Am Donnerstag 20 August 2009 13:44:15 schrieb Martijn van Steenbergen:
Goedemiddag café,
Consider the following function, using parsec-3.0.0:
la :: Parsec String () (Maybe Char) la = lookAhead (optionMaybe anyChar)
*Lookahead> parseTest (char 'a' <|> char 'b') "a" 'a' *Lookahead> parseTest (char 'a' <|> char 'b') "b" 'b' *Lookahead> parseTest (la *> char 'a' <|> char 'b') "a" 'a' *Lookahead> parseTest (la *> char 'a' <|> char 'b') "b" parse error at (line 1, column 2): unexpected "b" expecting "a"
The first three work fine and as expected, but the fourth example fails where I would expect success. I know <|> won't try the rhs if the lhs consumed input, but lookAhead's documentation promises not to consume any input. Is this a bug in Parsec or am I missing something?
Bad bug in Parsec (from the beginning, the same happens in parsec-2), I'd say. Desugared, we have lookAhead p = getParserState >>= \st -> p >>= \r -> setParserState st >>= \_ -> return r Due to the (>>=), whenever p consumes input, lookAhead will return (Consumed _) and there's no way to get rid of it, so (la *> char 'a') returns Consumed (Error something) on the input "b" and (<|>) doesn't try char 'b'. The code for lookAhead should look something like (parsec-2, to avoid 'returns' cluttering the code): lookAhead p = Parser $ \st -> case parserReply $ runP p st of Ok x s err -> Empty (Ok x st err) Error err -> Empty (Error err) Since exporting an 'unconsume' function wouldn't be desirable, lookAhead would have to move to Text.Parse(rCombinators.Parse)c.Prim. (not necessary in parsec-3 yet, since that exports all top level definitions from all modules so far).
Thanks,
Martijn.