RE: [PATCH] rtw88/pci: Rearrange the memory usage for skb in RX ISR

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> > > @@ -803,25 +812,14 @@ static void rtw_pci_rx_isr(struct rtw_dev
> *rtwdev,
> > > struct rtw_pci *rtwpci,
> > >                       skb_put(skb, pkt_stat.pkt_len);
> > >                       skb_reserve(skb, pkt_offset);
> > >
> > > -                     /* alloc a smaller skb to mac80211 */
> > > -                     new = dev_alloc_skb(pkt_stat.pkt_len);
> > > -                     if (!new) {
> > > -                             new = skb;
> > > -                     } else {
> > > -                             skb_put_data(new, skb->data,
> skb->len);
> > > -                             dev_kfree_skb_any(skb);
> > > -                     }
> >
> > I am not sure if it's fine to deliver every huge SKB to mac80211.
> > Because it will then be delivered to TCP/IP stack.
> > Hence I think either it should be tested to know if the performance
> > would be impacted or find out a more efficient way to send
> > smaller SKB to mac80211 stack.
> 
> I remember network stack only processes the skb with(in) pointers
> (skb->data) and the skb->len for data part.  It also checks real
> buffer boundary (head and end) of the skb to prevent memory overflow.
> Therefore, I think using the original skb is the most efficient way.
> 
> If I misunderstand something, please point out.
> 

It means if we still use a huge SKB (~8K) for every RX packet (~1.5K).
There is about 6.5K not used. And even more if we ping with large packet
size "eg. $ ping -s 65536", I am not sure if those huge SKBs will eat all of
the SKB mem pool, and then ping fails.

BTW, the original design of RTK_PCI_RX_BUF_SIZE to be (8192 + 24) is to
receive AMSDU packet in one SKB.
(Could probably enlarge it to RX VHT AMSDU ~11K)

Yan-Hsuan




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