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 <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))
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()).
The formula in libbpf should match it.
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.
>> 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);
>
