access-list 21 permit 0.0.0.0 238.184.62.0

Hello

Ok playing a while with my only 3 router that I have, I did this A Router advertising redistributed statics routes via RIP to a RIP/OSPF Router and then in this RIP/ OSPF Router apply a route map to select even routes and advertise it to other OSPF router.

So my question is What is the difference between this 2 acl? Ip access-list 21 permit 192.168.0.0 0.0.254.255 And Ip access-list 21 permit 0.0.0.0 238.184.62.0

I know that in the first case is wrong because the router put in the configuration Ip access-list 21 permit 192.168.1.0 0.0.254.255, but it work for this one and sure by luck

I have no clue how does this WildCard ACL was calculated.. Do you know the steps?

1w2d: RT: del 192.168.22.0 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: delete network route to 192.168.22.0 1w2d: RT: del 192.168.23.0 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: delete network route to 192.168.23.0 1w2d: RT: del 192.168.21.0 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: delete network route to 192.168.21.0 1w2d: RT: del 192.168.26.0 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: delete network route to 192.168.26.0 1w2d: RT: del 192.168.24.0 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: delete network route to 192.168.24.0 1w2d: RT: del 192.168.25.0 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: delete network route to 192.168.25.0 MJ# 1w2d: RT: add 192.168.22.0/24 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: add 192.168.24.0/24 via 10.80.80.2, ospf metric [110/20] 1w2d: RT: add 192.168.26.0/24 via 10.80.80.2, ospf metric [110/20]

On the RIP router you have this: S 192.168.25.0/24 is directly connected, Null0 S 192.168.24.0/24 is directly connected, Null0 S 192.168.26.0/24 is directly connected, Null0 S 192.168.21.0/24 is directly connected, Null0 S 192.168.23.0/24 is directly connected, Null0 S 192.168.22.0/24 is directly connected, Null0

This matches 192.168.x.y, where x is even and y is anything.

This matches x.y.z.0, where

x is any even value except 16 y is any multiple of 8 except 64 or 192 z is any even value below 192

To understand this, you have to look at the wildcards in binary. The 0 bits mean the corresponding bit in the address has to match exactly, the

1 bits mean the corresponding bits can be anything. So 238 is 11101110 in binary, and the address to match against is 00000000, so it matches anything whose binary representation is xxx0xxx0.

This matches 192.168.x.y, where x is odd and y is anything.

I don't understand your question. Wildcard ACLs are calculated by the administrator, not created by the router.