On Tue, Jun 2, 2009 at 1:36 PM, John Van Enk <[email protected]> wrote:
> What happens if you use `getRemainingLazyByteString' in your error > branch instead of `getLazyByteString'? > I actually am using getRemainingLazyByteString right now, and it still thinks I'm asking for a 20th byte. if I delete the other guarded branch of that function, it still fails saying I'm asking for the 20th byte. Dave > > On Tue, Jun 2, 2009 at 4:31 PM, David Leimbach <[email protected]> wrote: > > > > > > On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk <[email protected]> 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 <[email protected]> > 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 <[email protected]> > 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 > >> >> <[email protected]> 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 > >> > [email protected] > >> > http://www.haskell.org/mailman/listinfo/haskell-cafe > >> > > >> > > >> > >> > >> > >> -- > >> /jve > > > > > > > > -- > /jve >
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