On Mon, Feb 25, 2019 at 10:49:16AM +1000, David Gwynne wrote: > the mcl2k2 pool, aka the intel mbuf cluster pool, gets set up to allocate > at least 2048 + 2 bytes, which gets rounded up by 64 bytes to 2112 > bytes. this diff makes ix move the reception of packets to the end of > the 2112 byte allocation so there's space left at the front of the mbuf. > > this in turn makes it more likely that an m_prepend at another point in > the system will work without an extra mbuf allocation. eg, if you're > bridging or routing between vlans and vlans on svlans somewhere else, > this will be a bit faster with this diff. > > thoughts? ok?
I think using m_align() here may be benefitial. Since it does exactly that. Apart from that I have to agree, shifting the packet back makes a lot of sense. > Index: dev/pci/if_ix.c > =================================================================== > RCS file: /cvs/src/sys/dev/pci/if_ix.c,v > retrieving revision 1.152 > diff -u -p -r1.152 if_ix.c > --- dev/pci/if_ix.c 22 Jun 2017 02:44:37 -0000 1.152 > +++ dev/pci/if_ix.c 25 Feb 2019 00:40:47 -0000 > @@ -2445,7 +2445,7 @@ ixgbe_get_buf(struct rx_ring *rxr, int i > return (ENOBUFS); > > mp->m_len = mp->m_pkthdr.len = sc->rx_mbuf_sz; > - m_adj(mp, ETHER_ALIGN); > + m_adj(mp, mp->m_ext.ext_size - sc->rx_mbuf_sz); > > error = bus_dmamap_load_mbuf(rxr->rxdma.dma_tag, rxbuf->map, > mp, BUS_DMA_NOWAIT); > -- :wq Claudio
