Given a chain of images like [A] <- [B] <- [C] <- [D] <- [E], where
[E] is the active (topmost) image, if we want to remove [B] and [C] we
have to do it before linking [A] and [D].
When [A] is set as the backing image of [D] all its operations are
blocked, but removing [B] afterwards would undo those changes, since
[B] would still be connected to [A].
This would result in a situation where the backing image [A] has all
its operations unblocked.
Signed-off-by: Alberto Garcia <be...@igalia.com>
---
block.c | 3 ++-
1 file changed, 2 insertions(+), 1 deletion(-)
diff --git a/block.c b/block.c
index 9b707e3..2c36351 100644
--- a/block.c
+++ b/block.c
@@ -2624,13 +2624,14 @@ int bdrv_drop_intermediate(BlockDriverState *active,
BlockDriverState *top,
if (ret) {
goto exit;
}
- bdrv_set_backing_hd(new_top_bs, base_bs);
+ bdrv_set_backing_hd(new_top_bs, NULL);
QSIMPLEQ_FOREACH_SAFE(intermediate_state, &states_to_delete, entry, next) {
/* so that bdrv_close() does not recursively close the chain */
bdrv_set_backing_hd(intermediate_state->bs, NULL);
bdrv_unref(intermediate_state->bs);
}
+ bdrv_set_backing_hd(new_top_bs, base_bs);
ret = 0;
exit:
--
2.1.4
