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I have to connect two LANs: LAN1: and LAN2: I can't do simple routing, because 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 -j DNAT --to-destination
iptables -t nat -A POSTROUTING -s -j SNAT --to-source

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?

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Why can't you just route between the two networks??? – Khb Feb 4 '10 at 14:29
BECAUSE(second line): I can't do simple routing, because net is prohibited in LAN1 – TiFFolk Feb 4 '10 at 15:24

3 Answers 3

Like the first answer said, use -j NETMAP:

# iptables -t nat -A PREROUTING -d -j NETMAP --to
# iptables -t nat -A POSTROUTING -s -j NETMAP --to

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

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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

 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 
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You can do this with a small shell script

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

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

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/32, I have already done that but I want to find a more beautiful way, than 254*254*2 rules – TiFFolk Feb 4 '10 at 21:20

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