On Tue, Sep 13, 2011 at 9:31 AM, Zhi Yong Wu <zwu.ker...@gmail.com> wrote:
> On Tue, Sep 13, 2011 at 3:15 PM, Stefan Hajnoczi
> <stefa...@linux.vnet.ibm.com> wrote:
>> On Tue, Sep 13, 2011 at 10:38:28AM +0800, Zhi Yong Wu wrote:
>>> On Fri, Sep 9, 2011 at 6:38 PM, Stefan Hajnoczi
>>> <stefa...@linux.vnet.ibm.com> wrote:
>>> > On Fri, Sep 09, 2011 at 05:44:36PM +0800, Zhi Yong Wu wrote:
>>> >> Today, i did some basical I/O testing, and suddenly found that qemu
>>> >> write and rw speed is so low now, my qemu binary is built on commit
>>> >> 344eecf6995f4a0ad1d887cec922f6806f91a3f8.
>>> >>
>>> >> Do qemu have regression?
>>> >>
>>> >> The testing data is shown as below:
>>> >>
>>> >> 1.) write
>>> >>
>>> >> test: (g=0): rw=write, bs=512-512/512-512, ioengine=libaio, iodepth=1
>>> >
>>> > Please post your QEMU command-line. If your -drive is using
>>> > cache=writethrough then small writes are slow because they require the
>>> > physical disk to write and then synchronize its write cache. Typically
>>> > cache=none is a good setting to use for local disks.
>>> >
>>> > The block size of 512 bytes is too small. Ext4 uses a 4 KB block size,
>>> > so I think a 512 byte write from the guest could cause a 4 KB
>>> > read-modify-write operation on the host filesystem.
>>> >
>>> > You can check this by running btrace(8) on the host during the
>>> > benchmark. The blktrace output and the summary statistics will show
>>> > what I/O pattern the host is issuing.
>>> 8,2 0 1 0.000000000 337 A WS 425081504 + 8 <-
>>> (253,1) 42611360
>>
>> 8 blocks = 8 * 512 bytes = 4 KB
> How do you know each block size is 512 bytes?
The blkparse format specifier for blocks is 'n'. Here is the code to
print it from blkparse_fmt.c:
case 'n':
fprintf(ofp, strcat(format, "u"), t_sec(t));
And t_sec() is:
#define t_sec(t) ((t)->bytes >> 9)
So it divides the byte count by 512. Block size == sector size == 512 bytes.
You can get the blktrace source code here:
http://brick.kernel.dk/snaps/
Stefan
