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
