On Fri, Sep 9, 2011 at 2:48 PM, Zhi Yong Wu <zwu.ker...@gmail.com> 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. > 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. Stefan
