[Haskell-cafe] Success and one last issue with Data.Binary
John Van Enk
vanenkj at gmail.com
Tue Jun 2 16:28:43 EDT 2009
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
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
>
>
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--
/jve
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