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Message-ID: <20131008133045.GI28411@zion.uk.xensource.com>
Date: Tue, 8 Oct 2013 14:30:45 +0100
From: Wei Liu <wei.liu2@...rix.com>
To: Paul Durrant <paul.durrant@...rix.com>
CC: <xen-devel@...ts.xen.org>, <netdev@...r.kernel.org>,
Wei Liu <wei.liu2@...rix.com>,
David Vrabel <david.vrabel@...rix.com>,
Ian Campbell <ian.campbell@...rix.com>
Subject: Re: [PATCH net-next v2 2/5] xen-netback: Add support for IPv6
checksum offload from guest
On Tue, Oct 08, 2013 at 11:58:13AM +0100, Paul Durrant wrote:
[...]
> -/*
> - * This is the amount of packet we copy rather than map, so that the
> - * guest can't fiddle with the contents of the headers while we do
> - * packet processing on them (netfilter, routing, etc).
> +/* This is a miniumum size for the linear area to avoid lots of
> + * calls to __pskb_pull_tail() as we set up checksum offsets.
> */
> -#define PKT_PROT_LEN (ETH_HLEN + \
> - VLAN_HLEN + \
> - sizeof(struct iphdr) + MAX_IPOPTLEN + \
> - sizeof(struct tcphdr) + MAX_TCP_OPTION_SPACE)
> +#define PKT_PROT_LEN 128
>
Where does 128 come from?
> static u16 frag_get_pending_idx(skb_frag_t *frag)
> {
> @@ -1118,61 +1113,72 @@ static int xenvif_set_skb_gso(struct xenvif *vif,
> return 0;
> }
>
> -static int checksum_setup(struct xenvif *vif, struct sk_buff *skb)
> +static inline void maybe_pull_tail(struct sk_buff *skb, unsigned int len)
> {
> - struct iphdr *iph;
> + if (skb_is_nonlinear(skb) && skb_headlen(skb) < len) {
> + int target = min_t(int, skb->len, len);
> + __pskb_pull_tail(skb, target - skb_headlen(skb));
> + }
> +}
> +
> +static int checksum_setup_ip(struct xenvif *vif, struct sk_buff *skb,
> + int recalculate_partial_csum)
> +{
[...]
>
> if (recalculate_partial_csum) {
> struct tcphdr *tcph = tcp_hdr(skb);
> +
> + header_size = skb->network_header +
> + off +
> + sizeof(struct tcphdr) +
> + MAX_TCP_OPTION_SPACE;
> + maybe_pull_tail(skb, header_size);
> +
I presume this function is checksum_setup stripped down to handle IPv4
packet. What's the purpose of changing its behaviour? Why the pull_tail
here?
> +static int checksum_setup_ipv6(struct xenvif *vif, struct sk_buff *skb,
> + int recalculate_partial_csum)
> +{
> + int err = -EPROTO;
> + struct ipv6hdr *ipv6h = (void *)skb->data;
> + u8 nexthdr;
> + unsigned int header_size;
> + unsigned int off;
> + bool done;
> +
> + done = false;
> + off = sizeof(struct ipv6hdr);
> +
> + header_size = skb->network_header + off;
> + maybe_pull_tail(skb, header_size);
> +
> + nexthdr = ipv6h->nexthdr;
> +
> + while ((off <= sizeof(struct ipv6hdr) + ntohs(ipv6h->payload_len)) &&
> + !done) {
> + /* We only access at most the first 2 bytes of any option header
> + * to determine the next one.
> + */
> + header_size = skb->network_header + off + 2;
> + maybe_pull_tail(skb, header_size);
> +
Will this cause performance problem? Is it possible that you pull too
many times?
> + switch (nexthdr) {
> + case IPPROTO_FRAGMENT: {
> + struct frag_hdr *hp = (void *)(skb->data + off);
> +
> + nexthdr = hp->nexthdr;
> + off += 8;
> + break;
> + }
> + case IPPROTO_DSTOPTS:
> + case IPPROTO_HOPOPTS:
> + case IPPROTO_ROUTING: {
> + struct ipv6_opt_hdr *hp = (void *)(skb->data + off);
> +
> + nexthdr = hp->nexthdr;
> + off += ipv6_optlen(hp);
> + break;
> + }
> + case IPPROTO_AH: {
> + struct ip_auth_hdr *hp = (void *)(skb->data + off);
> +
> + nexthdr = hp->nexthdr;
> + off += (hp->hdrlen+2)<<2;
> + break;
> + }
> + default:
> + done = true;
> + break;
Can you make the logic explicit here?
case IPPROTO_TCP:
case IPPROTO_UDP:
done = true;
break;
default:
break;
Another minor suggestion is that change "done" to "found" because you're
trying to find the two type of headers.
> + switch (nexthdr) {
> + case IPPROTO_TCP:
> + if (!skb_partial_csum_set(skb, off,
> + offsetof(struct tcphdr, check)))
> + goto out;
> +
> + if (recalculate_partial_csum) {
> + struct tcphdr *tcph = tcp_hdr(skb);
> +
> + header_size = skb->network_header +
> + off +
> + sizeof(struct tcphdr) +
> + MAX_TCP_OPTION_SPACE;
> + maybe_pull_tail(skb, header_size);
> +
Same question as IPv4 counterpart, why do you need to pull here?
Wei.
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