# Re: Status of FST and mount times

```
On 2018年02月15日 10:15, Duncan wrote:
> 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)?```
```
My fault, off by 1 is really easy to screw things up.

So it's 400 * 493 ^ n.

And level 0 also fits into the calculation.

>  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.

Here level 0 means it's leaf. And I assume the average item size of each
And using 16K nodesize we have 16283, we get 407, I just round it to 400
to make calculation a little easier and more headroom for larger item.

So for level 0, we could have around 400 items.
For nodes (1 < level <= 7), since node ptr is fixed to 33 bytes, the
calculation is pretty simple now.

Thanks,
Qu

>

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