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Date:   Wed, 12 Jun 2019 12:21:34 +0200
From:   Lorenzo Bianconi <lorenzo.bianconi@redhat.com>
To:     Stanislaw Gruszka <sgruszka@redhat.com>
Cc:     Lorenzo Bianconi <lorenzo@kernel.org>, nbd@nbd.name,
        kvalo@codeaurora.org, linux-wireless@vger.kernel.org
Subject: Re: [PATCH v2 1/2] mt76: usb: fix rx A-MSDU support
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References: <cover.1559293385.git.lorenzo@kernel.org>
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On Jun 12, Stanislaw Gruszka wrote:
> On Wed, Jun 12, 2019 at 11:45:21AM +0200, Lorenzo Bianconi wrote:
> > > > +mt76u_build_rx_skb(u8 *data, int len, int buf_size,
> > > > +		   int *nsgs)
> > > > +{
> > > > +	int data_len =3D min(len, MT_SKB_HEAD_LEN);
>=20
> Oh, and this looks unneeded as well as for len < MT_SKB_HEAD_LEN=3D128
> we will go through fast path.

I guess if we remove data_len =3D min(len, MT_SKB_HEAD_LEN) and even *nsgs =
=3D 0 at
the end we are making some assumptions on the value of MT_SKB_HEAD_LEN and
buf_size. In the patch I just avoided them but maybe we can just assume that
MT_SKB_HEAD_LEN and buf_size will not changed in the future. What do you
think?

>=20
> > > mt7601u and iwlmvm just copy hdrlen + 8 and put the rest
> > > of the buffer in fragment, which supose to be more efficient,
> > > see comment in iwl_mvm_pass_packet_to_mac80211().
> >=20
> > Right here we copy 128B instead of 32 but I think it is good to have L3=
 and L4
> > header in the linear area of the skb since otherwise the stack will nee=
d to
> > align them
>=20
> Not sure if understand, I think aliment of L3 & L4 headers will be
> the same, assuming ieee80211 header is aligned the same in fragment
> buffer and in linear area. But if you think this is better to copy those
> to linear area I'm ok with that.

Sorry I have not been so clear. I mean in the stack before accessing a given
header we will run pskb_may_pull() that can end up copying the skb if there=
 is
not enough space in the skb->head

Regards,
Lorenzo

>=20
> Stanislaw

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