Inserting a character into the stream can be expensive if for example
the stream is a ByteString.
I tried the following crazy solution and it seems that it works:
succeed :: Parser ()
succeed = mkPT $ \st ->
return $ Consumed $ return $ Ok () st $ unknownError st
succeed is a parser that always succeeds without really consuming any
input but it also resets the error state.
I really have no understating of how mkPT works. Can someone tell me
if this is a bad idea?
On Wed, Aug 8, 2012 at 4:09 PM, Nick Vanderweit
I found a similar question asked in June 2009 on the haskell-beginners archives, titled "Clearing Parsec error messages." A hack that was proposed (http://www.haskell.org/pipermail/beginners/2009-June/001809.html) was to insert a dummy character into the stream, consume it, and then fail. Still, I'd like to see if there is a cleaner way to modify the error state in the Parsec monad.
NIck
On Wednesday, August 08, 2012 03:24:31 PM silly8888 wrote:
I am trying to create a parsec parser that parses an integer and then checks if that integer has the right size. If not, it generates an error. I tried the following:
8<--------------------------------------------------------------- import Text.Parsec import Text.Parsec.String
integer :: Parser Int integer = do s <- many1 digit let n = read s if n > 65535 then parserFail "integer overflow" else return n 8<---------------------------------------------------------------
The problem is that when I try this
parse integer "" "70000"
I get the following error:
Left (line 1, column 6): unexpected end of input expecting digit integer overflow
ie there are three error messages but I only want the last one. Is there something I can do about this?
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