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Message-ID: <fc6d27c7-6f00-4871-65d0-dffdcb2e1925@redhat.com>
Date: Wed, 20 Jan 2021 11:23:06 +0800
From: Jason Wang <jasowang@...hat.com>
To: Xuan Zhuo <xuanzhuo@...ux.alibaba.com>
Cc: "Michael S. Tsirkin" <mst@...hat.com>,
"David S. Miller" <davem@...emloft.net>,
Jakub Kicinski <kuba@...nel.org>,
Björn Töpel <bjorn.topel@...el.com>,
Magnus Karlsson <magnus.karlsson@...el.com>,
Jonathan Lemon <jonathan.lemon@...il.com>,
Alexei Starovoitov <ast@...nel.org>,
Daniel Borkmann <daniel@...earbox.net>,
Jesper Dangaard Brouer <hawk@...nel.org>,
John Fastabend <john.fastabend@...il.com>,
Andrii Nakryiko <andrii@...nel.org>,
Martin KaFai Lau <kafai@...com>,
Song Liu <songliubraving@...com>, Yonghong Song <yhs@...com>,
KP Singh <kpsingh@...nel.org>,
virtualization@...ts.linux-foundation.org, bpf@...r.kernel.org,
netdev@...r.kernel.org
Subject: Re: [PATCH net-next v2 2/7] virtio-net, xsk: distinguish XDP_TX and
XSK XMIT ctx
On 2021/1/18 下午7:54, Xuan Zhuo wrote:
> On Mon, 18 Jan 2021 14:45:16 +0800, Jason Wang <jasowang@...hat.com> wrote:
>> On 2021/1/16 上午10:59, Xuan Zhuo wrote:
>>> If support xsk, a new ptr will be recovered during the
>>> process of freeing the old ptr. In order to distinguish between ctx sent
>>> by XDP_TX and ctx sent by xsk, a struct is added here to distinguish
>>> between these two situations. virtnet_xdp_type.type It is used to
>>> distinguish different ctx, and virtnet_xdp_type.offset is used to record
>>> the offset between "true ctx" and virtnet_xdp_type.
>>>
>>> The newly added virtnet_xsk_hdr will be used for xsk.
>>>
>>> Signed-off-by: Xuan Zhuo <xuanzhuo@...ux.alibaba.com>
>>
>> Any reason that you can't simply encode the type in the pointer itself
>> as we used to do?
>>
>> #define VIRTIO_XSK_FLAG BIT(1)
>>
>> ?
> Since xdp socket does not use xdp_frame, we will encounter three data types when
> recycling: skb, xdp_frame, xdp socket.
Just to make sure we are in the same page. Currently, the pointer type
is encoded with 1 bit in the pointer. Can we simply use 2 bit to
distinguish skb, xdp, xsk?
Thanks
>
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