On Fri, Jul 19, 2024 at 9:56 AM Bailey Forrest <bcf@xxxxxxxxxx> wrote:
>
> On Fri, Jul 19, 2024 at 7:31 AM Praveen Kaligineedi
> <pkaligineedi@xxxxxxxxxx> wrote:
> >
> > On Thu, Jul 18, 2024 at 8:47 PM Willem de Bruijn
> > <willemdebruijn.kernel@xxxxxxxxx> wrote:
> > >
> > > On Thu, Jul 18, 2024 at 9:52 PM Praveen Kaligineedi
> > > <pkaligineedi@xxxxxxxxxx> wrote:
> > > >
> > > > On Thu, Jul 18, 2024 at 4:07 PM Willem de Bruijn
> > > > <willemdebruijn.kernel@xxxxxxxxx> wrote:
> > > >
> > > > > > + * segment, then it will count as two descriptors.
> > > > > > + */
> > > > > > + if (last_frag_size > GVE_TX_MAX_BUF_SIZE_DQO) {
> > > > > > + int last_frag_remain = last_frag_size %
> > > > > > + GVE_TX_MAX_BUF_SIZE_DQO;
> > > > > > +
> > > > > > + /* If the last frag was evenly divisible by
> > > > > > + * GVE_TX_MAX_BUF_SIZE_DQO, then it will not be
> > > > > > + * split in the current segment.
> > > > >
> > > > > Is this true even if the segment did not start at the start of the frag?
> > > > The comment probably is a bit confusing here. The current segment
> > > > we are tracking could have a portion in the previous frag. The code
> > > > assumed that the portion on the previous frag (if present) mapped to only
> > > > one descriptor. However, that portion could have been split across two
> > > > descriptors due to the restriction that each descriptor cannot exceed 16KB.
> > >
> > > >>> /* If the last frag was evenly divisible by
> > > >>> + * GVE_TX_MAX_BUF_SIZE_DQO, then it will not be
> > > >>> + * split in the current segment.
> > >
> > > This is true because the smallest multiple of 16KB is 32KB, and the
> > > largest gso_size at least for Ethernet will be 9K. But I don't think
> > > that that is what is used here as the basis for this statement?
> > >
> > The largest Ethernet gso_size (9K) is less than GVE_TX_MAX_BUF_SIZE_DQO
> > is an implicit assumption made in this patch and in that comment. Bailey,
> > please correct me if I am wrong..
>
> If last_frag_size is evenly divisible by GVE_TX_MAX_BUF_SIZE_DQO, it
> doesn't hit the edge case we're looking for.
>
> - If it's evenly divisible, then we know it will use exactly
> (last_frag_size / GVE_TX_MAX_BUF_SIZE_DQO) descriptors
This assumes that gso_segment start is aligned with skb_frag
start. That is not necessarily true, right?
If headlen minus protocol headers is 1B, then the first segment
will have two descriptors { 1B, 9KB - 1 }. And the next segment
can have skb_frag_size - ( 9KB - 1).
I think the statement is correct, but because every multiple
of 16KB is so much larger than the max gso_size of ~9KB,
that a single segment will never include more than two
skb_frags.
Quite possibly the code overestimates the number of
descriptors per segment now, but that is safe and only a
performance regression.
> - GVE_TX_MAX_BUF_SIZE_DQO > 9k, so we know each descriptor won't
> create a segment which exceeds the limit
For a net patch, it is generally better to make a small fix rather than rewrite.
That said, my sketch without looping over every segment:
while (off < skb->len) {
gso_size_left = shinfo->gso_size;
num_desc = 0;
while (gso_size_left) {
desc_len = min(gso_size_left, frag_size_left);
gso_size_left -= desc_len;
frag_size_left -= desc_len;
num_desc++;
if (num_desc > max_descs_per_seg)
return false;
if (!frag_size_left)
frag_size_left =
skb_frag_size(&shinfo->frags[frag_idx++]);
+ else
+ frag_size_left %= gso_size; /*
skip segments that fit in one desc */
}
}
