Messages in this thread | | | From | David Howells <> | Subject | Re: [PATCH 00/13] VFS: Filesystem information [ver #19] | Date | Wed, 01 Apr 2020 16:51:43 +0100 |
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Miklos Szeredi <miklos@szeredi.hu> wrote:
> For 30000 mounts, f= 146400us f2= 136766us p= 1406569us p2= > 221669us; p=9.6*f p=10.3*f2 p=6.3*p2
f = 146400us f2= 136766us p = 1406569us <--- Order of magnitude slower p2= 221669us
And more memory used because it's added a whole bunch of inodes and dentries to the cache. For each mount that's a pair for each dir and a pair for each file within the dir. So for the two files my test is reading, for 30000 mounts, that's 90000 dentries and 90000 inodes in mountfs alone.
(gdb) p sizeof(struct dentry) $1 = 216 (gdb) p sizeof(struct inode) $2 = 696 (gdb) p (216*696)*30000*3/1024/1024 $3 = 615
so 615 MiB of RAM added to the caches in an extreme case.
We're seeing customers with 10000+ mounts - that would be 205 MiB, just to read two values from each mount.
I presume you're not going through /proc/fdinfo each time as that would add another d+i - for >1GiB added to the caches for 30000 mounts.
David
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