From: Timur Tabi
> Sent: 11 October 2017 20:52
> The EMAC has a restriction that the upper 32 bits of the base addresses
> for the RFD and RRD rings must be the same. The ensure that restriction,
> we allocate twice the space for the RRD and locate it at an appropriate
> address.
>
> We also re-arrange the allocations so that invalid addresses are even
> less likely.
>
> Signed-off-by: Timur Tabi <ti...@codeaurora.org>
> ---
> drivers/net/ethernet/qualcomm/emac/emac-mac.c | 39
> ++++++++++++++++-----------
> 1 file changed, 24 insertions(+), 15 deletions(-)
>
> diff --git a/drivers/net/ethernet/qualcomm/emac/emac-mac.c
> b/drivers/net/ethernet/qualcomm/emac/emac-
> mac.c
> index 9cbb27263742..0f5ece5d9507 100644
> --- a/drivers/net/ethernet/qualcomm/emac/emac-mac.c
> +++ b/drivers/net/ethernet/qualcomm/emac/emac-mac.c
> @@ -734,6 +734,11 @@ static int emac_rx_descs_alloc(struct emac_adapter *adpt)
> rx_q->rrd.size = rx_q->rrd.count * (adpt->rrd_size * 4);
> rx_q->rfd.size = rx_q->rfd.count * (adpt->rfd_size * 4);
>
> + /* Check if the RRD and RFD are aligned properly, and if not, adjust. */
> + if (upper_32_bits(ring_header->dma_addr) !=
> + upper_32_bits(ring_header->dma_addr + ALIGN(rx_q->rrd.size, 8)))
> + ring_header->used = ALIGN(rx_q->rrd.size, 8);
> +
Isn't the memory allocated by a single kzalloc() call?
IIRC that guarantees it doesn't cross a power or 2 boundary less than
the size.
So if you allocate any size between 4k and 8k it won't cross an odd
4k boundary (etc).
So these checks are entirely pointless.
David
