umount_tree(): take all victims out of propagation graph at once

For each removed mount we need to calculate where the slaves will end up.
To avoid duplicating that work, do it for all mounts to be removed
at once, taking the mounts themselves out of propagation graph as
we go, then do all transfers; the duplicate work on finding destinations
is avoided since if we run into a mount that already had destination found,
we don't need to trace the rest of the way.  That's guaranteed
O(removed mounts) for finding destinations and removing from propagation
graph and O(surviving mounts that have master removed) for transfers.

Reviewed-by: Christian Brauner <brauner@kernel.org>
Signed-off-by: Al Viro <viro@zeniv.linux.org.uk>

1 files changed, 1 insertions, 0 deletions

diff --git a/fs/pnode.h b/fs/pnode.h
index 00ab153e3e9d..b029db225f33 100644
--- a/fs/pnode.h
+++ b/fs/pnode.h
@@ -42,6 +42,7 @@ static inline bool peers(const struct mount *m1, const struct mount *m2)
 }
 
 void change_mnt_propagation(struct mount *, int);
+void bulk_make_private(struct list_head *);
 int propagate_mnt(struct mount *, struct mountpoint *, struct mount *,
 		struct hlist_head *);
 void propagate_umount(struct list_head *);