[英]How to execute an action periodically in a GHCJS program?
应该通过Javascript使用setInterval
,还是使用一些基于线程的惯用解决方案?
使用setInterval
提出了亚历山大,Erik和Luite自己的一些挑战和评论让我尝试了线程。 这无缝地工作,代码非常干净,类似于以下内容:
import Control.Concurrent( forkIO, threadDelay )
import Control.Monad( forever )
... within an IO block
threadId <- forkIO $ forever $ do
threadDelay (60 * 1000 * 1000) -- one minute in microseconds, not milliseconds like in Javascript!
doWhateverYouLikeHere
Haskell具有轻量级线程的概念,因此这是以异步方式运行动作的惯用Haskell方式,就像使用Javascript setInterval
或setTimeout
。
如果你不关心动机,只需滚动到我的最佳解决方案runPeriodicallyConstantDrift
下面。 如果您更喜欢结果较差的简单解决方案,请参阅runPeriodicallySmallDrift
。
我的回答不是针对GHCJS的,并且没有经过GHCJS测试,只有GHC,但它说明了OP天真解决方案的问题 。
runPeriodicallyBigDrift
解决方案: runPeriodicallyBigDrift
这是我的OP解决方案的版本,用于下面的比较:
import Control.Concurrent ( threadDelay )
import Control.Monad ( forever )
-- | Run @action@ every @period@ seconds.
runPeriodicallyBigDrift :: Double -> IO () -> IO ()
runPeriodicallyBigDrift period action = forever $ do
action
threadDelay (round $ period * 10 ** 6)
假设“定期执行一个动作”意味着动作每隔几个周期开始一次, OP的解决方案threadDelay
问题,因为threadDelay
没有考虑动作本身所花费的时间。 在n次迭代之后,动作的开始时间将至少漂移到运行动作n次所需的时间!
runPeriodicallySmallDrift
解决方案: runPeriodicallySmallDrift
因此,如果我们实际上想要在每个时段开始新的操作,我们需要考虑操作所需的时间。 如果周期与生成线程所需的时间相比相对较大,那么这个简单的解决方案可能对您有用:
import Control.Concurrent ( threadDelay )
import Control.Concurrent.Async ( async, link )
import Control.Monad ( forever )
-- | Run @action@ every @period@ seconds.
runPeriodicallySmallDrift :: Double -> IO () -> IO ()
runPeriodicallySmallDrift period action = forever $ do
-- We reraise any errors raised by the action, but
-- we don't check that the action actually finished within one
-- period. If the action takes longer than one period, then
-- multiple actions will run concurrently.
link =<< async action
threadDelay (round $ period * 10 ** 6)
在我的实验中(下面有更多细节),在我的系统上生成一个线程需要大约0.001秒,因此在n次迭代之后runPeriodicallySmallDrift
的漂移大约是千分之一秒,这在某些用例中可以忽略不计。
runPeriodicallyConstantDrift
最后,假设我们只需要恒定漂移,这意味着漂移总是小于某个常数,并且不会随着周期动作的迭代次数而增长。 我们可以通过跟踪自开始以来的总时间来实现恒定漂移,并且当总时间是周期的n
倍时开始第n
次迭代:
import Control.Concurrent ( threadDelay )
import Data.Time.Clock.POSIX ( getPOSIXTime )
import Text.Printf ( printf )
-- | Run @action@ every @period@ seconds.
runPeriodicallyConstantDrift :: Double -> IO () -> IO ()
runPeriodicallyConstantDrift period action = do
start <- getPOSIXTime
go start 1
where
go start iteration = do
action
now <- getPOSIXTime
-- Current time.
let elapsed = realToFrac $ now - start
-- Time at which to run action again.
let target = iteration * period
-- How long until target time.
let delay = target - elapsed
-- Fail loudly if the action takes longer than one period. For
-- some use cases it may be OK for the action to take longer
-- than one period, in which case remove this check.
when (delay < 0 ) $ do
let msg = printf "runPeriodically: action took longer than one period: delay = %f, target = %f, elapsed = %f"
delay target elapsed
error msg
threadDelay (round $ delay * microsecondsInSecond)
go start (iteration + 1)
microsecondsInSecond = 10 ** 6
基于以下实验,漂移总是大约1/1000秒,与动作的迭代次数无关。
为了比较这些解决方案,我们创建了一个动作来跟踪它自己的漂移并告诉我们,并在上面的每个runPeriodically*
实现中运行它:
import Control.Concurrent ( threadDelay )
import Data.IORef ( newIORef, readIORef, writeIORef )
import Data.Time.Clock.POSIX ( getPOSIXTime )
import Text.Printf ( printf )
-- | Use a @runPeriodically@ implementation to run an action
-- periodically with period @period@. The action takes
-- (approximately) @runtime@ seconds to run.
