[CRIU] [PATCH 5/5] sk-unix: Don't fail if socket path lays on btrfs volume

[Pavel Emelyanov]

On 11/29/2013 01:34 PM, Cyrill Gorcunov wrote:
> 
> Because socket migh lay on btrfs volume (thus the device
> number reported by diag module won't be the same as obtained
> from stat(2)) we need to do an additional test and try
> to resolve device number with help of btrfs engine.
> 
> Signed-off-by: Cyrill Gorcunov <gorcunov at openvz.org>
> ---
>  sk-unix.c | 23 ++++++++++++++++-------
>  1 file changed, 16 insertions(+), 7 deletions(-)
> 


> 			if ((st.st_ino != uv->udiag_vfs_ino) ||
> -			    (st.st_dev != kdev_to_odev(uv->udiag_vfs_dev))) {
> +			    ((st.st_dev != kdev_to_odev(uv->udiag_vfs_dev) &&
> +			      !btrfs_dev_match(name, st.st_dev)))) {

Why do we do name matching? Let's imagine we have a btrfs on disk with 2 subvols
and have a unix socket on either of them.

Let the disk's device be D, subvol 1 id be V1, subvol 2 id be V2.
In this if (), let the st.st_ino is S and uv->udiag_vfs_dev is U.

If the socket is on the btrfs disk, then we'll have S != U, U == D and 
(S == V1 || S == V2), right? In order to fix the devices comparison we need
a btrfs_vol_to_dev() function that will convert V1 -> D, V2 -> D and any other
number to 0. This is valid, since D, V1 and V2 cannot match any other dev_t 
in the system.

Thus the proper if should look like this

if (S != U) {
	int btrfs_dev;

	/* S == V1 || S == V2 */
	btrfs_dev = btrfs_vol_to_dev(S);
	/* now btrfs_dev is D for btrfs and 0 otherwise */
	if (btrfs_dev != U)
		goto mismatch;
}

/* full match code here */

Thanks,
Pavel