[Haskell-cafe] parsec: parserFail & multiple error messages

[silly8888]

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
<nick.vanderweit at gmail.com> wrote:
> 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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>
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