Re: [PATCH 00/13] VFS: Filesystem information [ver #19]

[David Howells]

Miklos Szeredi <miklos@xxxxxxxxxx> 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