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 <maximmi@xxxxxxxxxxxx> > Reviewed-by: Tariq Toukan <tariqt@xxxxxxxxxxxx> > Acked-by: Saeed Mahameed <saeedm@xxxxxxxxxxxx> 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)) > 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);
