On Mon, Jun 04, 2007 at 12:03:57PM +0800, Wei Yongjun wrote:
So I want to know how the route announce a MTU larger then 576, such as
1280? RFC says ICMP error message return as much as we can without
exceeding 576 bytes.
I think there is a misunderstanding here. The RFC is talking about
how much of the payload may be included in the ICMP packet. It is
not talking about the length field in the original IP header. That
must be left untouched.
As you said, the RFC is talking about the payload of ICMP packet, it's
not greater then 576, if router announce a MTU larger then 1280, the
original IP header must be a size of 704? The format of ICMP message is
like this:
IPv4 header
ICMP header
payload
And ipv4 router alaways return as a packet size 576 because of this.
If MTU is greater then 576, HOST will used "mtu = guess_mtu(old_mtu)" to
get a MTU. Method of send ICMP Fragmentation Needed message is
difference from receive,
Which router is doing that?
The latest kernel 2.6.21.3 also doing so. The rule to send ICMP message
limit this.
Ref to net/ipv4/icmp.c
line 433 void icmp_send(struct sk_buff *skb_in, int type, int code,
__be32 info)
line 434 {
...
line 572 /* RFC says return as much as we can without exceeding 576
bytes. */
line 573
line 574 room = dst_mtu(&rt->u.dst);
line 575 if (room > 576)
line 576 room = 576;
line 577 room -= sizeof(struct iphdr) + icmp_param.replyopts.optlen;
line 578 room -= sizeof(struct icmphdr);
line 579
line 560 icmp_param.data_len = skb_in->len - icmp_param.offset;
line 561 if (icmp_param.data_len > room)
line 562 icmp_param.data_len = room;
line 563 icmp_param.head_len = sizeof(struct icmphdr);
line 576 do this.
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Wei Yongjun
Development Dept.I
Nanjing Fujitsu Nanda Software Tech. Co., Ltd.(FNST)
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Nanjing, 210029, China
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