On Wed, Feb 27, 2019 at 09:50:55AM +0800, Ming Lei wrote: > On Wed, Feb 27, 2019 at 07:45:50AM +1100, Dave Chinner wrote: > > On Tue, Feb 26, 2019 at 05:33:04PM +0800, Ming Lei wrote: > > > On Tue, Feb 26, 2019 at 03:58:26PM +1100, Dave Chinner wrote: > > > > On Mon, Feb 25, 2019 at 07:27:37PM -0800, Matthew Wilcox wrote: > > > > > On Tue, Feb 26, 2019 at 02:02:14PM +1100, Dave Chinner wrote: > > > > > > > Or what is the exact size of sub-page IO in xfs most of time? For > > > > > > > > > > > > Determined by mkfs parameters. Any power of 2 between 512 bytes and > > > > > > 64kB needs to be supported. e.g: > > > > > > > > > > > > # mkfs.xfs -s size=512 -b size=1k -i size=2k -n size=8k .... > > > > > > > > > > > > will have metadata that is sector sized (512 bytes), filesystem > > > > > > block sized (1k), directory block sized (8k) and inode cluster sized > > > > > > (32k), and will use all of them in large quantities. > > > > > > > > > > If XFS is going to use each of these in large quantities, then it doesn't > > > > > seem unreasonable for XFS to create a slab for each type of metadata? > > > > > > > > > > > > Well, that is the question, isn't it? How many other filesystems > > > > will want to make similar "don't use entire pages just for 4k of > > > > metadata" optimisations as 64k page size machines become more > > > > common? There are others that have the same "use slab for sector > > > > aligned IO" which will fall foul of the same problem that has been > > > > reported for XFS.... > > > > > > > > If nobody else cares/wants it, then it can be XFS only. But it's > > > > only fair we address the "will it be useful to others" question > > > > first..... > > > > > > This kind of slab cache should have been global, just like interface of > > > kmalloc(size). > > > > > > However, the alignment requirement depends on block device's block size, > > > then it becomes hard to implement as genera interface, for example: > > > > > > block size: 512, 1024, 2048, 4096 > > > slab size: 512*N, 0 < N < PAGE_SIZE/512 > > > > > > For 4k page size, 28(7*4) slabs need to be created, and 64k page size > > > needs to create 127*4 slabs. > > > > IDGI. Where's the 7/127 come from? > > > > We only require sector alignment at most, so as long as each slab > > object is aligned to it's size, we only need one slab for each block > > size. > > Each slab has fixed size, I remembered that you mentioned that the meta > data size can be 512 * N (1 <= N <= PAGE_SIZE / 512). > > https://marc.info/?l=linux-fsdevel&m=155115014513355&w=2 nggggh. *That* *is* *not* *what* *I* *said*. That is *what you said* and I said that was wrong and the actual sizes needed were: dgc> It is not. On a 64k page size machine, we use sub page slabs for dgc> metadata blocks of 2^N bytes where 9 <= N <= 15.. Please do the maths. I shoul dnot have to do it for you. Also, please don't confusing sector-in-LBA alignment with /memory buffer/ alignment. i.e. you're assuming that these 2kB IOs at different sector offsets like the following: 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 +-------+-------+-------+-------+-------+-------+-------+-------+ +ooooooooooooooooooooooooooooooo+ +ooooooooooooooooooooooooooooooo+ +ooooooooooooooooooooooooooooooo+ +ooooooooooooooooooooooooooooooo+ +ooooooooooooooooooooooooooooooo+ require a memory buffer alignment that matches the sector-in-LBA alignment. This is wrong - the memory alignment constraint is usually a hardware DMA engine limitation and has nothing to do with the LBA the IO will place/retreive the data in/from. i.e. An aligned 2kB slab will only allocate 2kB slab objects like so: 0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 +-------+-------+-------+-------+-------+-------+-------+-------+ +ooooooooooooooooooooooooooooooo+ +ooooooooooooooooooooooooooooooo+ and these memory buffers will always have 512 byte alignment. This meets the memory alignemnt requirements of all hardware (which is 512 byte alignment or smaller), and so we only need one slab per size. Cheers, Dave. -- Dave Chinner david@xxxxxxxxxxxxx
