Re: [Qemu-devel] Why qemu write/rw speed is so low?

[Zhi Yong Wu]

On Fri, Sep 9, 2011 at 9:54 PM, Stefan Hajnoczi <address@hidden> wrote:
> On Fri, Sep 9, 2011 at 2:48 PM, Zhi Yong Wu <address@hidden> wrote:
>> On Fri, Sep 9, 2011 at 6:38 PM, Stefan Hajnoczi
>> <address@hidden> 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.
>> Now i can not access my workstation in the office.
>> -drive if=virtio,cache=none,file=xxxx
>>
>>>
>>> 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 mean RCU? What is its work procedure? Can you explain in more
>> details if you are available?
>
> If the host file system manages space in 4 KB blocks, then a 512 byte
> to an unallocated part of the file causes the file system to find 4 KB
> of free space for this data.  Since the write is only 512 bytes and
> does not cover the entire 4 KB region, the file system initializes the
> remaining 3.5 KB with zeros and writes out the full 4 KB block.
>
> Now if a 512 byte write comes in for an allocated 4 KB block, then we
> need to read in the existing 4 KB, modify the 512 bytes in place, and
> write out the 4 KB block again.  This is read-modify-write.  In this
> worst-case scenario a 512 byte write turns into a 4 KB read followed
> by a 4 KB write.
A 512B write will lead to a 4KB read + 512B modify + a 4KB write.

got it. thanks.
>
> Stefan
>



-- 
Regards,

Zhi Yong Wu