Hi, probably somebody else has already come up with something better, but still... I surmise that you have two kinds of infix-operators, 1. dot-like operators, made up entirely of symbols (^!$%&/\,.:;#+-~* ...) 2. LaTeX-command-like operators, starting with a backslash and then followed by a nonempty sequence of letters (or possibly alphanumeric characters). Then the following helps: import Data.Char (isAlpha) lexer = lexer0{P.reservedOp = rOp} where lexer0 = P.makeTokenParser testdef resOp0 = P.reservedOp lexer0 resOp1 name = case name of ('\\':cs@(_:_)) | all isAlpha cs -> do string name notFollowedBy letter ("end of " ++ show name) _ -> fail (show name ++ " no good reservedOp") rOp name = lexeme $ try $ resOp0 name <|> resOp1 name lexeme p = do { x <- p; P.whiteSpace lexer0; return x } testdef = emptyDef { P.identStart = letter <|> char '_' , P.identLetter = alphaNum <|> char '_' , P.opStart = oneOf $ nub $ map (\s -> head s) $ P.reservedOpNames testdef -- , P.opLetter = oneOf (concat (P.reservedOpNames testdef)) , P.opLetter = oneOf symbs , P.reservedOpNames = [ ".", "\\in" ] } where symbs = filter (not . isAlpha) . concat $ P.reservedOpNames testdef --------------------------------------------------------------------- dafis@linux:~/Documents/haskell/Reto> cat input a.n dafis@linux:~/Documents/haskell/Reto> reto input InfixExpr OP_Dot (Ident "a") (Ident "n") If you have more complicated infix operators (e.g. \foo#bar:, :ouch:), it won't be so easy, anyway, you have to change the definition of reservedOp. Cheers, Daniel Am Freitag, 31. März 2006 01:15 schrieb Reto Kramer:
I ran into the following problem with parsec's handling of operators. I hope someone on the list can show me a trick that resolve my current issue. At the end of this message is the full code for the reproducer.
The language I'm parsing has infix operators of two forms. Some are special characters (e.g. a dot) and some are LaTeX like (e.g. \in). The letters that appear in the e.g. \in lead to problems with expression parsing.
The following is what I expect: $ cat input1 a.b $ ./test input1 InfixExpr OP_Dot (Ident "a") (Ident "b") Good!
Then I try (change the b to an n) $ cat input2 a.n $ ./test input2 Ident "a" OUTSCH! Changing the name of an identifier changed the expression!
Then I try (add a space right after the dot) $ cat input3 a. n $ ./test input2 InfixExpr OP_Dot (Ident "a") (Ident "n") Good!
What is going on?
The 'n' is part of the "\in" operator (see reservedOpNames in test.hs program at the end of message) and confuses the parsec expression parser's ability to determine the end of expressions/identifiers. Adding a space right after the '.' operator resolves the issue, but that's not a suitable option for my users.
Unfortunately I cannot remove the "\in" operator from the reservedOpNames list since otherwise "\in" is not recognized as an infix operator itself anymore.
QUESTION: how can I get parsec's expression parser to work with infix operators of the form "\in"?
Thanks, - Reto
-- complete reproducer (test.hs) -- compile with: ghc -o test -package parsec test.hs
module Main where
import List (nub) import System (getArgs) import Text.ParserCombinators.Parsec import Text.ParserCombinators.Parsec.Expr import Text.ParserCombinators.Parsec.Language( emptyDef ) import qualified Text.ParserCombinators.Parsec.Token as P
main :: IO () main = do{ args <- getArgs ; let fname = args !! 0 ; input <- readFile fname ; case parse spec fname input of Left err -> do{ putStr "parse error at " ; print err } Right x -> print x }
data Op = OP_Dot | OP_In deriving (Show, Eq, Ord) data Expr = InfixExpr Op Expr Expr
| Ident String | Number Integer deriving (Show, Eq, Ord)
spec :: Parser Expr spec = do { whiteSpace ; e <- expression ; return e }
expression :: Parser Expr expression = buildExpressionParser table basicExpr "expression"
op_infix :: Op -> Expr -> Expr -> Expr op_infix op a b = InfixExpr op a b
table :: OperatorTable Char () Expr table = [ [binary "." (op_infix OP_Dot) AssocLeft] , [binary "\\in" (op_infix OP_In) AssocLeft] ] binary name fun assoc = Infix (do{ reservedOp name; return fun }) assoc
basicExpr :: Parser Expr basicExpr = choice [ do{ i <- identifier ; return $ Ident i } ]
lexer = P.makeTokenParser testdef
testdef = emptyDef { P.identStart = letter <|> char '_' , P.identLetter = alphaNum <|> char '_' , P.opStart = oneOf $ nub $ map (\s -> head s) $ P.reservedOpNames testdef , P.opLetter = oneOf (concat (P.reservedOpNames testdef)) , P.reservedOpNames = [ ".", "\\in" ] }
dot = P.dot lexer parens = P.parens lexer braces = P.braces lexer squares = P.squares lexer semiSep = P.semiSep lexer semiSep1 = P.semiSep1 lexer commaSep = P.commaSep lexer commaSep1 = P.commaSep1 lexer brackets = P.brackets lexer whiteSpace = P.whiteSpace lexer symbol = P.symbol lexer identifier = P.identifier lexer reserved = P.reserved lexer reservedOp = P.reservedOp lexer integer = P.integer lexer natural = P.natural lexer charLiteral = P.charLiteral lexer stringLiteral = P.stringLiteral lexer
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