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

[David Leimbach]

On Tue, Jun 2, 2009 at 1:32 PM, John Van Enk <vanenkj at gmail.com> wrote:

> Perhaps there's some place in your code that's forcing the lazy read
> to consume more. Perhaps you could replace it with an explict (and
> strict) getBytes[1] in combination with remaining[2]?


Unfortunately, I'm using a Lazy ByteString network IO lib.  So I don't think
going to a strict ByteString is going to be possible.


>
>
> Is there a reason you want to use lazy byte strings rather than
> forcing full consumption? Do the 9P packets generally have a lot of
> trailing useless data?


Nope.  Just I noticed that there was a Network ByteString package that
utilized lazy bytestrings :-).

Even if that's why it's going for a 20th byte, shouldn't that be a bug?  :-)


>
>
> 1.
> http://hackage.haskell.org/packages/archive/binary/0.5.0.1/doc/html/Data-Binary-Get.html#v%3AgetBytes
> 2.
> http://hackage.haskell.org/packages/archive/binary/0.5.0.1/doc/html/Data-Binary-Get.html#v%3Aremaining
>
> On Tue, Jun 2, 2009 at 4:28 PM, David Leimbach <leimy2k at gmail.com> wrote:
> >
> >
> > On Tue, Jun 2, 2009 at 10:24 AM, 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.
> >
> > I've shown that I am not trying to decode more than I'm providing.  I've
> > asked, expliciitly, for 13 bytes, and then "remaining", and the library
> is
> > complaining about the 20th byte.
> >
> >>
> >> 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.
> >
> > The other branch is Rerror, which is a shorter message decode stream.
> >  Unfortunately, I can't get Debug.Trace to show anything to prove it's
> > taking this fork of the code.  I suppose I could unsafePerformIO :-)
> > Perhaps I just need a new version of "binary"??  I'll give it a go and
> try
> > your version.  But I need to decode over a dozen message types, so I will
> > need a case or guard or something.
> > Dave
> >>
> >>
> >> 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
>
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