我使用的是Pipes.Files,而后者又依赖于Pipes.Safe。我有这样的管道:
import Pipes
import Pipes.Files
import Pipes.Safe
import qualified Pipes.Prelude as P
allFiles :: (MonadIO m, MonadSafe m) => FilePath -> Producer FilePath m ()
allFiles path = find path (glob "*.hs" <> regular)
main :: IO ()
main = do
args <- parseArgsOrExit patterns =<< getArgs -- docopt stuff, not relevant
let ins = args `getAllArgs` argument "paths"
conf <- readConfig args
forM_ ins $ \path -> do
let source = allFiles path -- here lies the problem
>-> P.mapM (liftIO . analyze conf)
>-> P.map (filterResults conf)
>-> P.filter filterNulls
runSafeT $ runEffect $ exportStream conf source
问题是analyze
的类型是
analyze :: Config -> FilePath -> IO Result
allFiles
有MonadSafe m
和MonadIO m
限制。理论上liftIO
应该注意这一点,但我得到了这个错误:
Main.hs:45:22:
No instance for (Pipes.Safe.MonadSafe IO)
arising from a use of ‘allFiles’
In the first argument of ‘(>->)’, namely ‘allFiles path’
In the first argument of ‘(>->)’, namely
‘allFiles path >-> P.mapM (liftIO . analyze conf)’
In the first argument of ‘(>->)’, namely
‘allFiles path >-> P.mapM (liftIO . analyze conf)
>-> P.map (filterResults conf)’
Main.hs:49:20:
Couldn't match type ‘IO’ with ‘Pipes.Safe.SafeT IO’
Expected type: Pipes.Safe.SafeT IO ()
Actual type: IO ()
In the second argument of ‘($)’, namely
‘runEffect $ exportStream conf source’
In a stmt of a 'do' block:
runSafeT $ runEffect $ exportStream conf source
In the expression:
do { let source
= allFiles path >-> P.mapM (liftIO . analyze conf)
>-> P.map (filterResults conf)
>-> P.filter filterNulls;
runSafeT $ runEffect $ exportStream conf source }
还有第二个错误,我现在还不知道如何解释。
编辑:以下是其他类型的签名:
filterResults :: Config -> Result -> Result
filterNulls :: Result -> Bool
exportStream :: Config -> Producer Result IO () -> Effect IO ()