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[ahu@home ahu]$ ip address show
1: lo: <LOOPBACK,UP> mtu 3924 qdisc noqueue
link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00
inet 127.0.0.1/8 brd 127.255.255.255 scope host lo
2: dummy: <BROADCAST,NOARP> mtu 1500 qdisc noop
link/ether 00:00:00:00:00:00 brd ff:ff:ff:ff:ff:ff
3: eth0: <BROADCAST,MULTICAST,PROMISC,UP> mtu 1400 qdisc pfifo_fast qlen 100
link/ether 48:54:e8:2a:47:16 brd ff:ff:ff:ff:ff:ff
inet 10.0.0.1/8 brd 10.255.255.255 scope global eth0
4: eth1: <BROADCAST,MULTICAST,PROMISC,UP> mtu 1500 qdisc pfifo_fast qlen 100
link/ether 00:e0:4c:39:24:78 brd ff:ff:ff:ff:ff:ff
3764: ppp0: <POINTOPOINT,MULTICAST,NOARP,UP> mtu 1492 qdisc pfifo_fast qlen 10
inet 184.108.40.206 peer 220.127.116.11/32 scope global ppp0
Let's examine eth0 somewhat closer. It says that it is related to the inet address '10.0.0.1/8'. What does this mean? The /8 stands for the number of bits that are in the Network Address. There are 32 bits, so we have 24 bits left that are part of our network. The first 8 bits of 10.0.0.1 correspond to 10.0.0.0, our Network Address, and our netmask is 255.0.0.0.
Now the last two lines:
The other bits are connected to this interface, so 10.250.3.13 is directly available on eth0, as is 10.0.0.1 for example.
What does the last two lines << 10.250.3.13>> mean?
Last edited by your_shadow03; 06-18-2009 at 01:33 AM.