audit: audit_log_start running on auditd should not stop

The backlog cannot be consumed when audit_log_start is running on auditd
even if audit_log_start calls wait_for_auditd to consume it.
The situation is the deadlock because only auditd can consume the backlog.
If the other process needs to send the backlog, it can be also stopped
by the deadlock.

So, audit_log_start running on auditd should not stop.

You can see the deadlock with the following reproducer:
 # auditctl -a exit,always -S all
 # reboot
Signed-off-by: Toshiyuki Okajima <toshi.okajima@jp.fujitsu.com>
Reviewed-by: gaofeng@cn.fujitsu.com
Signed-off-by: Richard Guy Briggs <rgb@redhat.com>
Signed-off-by: Eric Paris <eparis@redhat.com>

kernel/audit.c

@@ -1319,7 +1319,8 @@ struct audit_buffer *audit_log_start(struct audit_context *ctx, gfp_t gfp_mask,
 	struct audit_buffer	*ab	= NULL;
 	struct timespec		t;
 	unsigned int		uninitialized_var(serial);
-	int reserve;
+	int reserve = 5; /* Allow atomic callers to go up to five
+			    entries over the normal backlog limit */
 	unsigned long timeout_start = jiffies;
 
 	if (audit_initialized != AUDIT_INITIALIZED)
@@ -1328,11 +1329,12 @@ struct audit_buffer *audit_log_start(struct audit_context *ctx, gfp_t gfp_mask,
 	if (unlikely(audit_filter_type(type)))
 		return NULL;
 
-	if (gfp_mask & __GFP_WAIT)
-		reserve = 0;
+	if (gfp_mask & __GFP_WAIT) {
+		if (audit_pid && audit_pid == current->pid)
+			gfp_mask &= ~__GFP_WAIT;
 		else
-		reserve = 5; /* Allow atomic callers to go up to five
-				entries over the normal backlog limit */
+			reserve = 0;
+	}
 
 	while (audit_backlog_limit
 	       && skb_queue_len(&audit_skb_queue) > audit_backlog_limit + reserve) {