audit: audit_log_start running on auditd should not stop (6dd80aba) · Commits · 戴 / test

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

+8 −6

Original line number	Diff line number	Diff line
		@@ -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) {

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