On 4/26/22 5:53 AM, Andrey Albershteyn wrote:
> On Tue, Apr 26, 2022 at 10:21:04AM +0200, Karel Zak wrote:
>> On Mon, Apr 25, 2022 at 10:38:02AM -0500, Eric Sandeen wrote:
>>> On 4/25/22 8:43 AM, Karel Zak wrote:
>>>>
>>>> Thanks Andrey,
>>>>
>>>> the code looks good.
>>>>
>>>> On Thu, Apr 21, 2022 at 03:09:44PM +0200, Andrey Albershteyn wrote:
>>>>> I had a look into other fs, like ext4 and btrfs, to implement FSSIZE
>>>>> for them, but I think I don't have enough expertize to do that as
>>>>> they have not so trivial metadata overhead calculation :)
>>>>
>>>> But it would be nice to have ext4 and btrfs too. Maybe Eric can help
>>>> you :-)
>>>>
>>>> Karel
>>>
>>> For ext4 something like e2fsprogs does is probably ok:
>>>
>>> static __u64 e2p_free_blocks_count(struct ext2_super_block *super)
>>> {
>>> return super->s_free_blocks_count |
>>> (ext2fs_has_feature_64bit(super) ?
>>> (__u64) super->s_free_blocks_hi << 32 : 0);
>>> }
Whoops, I mis-pasted, that should have been:
static __u64 e2p_blocks_count(struct ext2_super_block *super)
{
return super->s_blocks_count |
(ext2fs_has_feature_64bit(super) ?
(__u64) super->s_blocks_count_hi << 32 : 0);
}
(blocks count, not free blocks count)
> If we want to mimic statfs(2) + lsblk, then, I think this one won't
> be exactly it. I suppose s_free_blocks_count contains number of not
> used blocks which will vary with utilization... The statfs()
> calculates total number of blocks as total_blocks - (journalsize +
> each group free space). As far as I got it, both journalsize and
> group free space is not so trivial to calculate.
(Sorry for mixing this up by pasting e2p_free_blocks_count)
Ok, so I'm remembering what FSSIZE does in lsblk now :) it simply returns:
dev->fsstat.f_frsize * dev->fsstat.f_blocks
i.e. calculating the value based on statfs values.
This does get a little tricky for ext4, because statfs "f_blocks" is
affected by the minixdf/bsddf mount options.
ext4's f_blocks statfs calculation looks like this:
buf->f_blocks = ext4_blocks_count(es) - EXT4_C2B(sbi, overhead);
the default behavior, bsddf, subtracts out the overhead.
The overhead is a little tricky to calculate, but I /think/ it's precomputed and
stored in the superblock as well for (newer?) ext4, in the s_overhead_clusters
field.
So I *think* you can just do:
(block size) * (blocks count - overhead)
and get the same answer as statfs and lsblk. Does that work?
(Getting block count is a little tricky, because it may be split across
2 fields; s_blocks_count_lo for the lower 32 bits on smaller filesystems,
and s_blocks_count_hi only if EXT4_FEATURE_INCOMPAT_64BIT is set.)
(ext2 is messier, as a precomputed overhead value is not stored on disk.
Perhaps returning FSSIZE only if the type is ext4 might make sense.)
If I'm wrong about s_overhead_clusters being pre-calculated and stored, perhaps
just ignoring "overhead' altogether and treating FSSIZE in the "minixdf"
sense, and simply return s_blocks_count_lo & s_blocks_count_hi for an
answer.
Thanks,
-Eric
