The concurrent-machines package

[ Tags: benchmark, bsd3, concurrency, control, library ] [ Propose Tags ]

A simple use-case for this library is to run the stages of a pipelined streaming computation concurrently. If data is streaming through multiple processing stages, you might build a machine like

step1 >~> step2 >~> step3

The >~> operator connects the machines on either side with a one-element buffer. This means that data is pulled from upstream sources eagerly (perhaps pulling one more value than will be consumed by downstream), but it also means that each stage can be working simultaneously, increasing throughput of the entire pipeline.

A few small examples are available in the examples directory of the source repository.

[Skip to Readme]


Versions,,, 0.2.0, 0.2.1, 0.2.3,,,, 0.3.0, 0.3.1
Change log
Dependencies async (>=2.0.1 && <2.2), base (>=4.8 && <4.10), containers (==0.5.*), lifted-async (>=0.1 && <0.10), machines (>=0.5 && <0.7), monad-control (==1.0.*), semigroups (>=0.8 && <0.19), time (>=1.4 && <1.9), transformers (>=0.4 && <0.6), transformers-base (>=0.4 && <0.6) [details]
License BSD3
Copyright Copyright (C) 2014 Anthony Cowley
Author Anthony Cowley
Category Concurrency, Control
Source repository head: git clone
Uploaded Mon Jul 3 19:10:13 UTC 2017 by AnthonyCowley
Distributions NixOS:0.3.1
Downloads 878 total (30 in the last 30 days)
Rating 0.0 (0 ratings) [clear rating]
  • λ
  • λ
  • λ
Status Docs available [build log]
Last success reported on 2017-07-03 [all 1 reports]
Hackage Matrix CI





Build test with splot visual output


Use -f <flag> to enable a flag, or -f -<flag> to disable that flag. More info


Maintainer's Corner

For package maintainers and hackage trustees

Readme for concurrent-machines-0.3.1

[back to package description]

A simple example of a pipelined computation whose throughput is improved by concurrently running distinct processing stages is given in examples/Pipeline.hs.

import Data.Time.Clock (getCurrentTime, diffUTCTime)
import Control.Concurrent (threadDelay)
import Control.Monad.IO.Class (MonadIO, liftIO)
import Data.Machine.Concurrent

Suppose we have a worker that performs a computation on its input before producing output. This operation may take some time due to, say, network IO or just CPU load. Here we simulate an operation that takes some time with threadDelay.

worker :: String -> Double -> ProcessT IO () ()
worker name dt = repeatedly $ do _ <- await
                                 liftIO $ do
                                   putStrLn $ name ++ " working on its input"
                                   threadDelay dt'
                                 yield ()
  where dt' = floor $ dt * 1000000

We will use a little helper to time two variations of our test program.

timed :: MonadIO m => m a -> m (a, Double)
timed m = do t1 <- liftIO getCurrentTime
             r <- m
             t2 <- liftIO getCurrentTime
             return (r, realToFrac $ t2 `diffUTCTime` t1)

Now we will run a three-stage pipeline where each stage takes one second to process its input before yielding some output. A sequential execution strategy will thus take three seconds to pass an input through this pipeline. At the top-level, we will request three outputs, each of which will take three seconds to produce, resulting in a total execution time of approximately nine seconds.

main :: IO ()
main = do (r,dt) <- timed . runT . supply (repeat ()) $
            worker "A" 1 ~> worker "B" 1 ~> worker "C" 1 ~> taking 3
          putStrLn $ "Sequentially produced "++show r
          putStrLn $ "Sequential processing took "++show dt++"s"

If we instead run the same arrangement as a pipelined computation, we allow the independent stages to run concurrently, much like the stages in a pipelined CPU.

          (r',dt') <- timed . runT . supply (repeat ()) $
            worker "A" 1 >~> worker "B" 1 >~> worker "C" 1 >~> taking 3
          putStrLn $ "Pipeline produced "++show r'
          putStrLn $ "Pipeline processing took "++show dt'++"s"

With this arrangement, the first output we request takes three seconds to produce as we must wait for an input to pass through the entire length of the pipeline. However, successive outputs are following that first output through the pipeline so that we will produce more output at one second intervals. Therefore it takes is 3 + 2 = 5 seconds to produce three outputs: three seconds for the first output, and one second for each of the following two.

Build Status