Haskell/Polysemy print to StdOut

Question

我正在使用Haskell的Polysemy库，并且在尝试打印到StdOut时遇到了巨大的头痛。这是我的代码：

apiServer :: (Member Trace r, Member (Embed IO) r) => ServerT Api (Sem r)
apiServer = test :<|> values
where
test :: Sem r String
test = do
return "Server up!"
values :: Member Trace r => Sem r Value
values =
do
_ <- trace (show dummyValues)
return dummyValues
api :: Proxy Api
api = Proxy
serverHandler :: Handler [Char] :<|> Handler Value
serverHandler = hoistServer api interpretServer apiServer
where
interpretServer :: Sem '[Trace, Embed IO] a -> Handler a
interpretServer sem =
sem
|> traceToStdout
|> runM

错误为

• Couldn't match expected type: Handler a
with actual type: IO a
• In the expression: sem |> traceToStdout |> runM
In an equation for ‘interpretServer’:
interpretServer sem = sem |> traceToStdout |> runM
In an equation for ‘serverHandler’:
serverHandler
= hoistServer api interpretServer apiServer
where
interpretServer :: Sem '[Trace, Embed IO] a -> Handler a
interpretServer sem = sem |> traceToStdout |> runM
• Relevant bindings include
sem :: Sem '[Trace, Embed IO] a (bound at src/Lib.hs:58:21)
interpretServer :: Sem '[Trace, Embed IO] a -> Handler a
(bound at src/Lib.hs:58:5)

我知道我应该以某种方式解释Embed，但我不知道如何解释。我尝试了runEmbedded，但不知道在第一个arg中放什么。

Answer 1

我设法解决了这个问题，因为我知道先运行traceToStdout，然后运行runM会给我一个IO，而忽略它会给我带来Handler。

我可以通过在末尾添加liftIO来修复它，将IOMonad提升到HandlerMonad:中

serverHandler = hoistServer api interpretServer apiServer
where
interpretServer :: Sem '[Trace, Embed IO] a -> Handler a
interpretServer sem =
sem
|> traceToStdout
|> runM
|> liftIO