自动选择文件I / O的缓冲区大小

时间:2012-07-09 12:02:12

标签: java c++ c file-io buffer

我有一个非常基本的怀疑。通常,我必须编写使用缓冲文件I / O的应用程序,并且每当我面临选择缓冲区大小的困境时,我最终会通过非常讨厌的结果进行反复试验。我想知道是否有任何方法或算法可以根据Teracopy在Windows中处理文件时所基于的底层平台自动确定作业的最佳缓冲区大小。我主要使用Qt作为GUI。

如果可能的话,非常感谢C / C ++ / C#/ Java中的一个小例子!

谢谢!

2 个答案:

答案 0 :(得分:15)

在Java中,最佳值通常在L1高速缓存大小附近,通常为32 KB。在Java中,至少选择1024字节或1 MB没有太大区别(<20%)

如果您按顺序读取数据,通常您的操作系统足够智能,可以检测到这些并为您预取数据。

您可以做的是以下内容。该测试似乎显示了所用块大小的显着差异。

public static void main(String... args) throws IOException {
    for (int i = 512; i <= 2 * 1024 * 1024; i *= 2)
        readWrite(i);
}

private static void readWrite(int blockSize) throws IOException {
    ByteBuffer bb = ByteBuffer.allocateDirect(blockSize);
    long start = System.nanoTime();
    FileChannel out = new FileOutputStream("deleteme.dat").getChannel();
    for (int i = 0; i < (1024 << 20); i += blockSize) {
        bb.clear();
        while (bb.remaining() > 0)
            if (out.write(bb) < 1) throw new AssertionError();
    }
    out.close();
    long mid = System.nanoTime();
    FileChannel in = new FileInputStream("deleteme.dat").getChannel();
    for (int i = 0; i < (1024 << 20); i += blockSize) {
        bb.clear();
        while (bb.remaining() > 0)
            if (in.read(bb) < 1) throw new AssertionError();
    }
    in.close();
    long end = System.nanoTime();
    System.out.printf("With %.1f KB block size write speed %.1f MB/s, read speed %.1f MB/s%n",
            blockSize / 1024.0, 1024 * 1e9 / (mid - start), 1024 * 1e9 / (end - mid));
}

打印

With 0.5 KB block size write speed 96.6 MB/s, read speed 169.7 MB/s
With 1.0 KB block size write speed 154.2 MB/s, read speed 312.2 MB/s
With 2.0 KB block size write speed 201.5 MB/s, read speed 438.7 MB/s
With 4.0 KB block size write speed 288.0 MB/s, read speed 733.9 MB/s
With 8.0 KB block size write speed 318.4 MB/s, read speed 711.8 MB/s
With 16.0 KB block size write speed 540.6 MB/s, read speed 1263.7 MB/s
With 32.0 KB block size write speed 726.0 MB/s, read speed 1370.9 MB/s
With 64.0 KB block size write speed 801.8 MB/s, read speed 1536.5 MB/s
With 128.0 KB block size write speed 857.5 MB/s, read speed 1539.6 MB/s
With 256.0 KB block size write speed 794.0 MB/s, read speed 1781.0 MB/s
With 512.0 KB block size write speed 676.2 MB/s, read speed 1221.4 MB/s
With 1024.0 KB block size write speed 886.3 MB/s, read speed 1501.5 MB/s
With 2048.0 KB block size write speed 784.7 MB/s, read speed 1544.9 MB/s

此测试未显示的是硬盘驱动器仅支持60 MB / s读取和40 MB / s写入。您正在测试的是缓存内外的速度。如果这是您唯一的优先级,您将使用内存映射文件。

int blockSize = 32 * 1024;
ByteBuffer bb = ByteBuffer.allocateDirect(blockSize);
FileChannel out = new FileOutputStream("deleteme.dat").getChannel();
for (int i = 0; i < (1024 << 20); i += blockSize) {
    bb.clear();
    while (bb.remaining() > 0)
        if (out.write(bb) < 1) throw new AssertionError();
}
out.close();

long start = System.nanoTime();
FileChannel in = new FileInputStream("deleteme.dat").getChannel();
MappedByteBuffer map = in.map(FileChannel.MapMode.READ_ONLY, 0, in.size());
in.close();
long end = System.nanoTime();
System.out.printf("Mapped file at a rate of %.1f MB/s%n",
        1024 * 1e9 / (end - start));

打印

Mapped file at a rate of 589885.5 MB/s

这是如此之快,因为它只是将OS磁盘缓存中的数据直接映射到应用程序的内存中(因此不需要复制)

答案 1 :(得分:1)

我在C:

中看到了这段代码
#include <sys/types.h>
#include <sys/stat.h>
#include <unistd.h>
#include <stdio.h>

int main()
{
  struct stat fi;
  stat("/", &fi);
  printf("%d\n", fi.st_blksize);
  return 0;
}

返回最佳块大小。你需要用它来做那件事。我使用流源到16 *块大小的目标,以获得最佳性能。因为这个测试将通过一些硬件/操作系统在空闲时显示出最佳效果。但不是真实的情况。