Re: [Haskell-cafe] How to use notFollowedBy function in Parsec

Am Montag, 21. November 2005 03:27 schrieb Sara Kenedy: May I suggest endBy anyToken semi ? -- optionally replace semi by "char ';'", if you don't want to skip whitespace I think this is what you want --- stop at the first semicolon. If you want to ignore just a final semicolon, you might use endBy anyToken (optional semi >> eof), if you want to stop at the last semicolon, whatever comes thereafter, you have a problem, you'd need long lookahead. Cheers, Daniel

Thanks for your solution. However, when I try this,

...

str1 :: Parser String str1 = do str <- many anyToken notFollowedBy' semi return str

notFollowedBy' :: Show a => GenParser tok st a -> GenParser tok st () notFollowedBy' p = try $ join $ do a <- try p return (unexpected (show a)) <|> return (return ()) 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

When I compile, it still displays ";" at the end of the string.

Parser> run str1 "Hello ;" "Hello ;"

The reason, as I think, because anyToken accepts any kind of token, it considers ";" as token of its string. Thus, it does not understand notFollowedBy' ???

Do you have any ideas about this ??? Thanks.

On 11/19/05, Andrew Pimlott wrote:

...

On Sat, Nov 19, 2005 at 06:43:48PM -0500, Sara Kenedy wrote:

...

str1 :: Parser String str1 = do {str <- many anyToken; notFollowedBy semi; return str}

However, when I compile, there is an error.

ERROR "Test.hs":17 - Type error in application *** Expression : notFollowedBy semi *** Term : semi *** Type : GenParser Char () String *** Does not match : GenParser [Char] () [Char]

The problem is that notFollowedBy has type

notFollowedBy :: Show tok => GenParser tok st tok -> GenParser tok st ()

ie, the result type of the parser you pass to notFollowedBy has to be the same as the token type, in this case Char. (The reason for this type is obscure.) But semi has result type String. You could fix the type error by returning a dummy Char:

str1 = do {str <- many anyToken ; notFollowedBy (semi >> return undefined) ; return str}

I think this will even work; however notFollowedBy is a pretty squirrelly function. There was a discussion about it:

http://www.haskell.org/pipermail/haskell/2004-February/013621.html

Here is a version (which came out of that thread) with a nicer type, that probably also works more reliably (though I won't guarantee it):

notFollowedBy' :: Show a => GenParser tok st a -> GenParser tok st () notFollowedBy' p = try $ join $ do a <- try p return (unexpected (show a)) <|> return (return ())

Andrew

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