Qu Wenruo posted on Thu, 15 Feb 2018 09:42:27 +0800 as excerpted: > The easiest way to get a basic idea of how large your extent tree is > using debug tree: > > # btrfs-debug-tree -r -t extent <device> > > You would get something like: > btrfs-progs v4.15 extent tree key (EXTENT_TREE ROOT_ITEM 0) 30539776 > level 0 <<< > total bytes 10737418240 bytes used 393216 uuid > 651fcf0c-0ffd-4351-9721-84b1615f02e0 > > That level is would give you some basic idea of the size of your extent > tree. > > For level 0, it could contains about 400 items for average. > For level 1, it could contains up to 197K items. > ... > For leven n, it could contains up to 400 * 493 ^ (n - 1) items. > ( n <= 7 )
So for level 2 (which I see on a couple of mine here, ran it out of curiosity): 400 * 493 ^ (2 - 1) = 400 * 493 = 197200 197K for both level 1 and level 2? Doesn't look correct. Perhaps you meant a simple power of n, instead of (n-1)? That would yield ~97M for level 2, and would yield the given numbers for levels 0 and 1 as well, whereby using n-1 for level 0 yields less than a single entry, and 400 for level 1. Or the given numbers were for level 1 and 2, with level 0 not holding anything, not levels 0 and 1. But that wouldn't jive with your level 0 example, which I would assume could never happen if it couldn't hold even a single entry. -- Duncan - List replies preferred. No HTML msgs. "Every nonfree program has a lord, a master -- and if you use the program, he is your master." Richard Stallman -- To unsubscribe from this list: send the line "unsubscribe linux-btrfs" in the body of a message to majord...@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html
