From: Or Gerlitz [mailto:ogerlitz@mellanox.com]

On 10/30/2014 6:25 PM, David Laight wrote:

quoted

quoted

quoted

+static inline __wsum csum_ipv6_magic_nofold(const struct in6_addr *saddr,
+					    const struct in6_addr *daddr,
+					    __u32 len, unsigned short proto,
+					    __wsum sum)
+{
+	__wsum res = sum;
+
+	res = csum_add(res, (__force __wsum)saddr->in6_u.u6_addr32[0]);
+	res = csum_add(res, (__force __wsum)saddr->in6_u.u6_addr32[1]);
+	res = csum_add(res, (__force __wsum)saddr->in6_u.u6_addr32[2]);
+	res = csum_add(res, (__force __wsum)saddr->in6_u.u6_addr32[3]);
+	res = csum_add(res, (__force __wsum)daddr->in6_u.u6_addr32[0]);
+	res = csum_add(res, (__force __wsum)daddr->in6_u.u6_addr32[1]);
+	res = csum_add(res, (__force __wsum)daddr->in6_u.u6_addr32[2]);
+	res = csum_add(res, (__force __wsum)daddr->in6_u.u6_addr32[3]);

quoted

That probably generates a very long dependency chain.

Could you clarify this comment a bit?

csum_add() probably generates a 32bit 'add with carry' instruction.
So the above generates 8 instructions that have to be executed in series
(dependencies on the register and the carry flag).

On a 64bit cpu there are other options, eg adding 32bit values into
several 64bit registers, then adding those together and finally
collapsing the value to 32 then 16 bits.

Maybe __wsum does end up being 64bit (not looked), but gcc won't
generate a 'tree' of additions, it will still generate a dependency chain.

Hopefully the software 'checksum a buffer' function is written to
avoid these problems.

	David