2

我对 Haskell 中的协议缓冲区有疑问。我正在编写一个简单的 UDP 接收器并得到错误：

src/Main.hs:39:25:
    Ambiguous type variable `msg0' in the constraints:
      (Wire msg0)
        arising from a use of `messageGet' at src/Main.hs:39:25-34
      (Text.ProtocolBuffers.Reflections.ReflectDescriptor msg0)
        arising from a use of `messageGet' at src/Main.hs:39:25-34
    Probable fix: add a type signature that fixes these type variable(s)
    In the expression: (messageGet (B.pack mesg))
    In the second argument of `($)', namely
      `case (messageGet (B.pack mesg)) of {
         Left person -> putStrLn $ show person
         Right err -> error $ "Failed to parse address book." }'
    In a stmt of a 'do' block:
      return
      $ case (messageGet (B.pack mesg)) of {
          Left person -> putStrLn $ show person
          Right err -> error $ "Failed to parse address book." }

我怎样才能听从它的建议？（我只是在学习 Haskell。）

我的代码如下：

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, Wire)

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

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

import Network.Socket
import System.Posix.Directory
import System.Posix.Files
import System.Posix.IO
import System.Posix.Process
import System.Exit


echoserver :: IO ()
echoserver = do
           withSocketsDo $ do
                   sock <- socket AF_INET Datagram 0
                   bindSocket sock (SockAddrInet 4567 iNADDR_ANY)
                   socketRx sock

socketRx :: Socket -> IO ()
socketRx sock = do

         (mesg, recv_count, client) <- recvFrom sock 1500

         return $ case (messageGet (B.pack mesg)) of
                       Left person -> putStrLn $ show person
                       Right err   -> error $ "Failed to parse address book."
         socketRx sock



main::IO()
main = echoserver
4

2 回答 2

3

从http://hackage.haskell.org/packages/archive/protocol-buffers/2.0.9/doc/html/Text-ProtocolBuffers-WireMessage.html#v:messageGet的文档中，类型签名messageGet是

messageGet :: (ReflectDescriptor msg, Wire msg) => ByteString -> Either String (msg, ByteString)

返回值要么是String错误消息，要么是 amsg和一个残差ByteString。在您的代码中，您已经编写了

case messageGet (B.pack mesg) of
  Left person -> putStrLn $ show person
  Right err   -> error "Failed to parse address book."

如果它返回 a (msg, ByteString)，则该值绑定到变量err。由于err被忽略，因此msg无法确定实际类型，这就是错误消息告诉您的内容。事实上，任何类型的实例都Wire可以ReflectDescriptor工作，但程序对于每种类型的行为都会有所不同！由于编译器不知道你想要什么类型，你必须指定它。您可以通过注释的返回类型来指定它messageGet。

case messageGet (B.pack mesg) :: Either String (X, ByteString) of -- Use the actual message type in place of 'X'
  Left person -> putStrLn $ show person
  Right err   -> error "Failed to parse address book."

您可能还打算在代码中切换Left和Right情况。 Right是非错误情况（认为“错误”和“正确”）。切换案例不会自行消除错误消息。

于 2012-06-14T13:38:45.503 回答
1

像这样添加类型注释：

return $ case messageGet (B.pack mesg) :: Either Something SomethingElse of

（我不熟悉协议缓冲区，所以我不知道应该是什么类型——用你需要的实际类型替换它们。Something）SomethingElse

于 2012-06-14T13:32:36.630 回答