Re: [9fans] OUdphdr

["Russ Cox" <rsc@swtch.com>]

> the comment in this diff confuses me.  is OUdphdr now used
> for ip6 as opposed to Udphdr?  i guess i don't quite understand
> the emphasis on v6 since plan 9 represents all ip addresses as v6.
> 
> also, why are there two udp packet formats defined?
> 
> - erik
> 
>   enum 
> /n/sources/plan9//sys/include/ip.h:74,87 - ip.h:74,87
>   typedef struct OUdphdr OUdphdr;
>   struct OUdphdr
>   {
> - 	uchar	raddr[IPaddrlen];	/* V6 remote address */
> - 	uchar	laddr[IPaddrlen];	/* V6 local address */
> - 	uchar	rport[2];		/* remote port */
> - 	uchar	lport[2];		/* local port */
> + 	uchar	raddr[IPaddrlen];	/* remote address and port */
> + 	uchar	laddr[IPaddrlen];	/* local address and port */
> + 	uchar	rport[2];
> + 	uchar	lport[2];
>   };

Don't worry about the comment.  
It's just a clarification (since obviously
raddr doesn't have a port in it).
Nothing has actually changed in the format.

As the name suggests, OUdphdr is old and 
obsolete.  The new Udphdr added the interface
where the packet came in, which is needed for
things like dhcpd.  

There are two formats because the user space
programs and kernels couldn't be updated all
in one go.  Now that the kernels with the new
headers have propagated, the programs
that still use the old header format should really be
changed to use the new format.  Those programs
appear to be just ip/dhcpclient, ip/ipconfig,
ip/rip, and ip/udpecho.  But like a lot of Plan 9 
maintenance, it just never happened.

Russ