Question

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，我会看到CPU消耗不断增长。

如果我改变＆＃34; S.splitAt＆＃34; to＆＃34; splitAt＆＃34;在＆＃34; chunks＆＃34;的定义中，一切都很好。

有人可以猜到为什么会这样吗？

谢谢。

更新

在以下代码中，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

我清理了代码以避免混合普通旧列表和流融合列表。内存/ CPU泄漏仍然存在。要查看代码是否在旧列表上工作，只需删除Prelude import和＆＃34; .Stream＆＃34;在＆＃34; Data.List＆＃34;。

之后

Answer 1

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])

由此可以看出，由于它产生列表而不是流，因此阻碍了进一步融合。我认为，正确的做法是生成一个splitAt来生成流，或者更好地直接在流上编写chunks函数，并使用列表版本中的相应融合规则。

以下是关于流的splitAt我认为应该是好的。您当然需要将它与splitAt列表中的相应重写规则配对，如果这些重写规则变得棘手，可能直接编写chunks函数，尽管看起来有点棘手所以：

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'

使用Haskell和流融合不断增加CPU消耗

更新

1 个答案: