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TiFFolk
Asked: 2010-02-05 05:40:12 +0800 CST

Iptables massive 1:1 NAT

  • 772

I have to connect two LANs: LAN1: 10.10.0.0/16 and LAN2: 192.168.0.0/16. I can't do simple routing, because 192.168.0.0/16 net is prohibited in LAN1, so I am thinking of using Full cone nat (1:1) to translate 192.168.x.y/16 to 10.11.x.y/16. Each translation is done by this rules:

iptables -t nat -A PREROUTING -d 10.11.0.0/16 -j DNAT --to-destination 192.168.0.0/16
iptables -t nat -A POSTROUTING -s 192.168.0.0/16 -j SNAT --to-source 10.11.0.0/16

But I will have to enter 254*254*2 rules, what will, I think, result in enormous performance degradation. So, is there a way to write such one-to-one translation with minimum number of rules?

nat iptables

3 Answers

  1. I am not sure if it is present in all kernels, but what you may be looking for is the NETMAP target.

    From the iptables man page

    NETMAP
 This target allows you to statically map a whole network of 
 addresses onto another network of addresses. It can only be 
 used from rules in the nat table. 

 --to address[/mask]
     Network address to map to. The resulting address will be 
     constructed in the following way: All 'one' bits in the 
     mask are filled in from the new 'address'. All bits that 
     are zero in the mask are filled in from the original 
     address.
    • 16

  2. Like the first answer said, use -j NETMAP:

    # iptables -t nat -A PREROUTING -d 10.11.0.0/16 -j NETMAP --to 192.168.0.0/16
# iptables -t nat -A POSTROUTING -s 192.168.0.0/16 -j NETMAP --to 10.11.0.0/16

    It's probably a good idea to add -d 10.10.0.0/16 in the POSTROUTING row as well.

    • 8

  3. You can do this with a small shell script

    #!/bin/bash
for (( i=0 ; $i<256 ; i=$i+1 )) ; do
  for (( j=0 ; $j<256 ; j=$j+1 )) ; do
    iptables -t nat -A PREROUTING -d 10.11.$i.$j/16 -j DNAT --to-destination 192.168.$i.$j/16
    iptables -t nat -A POSTROUTING -s 192.168.$i.$j/16 -j SNAT --to-source 10.11.$i.$j/16
  done
done

    But I think thereis an error. I think it should be /32 and not /16.

    • 0