Question

我在“Real World Haskell”（第7章，第189页）中读到了IO缓冲，并尝试测试不同的缓冲大小如何影响性能。

import System.IO
import Data.Time.Clock
import Data.Char(toUpper)

main :: IO ()
main = do
  hInp <- openFile "bigFile.txt" ReadMode
  let bufferSize = truncate $ 2**10
  hSetBuffering hInp (BlockBuffering (Just bufferSize))
  bufferMode <- hGetBuffering hInp
  putStrLn $ "Current buffering mode: " ++ (show bufferMode)

  startTime <- getCurrentTime
  inp <- hGetContents hInp
  writeFile "processed.txt" (map toUpper inp)
  hClose hInp
  finishTime <- getCurrentTime
  print $ diffUTCTime finishTime startTime
  return ()

然后我创建了一个“bigFile.txt”

-rw-rw-r-- 1 user user 96M янв.  26 09:49 bigFile.txt

并针对此文件运行我的程序，缓冲区大小不同：

Current buffering mode: BlockBuffering (Just 32)
9.744967s   

Current buffering mode: BlockBuffering (Just 1024)
9.667924s                                      

Current buffering mode: BlockBuffering (Just 1048576)
9.494807s    

Current buffering mode: BlockBuffering (Just 1073741824)
9.792453s

但程序运行时间几乎相同。这是正常的，还是我做错了什么？

Answer 1

在现代操作系统上，由于1）内核执行预读，并且2）如果您已经读过，文件可能已经在页面缓存中，缓冲区大小可能对线性读取文件几乎没有影响最近的文件。

这是一个测量缓冲对写入的影响的程序。典型的结果是：

$ ./mkbigfile 32      -- 12.864733s
$ ./mkbigfile 64      --  9.668272s
$ ./mkbigfile 128     --  6.993664s
$ ./mkbigfile 512     --  4.130989s
$ ./mkbigfile 1024    --  3.536652s
$ ./mkbigfile 16384   --  3.055403s
$ ./mkbigfile 1000000 --  3.004879s

来源：

{-# LANGUAGE OverloadedStrings #-}

import qualified Data.ByteString as BS
import Data.ByteString (ByteString)
import Control.Monad
import System.IO
import System.Environment
import Data.Time.Clock

main = do
  (arg:_) <- getArgs
  let size = read arg
  let bs = "abcdefghijklmnopqrstuvwxyz"
      n = 96000000 `div` (length bs)
  h <- openFile "bigFile.txt" WriteMode
  hSetBuffering h (BlockBuffering (Just size))
  startTime <- getCurrentTime
  replicateM_ n $ hPutStrLn h bs
  hClose h
  finishTime <- getCurrentTime
  print $ diffUTCTime finishTime startTime
  return ()

如何通过缓冲加速Haskell IO？

1 个答案: