With xfs file system that has about 1 million files, would the default value
for the directory structure be sufficient? We can remove the "-n
size=<value>'" option and just use the default.
Thanks,
-Al
-----Original Message-----
From: Dave Chinner [mailto:david@xxxxxxxxxxxxx]
Sent: Monday, October 12, 2015 5:23 PM
To: Al Lau (alau2)
Cc: xfs@xxxxxxxxxxx
Subject: Re: mkfs.xfs -n size=65536
On Fri, Oct 09, 2015 at 10:40:00PM +0000, Al Lau (alau2) wrote:
> I am looking for more details on the "-n size=65536" option in
> mkfs.xfs. The question is the memory allocation this option
> generates. The system is Redhat EL 7.0 (3.10.0-229.1.2.el7.x86_64).
>
> We have been getting this memory allocation deadlock message in the
> /var/log/messages file. The file system is used for ceph OSD and it
> has about 531894 files.
So, if you only have half a million files being stored, why would you optimised
the directory structure for tens of millions of files in a single directory?
> Oct 6 07:11:09 abc-ceph1-xyz kernel: XFS: possible memory allocation
> deadlock in kmem_alloc (mode:0x8250)
mode = ___GFP_WAIT | ___GFP_IO | ___GFP_NOWARN | ___GFP_ZERO
= GFP_NOFS | GFP_ZERO | GFP_NOWARN
which means it's come through kmem_zalloc() and so is a heap allocation and
hence probably quite small.
Hence I doubt that has anything to do with the directory block size, as the
directory blocks are allocated as single pages through a completely allocation
different path and them virtually mapped...
Cheers,
Dave.
--
Dave Chinner
david@xxxxxxxxxxxxx
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