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@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@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