[PATCH 00/13] VFS: Filesystem information [ver #19]
dhowells at redhat.com
Wed Apr 1 15:51:43 UTC 2020
Miklos Szeredi <miklos at 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
p = 1406569us <--- Order of magnitude slower
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.
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