On Mon, 24 Dec 2007 18:39:50 +0000
Al Viro <[EMAIL PROTECTED]> wrote:
> On Mon, Dec 24, 2007 at 02:15:40PM +0100, Andi Kleen wrote:
> > Al Viro <[EMAIL PROTECTED]> writes:
> > >
> > > Checksum is fixed-endian and we want it that way; IOW, what we end up
> > > storing in skb->csum should be fixed-endian as well.
> >
> > AFAIK skb->csum is always native endian because it normally
> > needs to be manipulated further even for RX.
>
> No. It needs to be manipulated, but that's exactly why it can't be
> (and isn't) kept host-endian. Large part of the reason why checksums are
> done the way they are done (operations mod 0xffff, etc.) is that
> they can be implemented via native arithmetics without any conversions;
> e.g. if you do
>
> add(u8 a[2], u8 b[2], u8 sum[2])
> {
> u32 x = *(u16 *)a + *(u16 *)b;
> if (x > 0xffff)
> x -= 0xffff;
> *(u16 *)sum = x;
> }
>
> you will get the same behaviour on big- and little-endian boxen, even though
> the intermediate integer values will be of course different.
>
> skb->csum *must* be stored in the same order on l-e and b-e boxen; that
> way you don't need to convert it or raw data when updating the sucker [*].
>
> [*] it's slightly more complicated since skb->csum is 4-byte, not 2-byte
> and the real invariant is "checksum of 4-octet array at &skb->csum must
> not depend on host" (so e.g XX YY 00 00 and 00 00 XX YY are equivalent -
> checksum doesn't change from reordering octet pairs; XX YY 00 00 and
> 00 00 YY XX are very definitely *NOT* equivalent; odd and even bytes
> can't be exchanged).
Did you test this on real hardware?
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