[Haskell-cafe] Success and one last issue with Data.Binary

[David Leimbach]

On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk <vanenkj at gmail.com> wrote:

> I think Thomas' point was that some other branch in `getSpecific' is
> running. Is there a chance we can see the rest of `getSpecific'?


Sure:  (In the meantime, I'll try the suggested code from before)

 get = do s <- getWord32le
             mtype <- getWord8
             getSpecific s mtype
        where
          getSpecific s mt
                      | mt == mtRversion = do t <- getWord16le
                                              ms <- getWord32le
                                              ss <- getWord16le
                                              v <-
getRemainingLazyByteString
                                              return $ MessageClient $
Rversion {size=s,

    mtype=mt,

    tag=t,

    msize=ms,

    ssize=ss,

    version=v}
                      | mt == mtRerror = do t <- getWord16le
                                            ss <- getWord16le
                                            e <- getLazyByteString $
fromIntegral ss
                                            return $ MessageClient $ Rerror
{size=s,

mtype=mt,

tag=t,

ssize=ss,

ename=e}



>
>
> On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach <leimy2k at gmail.com> wrote:
> > The thing is I have 19 bytes in the hex string I provided:
> > 1300000065ffff000400000600395032303030
> > That's 38 characters or 19 bytes.
> > The last 4 are 9P2000
> > 13000000  = 4 bytes for 32bit message payload,  This is little endian for
> 19
> > bytes total.
> > 65 = 1 byte for message type.  65 is "Rversion" or the response type for
> a
> > Tversion request
> > ffff = 2 bytes for 16bit message "tag".
> >
> > 00040000 = 4 bytes for the 32 bit maximum message payload size I'm
> > negotiating with the 9P server.  This is little endian for 1024
> > 0600 =  2 bytes for 16 bit value for the length of the "string" I'm
> sending.
> >  The strings are *NOT* null terminated in 9p, and this is little endian
> for
> > 6 bytes remaining.
> > 5032303030 = 6 bytes the ASCII or UTF-8 string "9P2000" which is 6 bytes
> > 4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
> > As far as I can see, my "get" code does NOT ask for a 20th byte, so why
> am I
> > getting that error?
> > get = do s <- getWord32le  -- 4
> >              mtype <- getWord8  -- 1
> >              getSpecific s mtype
> >         where
> >           getSpecific s mt
> >                       | mt == mtRversion = do t <- getWord16le -- 2
> >                                               ms <- getWord32le  -- 4
> >                                               ss <- getWord16le -- 2
> >                                               v <-
> > getRemainingLazyByteString  -- remaining should be 6 bytes.
> >                                               return $ MessageClient $
> > Rversion {size=s,
> >
> >                         mtype=mt,
> >
> >                         tag=t,
> >
> >                         msize=ms,
> >
> >                         ssize=ss,
> >
> >                         version=v}
> > Should I file a bug?  I don't believe I should be seeing an error message
> > claiming a failure at the 20th byte when I've never asked for one.
> > Dave
> >
> > On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk <vanenkj at gmail.com> wrote:
> >>
> >> Thomas,
> >>
> >> You're correct. For some reason, I based my advice on the thought that
> >> 19 was the minimum size instead of 13.
> >>
> >> On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
> >> <thomas.dubuisson at gmail.com> wrote:
> >> >> I think getRemainingLazyByteString expects at least one byte
> >> > No, it works with an empty bytestring.  Or, my tests do with binary
> >> > 0.5.0.1.
> >> >
> >> > The specific error means you are requiring more data than providing.
> >> > First check the length of the bytestring you pass in to the to level
> >> > decode (or 'get') routine and walk though that to figure out how much
> >> > it should be consuming.  I notice you have a guard on the
> >> > 'getSpecific' function, hopefully you're sure the case you gave us is
> >> > the branch being taken.
> >> >
> >> > I think the issue isn't with the code provided.  I cleaned up the code
> >> > (which did change behavior due to the guard and data declarations that
> >> > weren't in the mailling) and it works fine all the way down to the
> >> > expected minimum of 13 bytes.
> >> >
> >> >
> >> >> import Data.ByteString.Lazy
> >> >> import Data.Binary
> >> >> import Data.Binary.Get
> >> >>
> >> >> data RV =
> >> >> Rversion {     size   :: Word32,
> >> >>                mtype  :: Word8,
> >> >>                tag    :: Word16,
> >> >>                msize  :: Word32,
> >> >>                ssize  :: Word16,
> >> >>                version :: ByteString}
> >> >>       deriving (Eq, Ord, Show)
> >> >
> >> >> instance Binary RV where
> >> >>  get = do s <- getWord32le
> >> >>          mtype <- getWord8
> >> >>          getSpecific s mtype
> >> >>   where
> >> >>    getSpecific s mt = do t <- getWord16le
> >> >>                          ms <- getWord32le
> >> >>                          ss <- getWord16le
> >> >>                          v <- getRemainingLazyByteString
> >> >>                          return $ Rversion {size=s,
> >> >>                                             mtype=mt,
> >> >>                                             tag=t,
> >> >>                                             msize=ms,
> >> >>                                             ssize=ss,
> >> >>                                             version=v }
> >> >>  put _ = undefined
> >> >
> >>
> >>
> >>
> >> --
> >> /jve
> >
> >
> > _______________________________________________
> > Haskell-Cafe mailing list
> > Haskell-Cafe at haskell.org
> > http://www.haskell.org/mailman/listinfo/haskell-cafe
> >
> >
>
>
>
> --
> /jve
>
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