我正在尝试执行以下操作：
sourceIRC
:: (MonadBaseControl IO m, MonadLogger m)
=> NetworkSettings
-> Producer (ConnectionT m) Message
sourceIRC networkSettings = do
withConnectionForever networkSettings $ \socket -> do
bracket (liftBase $ Network.socketToHandle socket IO.ReadWriteMode)
(\handle -> liftBase $ IO.hClose handle)
(\handle -> do
mvar <- newMVar False
bracket (fork $ do
threadDelay 5000000
_ <- swapMVar mvar True
return ())
(killThread)
(\_ -> runConnectionT handle (sourceHandle handle))
takeMVar mvar)
如您所见，我正在尝试Producer根据原语创建 a withConnectionForever。该原语的类型为：
withConnectionForever
:: (MonadBaseControl IO m, MonadLogger m)
=> NetworkSettings
-> (Network.Socket -> m Bool)
-> m ()
可以想象，我在编译时收到一条错误消息！这是：
Haskell/IRC.hs:128:54:
Couldn't match expected type `ConnectionT m0 a0'
with actual type `ConduitM i0 ByteString.ByteString m1 ()'
In the return type of a call of `sourceHandle'
In the second argument of `runConnectionT', namely
`(sourceHandle handle)'
In the expression: runConnectionT handle (sourceHandle handle)
现在，我知道调用的类型withConnectionForever显然不是一个管道，但我希望它可以成为一个管道，因为管道也是一个单子并且withConnectionForever使用自由单子而不是硬编码一。我对消息试图传达的内容的理解是，这并没有发生，我想知道为什么以及我能做些什么。
为了完整起见，这里是原语的来源：
withConnectionForever
:: (MonadBaseControl IO m, MonadLogger m)
=> NetworkSettings
-> (Network.Socket -> m Bool)
-> m ()
withConnectionForever networkSettings action = do
let loop nFailures = do
maybeSocket <- newConnection networkSettings
case maybeSocket of
Nothing -> return ()
Just socket -> do
terminatedNormally <- action socket
if terminatedNormally
then loop 0
else do
exponentTwiddle <- liftBase $ Random.randomRIO (0, 100)
let exponent =
1.25 + fromIntegral (exponentTwiddle - 50) / 100.0
delay = floor $ 1000000.0 *
((0.5 ** (fromIntegral nFailures * negate exponent))
- 1.0 :: Double)
$(logInfo) (Text.concat
["Abnormal disconnection from the network ",
networkSettingsName networkSettings,
"; pausing attempts for ",
Text.pack $ show $ fromIntegral delay / 1000000.0,
" seconds..."])
liftBase $ threadDelay delay
loop (nFailures + 1)
loop 0
我真的不想重写原语，除非它可以以微创方式完成，但我想那是在桌面上。
提前致谢！