On tor, mar 10, 2022 at 15:54, Andrew Lunn [off-list ref] wrote:

quoted

+		if (mv88e6xxx_port_is_locked(chip, chip->ports[spid].port))
+			err = mv88e6xxx_switchdev_handle_atu_miss_violation(chip,
+									    chip->ports[spid].port,
+									    &entry,
+									    fid);

quoted

+static int mv88e6xxx_find_vid_on_matching_fid(struct mv88e6xxx_chip *chip,
+					      const struct mv88e6xxx_vtu_entry *entry,
+					      void *priv)
+{
+	struct mv88e6xxx_fid_search_ctx *ctx = priv;
+
+	if (ctx->fid_search == entry->fid) {
+		ctx->vid_found = entry->vid;
+		return 1;
+	}
+	return 0;
+}
+
+int mv88e6xxx_switchdev_handle_atu_miss_violation(struct mv88e6xxx_chip *chip,
+						  int port,
+						  struct mv88e6xxx_atu_entry *entry,
+						  u16 fid)
+{
+	struct switchdev_notifier_fdb_info info = {
+		.addr = entry->mac,
+		.vid = 0,
+		.added_by_user = false,
+		.is_local = false,
+		.offloaded = true,
+		.locked = true,
+	};
+	struct mv88e6xxx_fid_search_ctx ctx;
+	struct netlink_ext_ack *extack;
+	struct net_device *brport;
+	struct dsa_port *dp;
+	int err;
+
+	ctx.fid_search = fid;
+	err = mv88e6xxx_vtu_walk(chip, mv88e6xxx_find_vid_on_matching_fid, &ctx);

I could be reading this code wrong, but it looks like you assume there
is a single new entry in the ATU. But interrupts on these devices are
slow. It would be easy for two or more devices to pop into existence
at the same time. Don't you need to walk the whole ATU to find all the
new entries? Have you tried this with a traffic generating populating
the ATU with new entries at different rates, up to line rate? Do you
get notifications for them all?

    Andrew

We have not tried your said test, but if a packet doesn't manage to
trigger a ATU miss violation interrupt, not much will happen as far as I
see. The device sending the packet will not get access, but if it sends
again (maybe after a short while), it can still trigger the ATU miss
violation interrupt and get access.
I think that the normal behaviour for a device would be to try and
connect, and if that is not successfull inside a short time, it will wait
for a timeout before trying again.