使用 State 的管道“运行”

Question

I have a producer:

p:: Producer Message IO r .

I can process all messages using:

runEffect $ for p processMessage

where

processMessage:: Message -> Effect IO () .

How can I implement stateful processing using something like:

processMessage:: Message -> Effect (StateT MyState IO) () ?

Answer 1

Short answer:

1. Modify your producer to be agnostic to the monad it's running in
2. Your processMessage is fine
3. The runEffect returns StateT MyState IO () , you need to evaluate it

Longer answer with a dummy example:

Your producer is locked into the IO monad, you need to modify it to be either in MonadIO m or in the explicit state monad.

import Control.Monad.State
import Pipes

type Message = Int

p :: MonadIO m => Producer Message m ()
p = each [1..10]

The signature of your processMessage is already fine. I'm following your signature and adding some simple logic to exercise the IO and State functionality

processMessage :: Message -> Effect (StateT MyState IO) ()
processMessage msg = do
  modify (+ msg)
  liftIO (print msg)

Then the final step. runEffect:: Monad m => Effect m r -> m r , if you substitute the m with a concrete type, this ends up being runEffect:: Effect (StateT MyState IO) () -> StateT MyState IO () , meaning that you'll be left with the state monad which still needs to be executed. There's three variants for executing the state monad, runStateT , evalStateT and execStateT . I chose execStateT:: StateT MyState IO () -> IO MyState variant in this example, but choose whichever you need in your case.

main :: IO ()
main = do
  st <- execStateT (runEffect $ for p processMessage) 0
  putStrLn $ "End state: " <> show st