Question

我已经构建了一个小型UDP / protobuf发送器和接收器。我花了一个上午的时间试图找出protobuf解码产生错误的原因，结果却发现发送器（Spoke.hs）发送的数据不正确。

代码使用unpack将Lazy.ByteStrings转换为网络包将发送的字符串。我在Hoogle找到了unpack。它可能不是我正在寻找的函数，但它的描述看起来合适：“O（n）将ByteString转换为字符串。”

Spoke.hs产生以下输出：

chris@gigabyte:~/Dropbox/haskell-workspace/hub/dist/build/spoke$ ./spoke
45
45
["a","8","4a","6f","68","6e","20","44","6f","65","10","d2","9","1a","10","6a","64","6f","65","40","65","78","61","6d","70","6c","65","2e","63","6f","6d","22","c","a","8","35","35","35","2d","34","33","32","31","10","1"]

虽然wireshark告诉我数据包中的数据是：

0a:08:4a:6f:68:6e:20:44:6f:65:10:c3:92:09:1a:10:6a:64:6f:65:40:65:78:61:6d:70:6c:65:2e:63:6f:6d:22:0c:0a:08:35:35:35:2d:34:33:32:31:10

Spoke.hs和Wireshark的长度（45）相同。

Wireshark缺少最后一个字节（值0x01），中心值流不同（在Wireshark中大一个字节）。

Spoke.hs中的

"65","10","d2","9"与Wireshark中的65:10:c3:92:09。

由于0x10是DLE，它让我感到震惊，可能还有一些逃脱，但我不知道为什么。

我对Wireshark有多年的信任，只有几十小时的Haskell经验，所以我认为这是错误的代码。

任何建议表示赞赏。

-- Spoke.hs:

module Main where

import Data.Bits
import Network.Socket -- hiding (send, sendTo, recv, recvFrom)
-- import Network.Socket.ByteString
import Network.BSD
import Data.List
import qualified Data.ByteString.Lazy.Char8 as B
import Text.ProtocolBuffers.Header (defaultValue, uFromString)
import Text.ProtocolBuffers.WireMessage (messageGet, messagePut)
import Data.Char (ord, intToDigit)
import Numeric

import Data.Sequence ((><), fromList)

import AddressBookProtos.AddressBook
import AddressBookProtos.Person
import AddressBookProtos.Person.PhoneNumber
import AddressBookProtos.Person.PhoneType

data UDPHandle = 
     UDPHandle {udpSocket  :: Socket,
                udpAddress :: SockAddr}
opensocket :: HostName             -- ^ Remote hostname, or localhost
           -> String               -- ^ Port number or name
           -> IO UDPHandle         -- ^ Handle to use for logging
opensocket hostname port =
    do -- Look up the hostname and port.  Either raises an exception
       -- or returns a nonempty list.  First element in that list
       -- is supposed to be the best option.
       addrinfos <- getAddrInfo Nothing (Just hostname) (Just port)
       let serveraddr = head addrinfos

       -- Establish a socket for communication
       sock <- socket (addrFamily serveraddr) Datagram defaultProtocol

       -- Save off the socket, and server address in a handle
       return $ UDPHandle sock (addrAddress serveraddr)

john = Person {
  AddressBookProtos.Person.id = 1234,
  name = uFromString "John Doe",
  email = Just $ uFromString "jdoe@example.com",
  phone = fromList [
    PhoneNumber {
      number = uFromString "555-4321",
      type' = Just HOME
    }
  ]
}

johnStr = B.unpack (messagePut john)

charToHex x = showIntAtBase 16 intToDigit (ord x) ""

main::IO()
main = 
    do udpHandle <- opensocket "localhost" "4567"
       sent <- sendTo (udpSocket udpHandle) johnStr (udpAddress udpHandle)
       putStrLn $ show $ length johnStr
       putStrLn $ show sent
       putStrLn $ show $ map charToHex johnStr
       return ()

Answer 1

我在bytestring包中看到的文档将解压缩为将ByteString转换为[Word8]，这与String不同。我希望ByteString和String之间存在一些字节差异，因为String是Unicode数据，而ByteString只是一个有效的字节数组，但unpack不应该能够首先产生String。

所以你可能会在这里犯下Unicode转换，或者当底层数据真的没有并且很少结束时，至少有些东西将它解释为Unicode。

为什么Haskell / unpack搞乱了我的字节？

1 个答案: