On Thu, 13 Jun 2019 14:01:39 +0000, Maxim Mikityanskiy wrote:
> On 2019-06-12 23:23, Jakub Kicinski wrote:
> > On Wed, 12 Jun 2019 15:56:48 +0000, Maxim Mikityanskiy wrote:
> >> Currently, libbpf uses the number of combined channels as the maximum
> >> queue number. However, the kernel has a different limitation:
> >>
> >> - xdp_reg_umem_at_qid() allows up to max(RX queues, TX queues).
> >>
> >> - ethtool_set_channels() checks for UMEMs in queues up to
> >> combined_count + max(rx_count, tx_count).
> >>
> >> libbpf shouldn't limit applications to a lower max queue number. Account
> >> for non-combined RX and TX channels when calculating the max queue
> >> number. Use the same formula that is used in ethtool.
> >>
> >> Signed-off-by: Maxim Mikityanskiy <maxi...@mellanox.com>
> >> Reviewed-by: Tariq Toukan <tar...@mellanox.com>
> >> Acked-by: Saeed Mahameed <sae...@mellanox.com>
> >
> > I don't think this is correct. max_tx tells you how many TX channels
> > there can be, you can't add that to combined. Correct calculations is:
> >
> > max_num_chans = max(max_combined, max(max_rx, max_tx))
>
> First of all, I'm aligning with the formula in the kernel, which is:
>
> curr.combined_count + max(curr.rx_count, curr.tx_count);
>
> (see net/core/ethtool.c, ethtool_set_channels()).
curr != max. ethtool code you're pointing me to (and which I wrote)
uses the current allocation, not the max values.
> The formula in libbpf should match it.
The formula should be based on understanding what we're doing,
not copying some not-really-equivalent code from somewhere :)
Combined is a basically a queue pair, RX is an RX ring with a dedicated
IRQ, and TX is a TX ring with a dedicated IRQ. If driver supports both
combined and single purpose interrupt vectors it will most likely set
max_rx = num_hw_rx
max_tx = num_hw_tx
max_combined = min(rx, tx)
Like with most ethtool APIs there are some variations to this.
> Second, the existing drivers have either combined channels or separate
> rx and tx channels. So, for the first kind of drivers, max_tx doesn't
> tell how many TX channels there can be, it just says 0, and max_combined
> tells how many TX and RX channels are supported. As max_tx doesn't
> include max_combined (and vice versa), we should add them up.
By existing drivers you mean Intel drivers which implement AF_XDP,
and your driver? Both Intel and MLX drivers seem to only set
max_combined.
If you mean all drivers across the kernel, then I believe the best
formula is what I gave you.
> >> tools/lib/bpf/xsk.c | 6 +++---
> >> 1 file changed, 3 insertions(+), 3 deletions(-)
> >>
> >> diff --git a/tools/lib/bpf/xsk.c b/tools/lib/bpf/xsk.c
> >> index bf15a80a37c2..86107857e1f0 100644
> >> --- a/tools/lib/bpf/xsk.c
> >> +++ b/tools/lib/bpf/xsk.c
> >> @@ -334,13 +334,13 @@ static int xsk_get_max_queues(struct xsk_socket *xsk)
> >> goto out;
> >> }
> >>
> >> - if (channels.max_combined == 0 || errno == EOPNOTSUPP)
> >> + ret = channels.max_combined + max(channels.max_rx, channels.max_tx);
> >> +
> >> + if (ret == 0 || errno == EOPNOTSUPP)
> >> /* If the device says it has no channels, then all traffic
> >> * is sent to a single stream, so max queues = 1.
> >> */
> >> ret = 1;
> >> - else
> >> - ret = channels.max_combined;
> >>
> >> out:
> >> close(fd);
