btrfs: do not ASSERT() if the newly created subvolume already got read

In the Linux kernel, the following vulnerability has been resolved:

btrfs: do not ASSERT() if the newly created subvolume already got read

[BUG]

There is a syzbot crash, triggered by the ASSERT() during subvolume

creation:

assertion failed: !anon_dev, in fs/btrfs/disk-io.c:1319

------------[ cut here ]------------

kernel BUG at fs/btrfs/disk-io.c:1319!

invalid opcode: 0000 [#1] PREEMPT SMP KASAN

RIP: 0010:btrfs_get_root_ref.part.0+0x9aa/0xa60

<TASK>

btrfs_get_new_fs_root+0xd3/0xf0

create_subvol+0xd02/0x1650

btrfs_mksubvol+0xe95/0x12b0

__btrfs_ioctl_snap_create+0x2f9/0x4f0

btrfs_ioctl_snap_create+0x16b/0x200

btrfs_ioctl+0x35f0/0x5cf0

__x64_sys_ioctl+0x19d/0x210

do_syscall_64+0x3f/0xe0

entry_SYSCALL_64_after_hwframe+0x63/0x6b

---[ end trace 0000000000000000 ]---

[CAUSE]

During create_subvol(), after inserting root item for the newly created

subvolume, we would trigger btrfs_get_new_fs_root() to get the

btrfs_root of that subvolume.

The idea here is, we have preallocated an anonymous device number for

the subvolume, thus we can assign it to the new subvolume.

But there is really nothing preventing things like backref walk to read

the new subvolume.

If that happens before we call btrfs_get_new_fs_root(), the subvolume

would be read out, with a new anonymous device number assigned already.

In that case, we would trigger ASSERT(), as we really expect no one to

read out that subvolume (which is not yet accessible from the fs).

But things like backref walk is still possible to trigger the read on

the subvolume.

Thus our assumption on the ASSERT() is not correct in the first place.

[FIX]

Fix it by removing the ASSERT(), and just free the @anon_dev, reset it

to 0, and continue.

If the subvolume tree is read out by something else, it should have

already get a new anon_dev assigned thus we only need to free the

preallocated one.