[Qemu-devel] [PULL 13/16] blockjob: update nodes head while removing all bdrv

[Max Reitz <mreitz@redhat.com>]

From: Sergio Lopez <slp@redhat.com>

block_job_remove_all_bdrv() iterates through job->nodes, calling
bdrv_root_unref_child() for each entry. The call to the latter may
reach child_job_[can_]set_aio_ctx(), which will also attempt to
traverse job->nodes, potentially finding entries that where freed
on previous iterations.

To avoid this situation, update job->nodes head on each iteration to
ensure that already freed entries are no longer linked to the list.

RHBZ: https://bugzilla.redhat.com/show_bug.cgi?id=1746631
Signed-off-by: Sergio Lopez <slp@redhat.com>
Cc: qemu-stable@nongnu.org
Signed-off-by: Max Reitz <mreitz@redhat.com>
Message-id: 20190911100316.32282-1-mreitz@redhat.com
Reviewed-by: Sergio Lopez <slp@redhat.com>
Signed-off-by: Max Reitz <mreitz@redhat.com>
---
 blockjob.c | 17 +++++++++++++----
 1 file changed, 13 insertions(+), 4 deletions(-)

diff --git a/blockjob.c b/blockjob.c
index 2abed0f551..c6e20e2fcd 100644
--- a/blockjob.c
+++ b/blockjob.c
@@ -175,14 +175,23 @@ static const BdrvChildRole child_job = {
 
 void block_job_remove_all_bdrv(BlockJob *job)
 {
-    GSList *l;
-    for (l = job->nodes; l; l = l->next) {
+    /*
+     * bdrv_root_unref_child() may reach child_job_[can_]set_aio_ctx(),
+     * which will also traverse job->nodes, so consume the list one by
+     * one to make sure that such a concurrent access does not attempt
+     * to process an already freed BdrvChild.
+     */
+    while (job->nodes) {
+        GSList *l = job->nodes;
         BdrvChild *c = l->data;
+
+        job->nodes = l->next;
+
         bdrv_op_unblock_all(c->bs, job->blocker);
         bdrv_root_unref_child(c);
+
+        g_slist_free_1(l);
     }
-    g_slist_free(job->nodes);
-    job->nodes = NULL;
 }
 
 bool block_job_has_bdrv(BlockJob *job, BlockDriverState *bs)
-- 
2.21.0