On tor, mar 10, 2022 at 15:54, Andrew Lunn <andrew@xxxxxxx> wrote:
>> + if (mv88e6xxx_port_is_locked(chip, chip->ports[spid].port))
>> + err = mv88e6xxx_switchdev_handle_atu_miss_violation(chip,
>> + chip->ports[spid].port,
>> + &entry,
>> + fid);
>
>> +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.
