On Jun 7, 2005, at 5:29 PM, Julian Elischer wrote:
I agree with your suggestion, but how can you have an ip_vhl of
0x42? Doesn't a valid IP packet need to have a header length of
at least 5 (5 << 2 == 20 bytes)?
huh?
the first byte of an IP packet is not the length.. the first byte
you see on a
packet trace will be the 4 bit version followed by a 4 bit HEADER
length,
followed by 8 bits of TOS (type of Service) and 16 bits of total
packet length.
Agreed, but note that I said "header length" above, too. The
smallest possible IP packet, without any IP options set, has an IP
header length of 20, yes?
http://www.ietf.org/rfc/rfc0791.txt, page 10, says:
" IHL: 4 bits
Internet Header Length is the length of the internet header in 32
bit words, and thus points to the beginning of the data. Note that
the minimum value for a correct header is 5."
...and from <netinet/ip.h>:
#define IPVERSION 4
/*
* Structure of an internet header, naked of options.
*/
struct ip {
#ifdef _IP_VHL
u_char ip_vhl; /* version << 4 | header
length >> 2 */
#else
#if BYTE_ORDER == LITTLE_ENDIAN
u_int ip_hl:4, /* header length */
ip_v:4; /* version */
#endif
#if BYTE_ORDER == BIG_ENDIAN
u_int ip_v:4, /* version */
ip_hl:4; /* header length */
#endif
[ ... ]
#define IP_VHL_BORING 0x45
0x45 means ip_v == 4 (IPv4) and ip_hl == 5 (five 32-bit words, or 20
bytes >> 2).
(it's in big-endian format.. beware)
Sure. Network wire order is always big endian. :-)
--
-Chuck
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