testRunPeriodically :: (Double -> IO () -> IO ()) -> Double -> Double -> IO ()
testRunPeriodically runPeriodically runtime period = do
iterationRef <- newIORef 0
start <- getPOSIXTime
startRef <- newIORef start
runPeriodically period $ action startRef iterationRef
where
action startRef iterationRef = do
now <- getPOSIXTime
start <- readIORef startRef
iteration <- readIORef iterationRef
writeIORef iterationRef (iteration + 1)
let drift = (iteration * period) - (realToFrac $ now - start)
printf "test: iteration = %.0f, drift = %f\n" iteration drift
threadDelay (round $ runtime * 10**6)
以下是测试结果。 在每种情况下,测试运行0.05秒的动作,并使用两倍的时间,即0.1秒。
对于runPeriodicallyBigDrift
,n次迭代后的漂移大约是单次迭代的运行时间的n倍,如预期的那样。 在100次迭代之后,漂移为-5.15,并且仅从动作的运行时间预测的漂移为-5.00:
ghci> testRunPeriodically runPeriodicallyBigDrift 0.05 0.1
...
test: iteration = 98, drift = -5.045410253
test: iteration = 99, drift = -5.096661091
test: iteration = 100, drift = -5.148137684
test: iteration = 101, drift = -5.199764033999999
test: iteration = 102, drift = -5.250980596
...
对于runPeriodicallySmallDrift
,n次迭代后的漂移大约为0.001秒,可能是在我的系统上生成线程所需的时间:
ghci> testRunPeriodically runPeriodicallySmallDrift 0.05 0.1
...
test: iteration = 98, drift = -0.08820333399999924
test: iteration = 99, drift = -0.08908210599999933
test: iteration = 100, drift = -0.09006684400000076
test: iteration = 101, drift = -0.09110764399999915
test: iteration = 102, drift = -0.09227584299999947
...
对于runPeriodicallyConstantDrift
,漂移在大约0.001秒内保持不变(加上噪声):
ghci> testRunPeriodically runPeriodicallyConstantDrift 0.05 0.1
...
test: iteration = 98, drift = -0.0009586619999986112
test: iteration = 99, drift = -0.0011010979999994674
test: iteration = 100, drift = -0.0011610369999992542
test: iteration = 101, drift = -0.0004908619999977049
test: iteration = 102, drift = -0.0009897379999994627
...
如果我们关心恒定漂移的水平,那么更复杂的解决方案可以跟踪平均恒定漂移并对其进行调整。
在实践中,我意识到我的一些循环具有从一次迭代传递到下一次迭代的状态。 这是runPeriodicallyConstantDrift
的一个小概括,以支持:
import Control.Concurrent ( threadDelay )
import Data.IORef ( newIORef, readIORef, writeIORef )
import Data.Time.Clock.POSIX ( getPOSIXTime )
import Text.Printf ( printf )
-- | Run a stateful @action@ every @period@ seconds.
--
-- Achieves uniformly bounded drift (i.e. independent of the number of
-- iterations of the action) of about 0.001 seconds,
runPeriodicallyWithState :: Double -> st -> (st -> IO st) -> IO ()
runPeriodicallyWithState period st0 action = do
start <- getPOSIXTime
go start 1 st0
where
go start iteration st = do
st' <- action st
now <- getPOSIXTime
let elapsed = realToFrac $ now - start
let target = iteration * period
let delay = target - elapsed
-- Warn if the action takes longer than one period. Originally I
-- was failing in this case, but in my use case we sometimes,
-- but very infrequently, take longer than the period, and I
-- don't actually want to crash in that case.
when (delay < 0 ) $ do
printf "WARNING: runPeriodically: action took longer than one period: delay = %f, target = %f, elapsed = %f"
delay target elapsed
threadDelay (round $ delay * microsecondsInSecond)
go start (iteration + 1) st'
microsecondsInSecond = 10 ** 6
-- | Run a stateless @action@ every @period@ seconds.
--
-- Achieves uniformly bounded drift (i.e. independent of the number of
-- iterations of the action) of about 0.001 seconds,
runPeriodically :: Double -> IO () -> IO ()
runPeriodically period action =
runPeriodicallyWithState period () (const action)
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