Subject: Re: [RFC] skge csum problems

From: Al Viro <[email-masked]>
Date: Mon, 24 Dec 2007 18:39:50 +0000

On Mon, Dec 24, 2007 at 02:15:40PM +0100, Andi Kleen wrote:
> Al Viro <[email-masked]> 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).