haskell - Haskell和流融合不断增加CPU消耗

这是一个简短的Haskell程序，可产生440 Hz的声音。它使用pulseaudio作为音频后端。

import GHC.Float
import Control.Arrow
import Sound.Pulse.Simple
import qualified Data.List.Stream as S
import Data.List

type Time = Double
type Frequency = Double
type Sample = Double
type CV = Double

chunksize = 441 * 2
sampleRate :: (Fractional a) => a
sampleRate = 44100

integral :: [Double] -> [Double]
integral = scanl1  (\acc x -> acc + x / sampleRate)

chunks :: Int -> [a] -> [[a]]
chunks n = S.takeWhile (not . S.null) . S.unfoldr (Just . S.splitAt n)

pulseaudioOutput :: [Sample] -> IO ()
pulseaudioOutput sx = do

    pa <- simpleNew Nothing "Synths" Play Nothing "Synths PCM output"
         (SampleSpec (F32 LittleEndian) 44100 1) Nothing Nothing

    mapM_ (simpleWrite pa . S.map double2Float) $ chunks 1000 sx

    simpleDrain pa
    simpleFree pa

oscSine :: Frequency -> [CV] ->  [Sample]
oscSine f = S.map sin <<< integral <<< S.map ((2 * pi * f *) . (2**))

music ::[Sample]
music = oscSine 440 (S.repeat 0)

main = do
    pulseaudioOutput music

如果我编译并运行它，我将看到不断增加的CPU消耗。

如果在“块”的定义中将“S.splitAt”更改为“splitAt”，一切都很好。

谁能猜出为什么会这样？

谢谢你。

更新

在下面的代码中，所有三个版本的块都可以产生上述行为:

import GHC.Float
import Control.Arrow
import Sound.Pulse.Simple
import Data.List.Stream

import Prelude hiding ( unfoldr
                      , map
                      , null
                      , scanl1
                      , takeWhile
                      , repeat
                      , splitAt
                      , drop
                      , take
                      )

type Time = Double
type Frequency = Double
type Sample = Double
type CV = Double

chunksize = 441 * 2
sampleRate :: (Fractional a) => a
sampleRate = 44100

integral :: [Double] -> [Double]
integral = scanl1  (\acc x -> acc + x / sampleRate)

chunks :: Int -> [a] -> [[a]]
--chunks n = takeWhile (not . null) . unfoldr (Just . splitAt n)
--chunks n xs = take n xs : chunks n (drop n xs)
chunks n xs = h : chunks n t
    where
        (h, t) = splitAt n xs

pulseaudioOutput :: [Sample] -> IO ()
pulseaudioOutput sx = do

    pa <- simpleNew Nothing "Synths" Play Nothing "Synths PCM output"
         (SampleSpec (F32 LittleEndian) 44100 1) Nothing Nothing

    mapM_ (simpleWrite pa . map double2Float) $ chunks 1000 sx

    simpleDrain pa
    simpleFree pa

oscSine :: Frequency -> [CV] ->  [Sample]
oscSine f = map sin <<< integral <<< map ((2 * pi * f *) . (2**))

music ::[Sample]
music = oscSine 440 (repeat 0)

main = do
    pulseaudioOutput music

我清理了代码，避免将普通的旧列表和流融合列表混在一起。内存/CPU泄漏仍然存在。要查看代码是否适用于旧列表，只需删除“Data.List”之后的Prelude导入和“.Stream”即可。

最佳答案

由融合规则(http://hackage.haskell.org/package/stream-fusion-0.1.2.5/docs/Data-Stream.html#g:12)替换的流上的splitAt具有以下签名:

splitAt :: Int -> Stream a -> ([a], [a])

由此可见，由于它产生列表而不是流，因此阻碍了进一步的融合。我认为，正确的做法是生成可以生成流的splitAt，或者更好的方法是使用列表版本中的适当融合规则直接在流上编写chunks函数。

这是我认为应该不错的流上的splitAt。您当然需要将其与列表中splitAt中的适当重写规则配对，如果这些重写规则变得棘手，则也许直接编写chunks函数，尽管这样做似乎也有些棘手:

splitAt :: Int -> Stream a -> (Stream a, Stream a)
splitAt n0 (Stream next s0)
  | n0 < 0    = (nilStream, (Stream next s0))
  | otherwise = loop_splitAt n0 s0
  where
    nilStream = Stream (const Done) s0
    loop_splitAt  0 !s = (nilStream, (Stream next s))
    loop_splitAt !n !s = case next s of
      Done            -> (nilStream, nilStream)
      Skip    s'      -> loop_splitAt n s'
      Yield x s'      -> (cons x xs', xs'')
        where
          (xs', xs'') = loop_splitAt (n-1) s'