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

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 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 wrote:

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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.

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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)

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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