From: Vladimir Oltean olteanv@gmail.com
[ Upstream commit b6f2494d311a19b33b19708543e7ef6dea1de459 ]
Sometimes the PTP synchronization on the switch 'jumps':
ptp4l[11241.155]: rms 8 max 16 freq -21732 +/- 11 delay 742 +/- 0 ptp4l[11243.157]: rms 7 max 17 freq -21731 +/- 10 delay 744 +/- 0 ptp4l[11245.160]: rms 33592410 max 134217731 freq +192422 +/- 8530253 delay 743 +/- 0 ptp4l[11247.163]: rms 811631 max 964131 freq +10326 +/- 557785 delay 743 +/- 0 ptp4l[11249.166]: rms 261936 max 533876 freq -304323 +/- 126371 delay 744 +/- 0 ptp4l[11251.169]: rms 48700 max 57740 freq -20218 +/- 30532 delay 744 +/- 0 ptp4l[11253.171]: rms 14570 max 30163 freq -5568 +/- 7563 delay 742 +/- 0 ptp4l[11255.174]: rms 2914 max 3440 freq -22001 +/- 1667 delay 744 +/- 1 ptp4l[11257.177]: rms 811 max 1710 freq -22653 +/- 451 delay 744 +/- 1 ptp4l[11259.180]: rms 177 max 218 freq -21695 +/- 89 delay 741 +/- 0 ptp4l[11261.182]: rms 45 max 92 freq -21677 +/- 32 delay 742 +/- 0 ptp4l[11263.186]: rms 14 max 32 freq -21733 +/- 11 delay 742 +/- 0 ptp4l[11265.188]: rms 9 max 14 freq -21725 +/- 12 delay 742 +/- 0 ptp4l[11267.191]: rms 9 max 16 freq -21727 +/- 13 delay 742 +/- 0 ptp4l[11269.194]: rms 6 max 15 freq -21726 +/- 9 delay 743 +/- 0 ptp4l[11271.197]: rms 8 max 15 freq -21728 +/- 11 delay 743 +/- 0 ptp4l[11273.200]: rms 6 max 12 freq -21727 +/- 8 delay 743 +/- 0 ptp4l[11275.202]: rms 9 max 17 freq -21720 +/- 11 delay 742 +/- 0 ptp4l[11277.205]: rms 9 max 18 freq -21725 +/- 12 delay 742 +/- 0
Background: the switch only offers partial RX timestamps (24 bits) and it is up to the driver to read the PTP clock to fill those timestamps up to 64 bits. But the PTP clock readout needs to happen quickly enough (in 0.135 seconds, in fact), otherwise the PTP clock will wrap around 24 bits, condition which cannot be detected.
Looking at the 'max 134217731' value on output line 3, one can see that in hex it is 0x8000003. Because the PTP clock resolution is 8 ns, that means 0x1000000 in ticks, which is exactly 2^24. So indeed this is a PTP clock wraparound, but the reason might be surprising.
What is going on is that sja1105_tstamp_reconstruct(priv, now, ts) expects a "now" time that is later than the "ts" was snapshotted at. This, of course, is obvious: we read the PTP time _after_ the partial RX timestamp was received. However, the workqueue is processing frames from a skb queue and reuses the same PTP time, read once at the beginning. Normally the skb queue only contains one frame and all goes well. But when the skb queue contains two frames, the second frame that gets dequeued might have been partially timestamped by the RX MAC _after_ we had read our PTP time initially.
The code was originally like that due to concerns that SPI access for PTP time readout is a slow process, and we are time-constrained anyway (aka: premature optimization). But some timing analysis reveals that the time spent until the RX timestamp is completely reconstructed is 1 order of magnitude lower than the 0.135 s deadline even under worst-case conditions. So we can afford to read the PTP time for each frame in the RX timestamping queue, which of course ensures that the full PTP time is in the partial timestamp's future.
Fixes: f3097be21bf1 ("net: dsa: sja1105: Add a state machine for RX timestamping") Signed-off-by: Vladimir Oltean olteanv@gmail.com Signed-off-by: David S. Miller davem@davemloft.net Signed-off-by: Greg Kroah-Hartman gregkh@linuxfoundation.org --- drivers/net/dsa/sja1105/sja1105_main.c | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-)
--- a/drivers/net/dsa/sja1105/sja1105_main.c +++ b/drivers/net/dsa/sja1105/sja1105_main.c @@ -1986,12 +1986,12 @@ static void sja1105_rxtstamp_work(struct
mutex_lock(&priv->ptp_lock);
- now = priv->tstamp_cc.read(&priv->tstamp_cc); - while ((skb = skb_dequeue(&data->skb_rxtstamp_queue)) != NULL) { struct skb_shared_hwtstamps *shwt = skb_hwtstamps(skb); u64 ts;
+ now = priv->tstamp_cc.read(&priv->tstamp_cc); + *shwt = (struct skb_shared_hwtstamps) {0};
ts = SJA1105_SKB_CB(skb)->meta_tstamp;