Priority 1: +,-
Priority 2: *, /
Priority 3: ^
expr ::= factor (op factor)*
factor ::= number | '(' expr ')'
op ::= '+' | '-' | '*' | '/' | '^'
number ::= ('0' | '1' | ... | '9')+
-}
lexer :: P.TokenParser ()
lexer = P.makeTokenParser
(emptyDef
{ reservedOpNames = ["*","/","+","-", "^"]
})
whiteSpace= P.whiteSpace lexer
natural = P.natural lexer
parens = P.parens lexer
reservedOp = P.reservedOp lexer
gram = do whiteSpace
x <- expr
eof
return x
expr :: Parser Integer
expr = buildExpressionParser table factor
<?> "expression"
-- Earlier in the list means a higher priority
table = [[op "^" (^) AssocRight]
,[op "*" (*) AssocLeft, op "/" div AssocLeft]
,[op "+" (+) AssocLeft, op "-" (-) AssocLeft]
]
where op s f assoc = Infix (do{ reservedOp s; return f} <?>
"operator") assoc
factor = natural
<|>
(parens expr
"factor")
run :: Show a => Parser a -> String -> IO ()
run p input
= case (parse p "" input) of
Left err -> do{ putStr "parse error at "
; print err
}
Right x -> print x
exampleRun = run gram "(10 ^3 - (1 + 3))"
This grammar parses the EBNF like grammar included in the source.
As you can see parens is an accessor function of the TokenParser record. A
TokenParser record is thus produced by calling makeTokenParser and supplying
that function with a language definition. I used the emptyDef (empty
definition) as an argument for the makeTokenParser and only updated the
operatornames. After constructing a TokenParser parsers such as parens are
accessed by their accessor functions. For ease of use you can use a
qualified import for the token parser module and define the parens parser
and other parsers at toplevel like I did.
I only tested this with Parsec 2.1.0.1 but the idea should be the same. You
can read some more explanations and examples in my bachelor paper if you'd
like[1][2].
Good luck,
Bas van Gijzel
(server is currently down :( )
[1] Comparing Parser Construction Techniques:
http://referaat.cs.utwente.nl/new/papers.php?confid=12
[2] Parser Code:
http://fmt.cs.utwente.nl/~michaelw/projects/vgijzel/ParserCode.ziphttp://fmt.cs.utwente.nl/%7Emichaelw/projects/vgijzel/ParserCode.zip
On Sat, Jul 11, 2009 at 00:27, Paul Sujkov
Hi haskellers,
I'm trying to use buildExpressionParser parser generator from ParsecExpr module of Parsec ( http://legacy.cs.uu.nl/daan/download/parsec/parsec.html#ParsecExpr). It works well, except for the "parens" token parser ( http://legacy.cs.uu.nl/daan/download/parsec/parsec.html#parens). This code (sample from Expressions part of the manual) typechecks fine:
expr :: Parser Integer expr = buildExpressionParser table factor "expression"
table = [[op "*" (*) AssocLeft, op "/" div AssocLeft] ,[op "+" (+) AssocLeft, op "-" (-) AssocLeft] ] where op s f assoc = Infix (do{ string s; return f}) assoc
factor = do{ char '(' ; x <- expr ; char ')' ; return x } <|> number "simple expression"
but if I try to use parens:
factor = parens expr <|> number "simple expression"
it fails to typecheck:
Couldn't match expected type `GenTokenParser s u m' against inferred type `Parser Integer' In the first argument of `parens', namely `expr' In the first argument of `(<|>)', namely `parens expr' In the first argument of `()', namely `parens expr <|> number' Failed, modules loaded: none.
the type of expr infers to GenParser Char () Integer, and the expected type for the parens is GenTokenParser s u m (however, manual introduces it with the type CharParser st a expected)
It seems pretty weird for me, as there are numerous examples of using parens with the buildExpressionParser (e.g. http://blog.moertel.com/articles/2005/08/27/power-parsing-with-haskell-and-p...) and nobody comments such an error. I know I'm missing something very simple here, maybe someone could help me with it? Thanks in advace
I'm using GHC 6.10.1 and Parsec 3.0.0
-- Regards, Paul Sujkov
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