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Date: Mon, 11 Mar 2024 01:49:06 -0400
From: Kent Overstreet <kent.overstreet@...ux.dev>
To: Miklos Szeredi <miklos@...redi.hu>
Cc: Dave Chinner <david@...morbit.com>, 
	"Darrick J. Wong" <djwong@...nel.org>, Neal Gompa <neal@...pa.dev>, linux-fsdevel@...r.kernel.org, 
	linux-bcachefs@...r.kernel.org, linux-btrfs@...r.kernel.org, linux-kernel@...r.kernel.org, 
	Josef Bacik <josef@...icpanda.com>, Miklos Szeredi <mszeredi@...hat.com>, 
	Christian Brauner <brauner@...nel.org>, David Howells <dhowells@...hat.com>
Subject: Re: [PATCH v2] statx: stx_subvol

On Mon, Mar 11, 2024 at 06:30:21AM +0100, Miklos Szeredi wrote:
> On Mon, 11 Mar 2024 at 03:17, Dave Chinner <david@...morbit.com> wrote:
> >
> > On Fri, Mar 08, 2024 at 08:56:33AM -0800, Darrick J. Wong wrote:
> > > Should the XFS data and rt volumes be reported with different stx_vol
> > > values?
> >
> > No, because all the inodes are on the data volume and the same inode
> > can have data on the data volume or the rt volume. i.e. "data on rt,
> > truncate, clear rt, copy data back into data dev".  It's still the
> > same inode, and may have exactly the same data, so why should change
> > stx_vol and make it appear to userspace as being a different inode?
> 
> Because stx_vol must not be used by userspace to distinguish between
> unique inodes.  To determine if two inodes are distinct within a
> filesystem (which may have many volumes) it should query the file
> handle and compare that.
> 
> If we'll have a filesystem that has a different stx_vol but the same
> fh, all the better.

I agree that stx_vol should not be used for uniqueness testing, but
that's a non sequitar here; Dave's talking about the fact that volume
isn't a constatn for a given inode on XFS. And that's a good point;
volumes on XFS don't map to the filesystem path heirarchy in a nice
clean way like on btrfs and bcachefs (and presumably ZFS).

Subvolumes on btrfs and bcachefs form a tree, and that's something we
should document about stx_subvol - recursively enumerable things are
quite nice to work with.

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