From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S1754593AbeEHHRB (ORCPT ); Tue, 8 May 2018 03:17:01 -0400 Received: from mx3-rdu2.redhat.com ([66.187.233.73]:60656 "EHLO mx1.redhat.com" rhost-flags-OK-OK-OK-FAIL) by vger.kernel.org with ESMTP id S1754287AbeEHHRA (ORCPT ); Tue, 8 May 2018 03:17:00 -0400 Subject: Re: [RFC v3 4/5] virtio_ring: add event idx support in packed ring To: Tiwei Bie Cc: "Michael S. Tsirkin" , virtualization@lists.linux-foundation.org, linux-kernel@vger.kernel.org, netdev@vger.kernel.org, wexu@redhat.com, jfreimann@redhat.com References: <20180502164828-mutt-send-email-mst@kernel.org> <20180502151255.h3x6rhszxa3euinl@debian> <20180502184015-mutt-send-email-mst@kernel.org> <20180503011116.qvoyblcpklinrk26@debian> <20180503044218-mutt-send-email-mst@kernel.org> <20180503020949.5u3qz32gsk33z6vk@debian> <9f0b4e37-63ff-42f9-f2e6-3747a19a0206@redhat.com> <20180503135430.lbtvn4p4lyu3ksqo@debian> <12ede490-f674-2b89-d639-266b5fe15466@redhat.com> <20180508064409.kcn6amhsxu7nkuuc@debian> From: Jason Wang Message-ID: <34f2c690-7cb2-f9ea-2ce9-40f4ccb594c9@redhat.com> Date: Tue, 8 May 2018 15:16:53 +0800 User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:52.0) Gecko/20100101 Thunderbird/52.7.0 MIME-Version: 1.0 In-Reply-To: <20180508064409.kcn6amhsxu7nkuuc@debian> Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Content-Language: en-US Sender: linux-kernel-owner@vger.kernel.org List-ID: X-Mailing-List: linux-kernel@vger.kernel.org On 2018年05月08日 14:44, Tiwei Bie wrote: > On Tue, May 08, 2018 at 01:40:40PM +0800, Jason Wang wrote: >> On 2018年05月08日 11:05, Jason Wang wrote: >>>> Because in virtqueue_enable_cb_delayed(), we may set an >>>> event_off which is bigger than new and both of them have >>>> wrapped. And in this case, although new is smaller than >>>> event_off (i.e. the third param -- old), new shouldn't >>>> add vq->num, and actually we are expecting a very big >>>> idx diff. >>> Yes, so to calculate distance correctly between event and new, we just >>> need to compare the warp counter and return false if it doesn't match >>> without the need to try to add vq.num here. >>> >>> Thanks >> Sorry, looks like the following should work, we need add vq.num if >> used_wrap_counter does not match: >> >> static bool vhost_vring_packed_need_event(struct vhost_virtqueue *vq, >>                       __u16 off_wrap, __u16 new, >>                       __u16 old) >> { >>     bool wrap = off_wrap >> 15; >>     int off = off_wrap & ~(1 << 15); >>     __u16 d1, d2; >> >>     if (wrap != vq->used_wrap_counter) >>         d1 = new + vq->num - off - 1; > Just to draw your attention (maybe you have already > noticed this). I miss this, thanks! > > In this case (i.e. wrap != vq->used_wrap_counter), > it's also possible that (off < new) is true. Because, > > when virtqueue_enable_cb_delayed_packed() is used, > `off` is calculated in driver in a way like this: > > off = vq->last_used_idx + bufs; > if (off >= vq->vring_packed.num) { > off -= vq->vring_packed.num; > wrap_counter ^= 1; > } > > And when `new` (in vhost) is close to vq->num. The > vq->last_used_idx + bufs (in driver) can be bigger > than vq->vring_packed.num, and: > > 1. `off` will wrap; > 2. wrap counters won't match; > 3. off < new; > > And d1 (i.e. new + vq->num - off - 1) will be a value > bigger than vq->num. I'm okay with this, although it's > a bit weird. So I'm considering something more compact by reusing vring_need_event() by pretending a larger queue size and adding vq->num back when necessary: static bool vhost_vring_packed_need_event(struct vhost_virtqueue *vq,                       __u16 off_wrap, __u16 new,                       __u16 old) {     bool wrap = vq->used_wrap_counter;     int off = off_wrap & ~(1 << 15);     __u16 d1, d2;     if (new < old) {         new += vq->num;         wrap ^= 1;     }     if (wrap != off_wrap >> 15)         off += vq->num;     return vring_need_event(off, new, old); } > > Best regards, > Tiwei Bie > >>     else >>         d1 = new - off - 1; >> >>     if (new > old) >>         d2 = new - old; >>     else >>         d2 = new + vq->num - old; >> >>     return d1 < d2; >> } >> >> Thanks >>