Lets suppose we have a computer with three NICs:

NIC_1: 172.16.1.1/24

NIC_2: 192.168.1.1/24

NIC_3: 103.55.34.32/24

We send a packet with the destination IP address being 255.255.255.255.

1) if the default gw is on the 103.55.34/24 network, will the packet be sent through NIC_3 or will the packet be sent through every NIC?

2) If we make an assumption that the whole internet is a one big network and this network network address is 0.0.0.0/0, the 255.255.255.255 address looks like a broadcast address for the whole network (for the internet), right?

3) What would be the meaning of a network address like this: 1.2.0.0/0? (is this correct and if not, why not)

link|improve this question

1  
1.2.0.0/0 cannot be a network address because a network address has all the host bits set to zero. Since /0 means all the bits are host bits, the only valid /0 network address is 0.0.0.0/0. – David Schwartz Jan 9 at 1:24
feedback

1 Answer

up vote 5 down vote accepted
  1. It will be sent through the first interface in the route table that matches the destination, evaluated in the order static route>route>gateway>metric. If you type route print you'll see something similar to the following:

    Network Destination        Netmask          Gateway       Interface  Metric
    [Loads of entries for your various NICs, routes and static routes]
    255.255.255.255  255.255.255.255    103.55.34.254   103.55.34.32       1
    Default Gateway:       103.55.34.254
    

    Indicating that the broadcast/multicast test ping will use your NIC_3

  2. Assuming you had a network 0.0.0.0/0, then yes 255.255.255.255 WOULD be the broadcast address. The Internet however is not one big network, it's a series of interconnected completely separate networks (hence inter-net!)

link|improve this answer
3  
See my answer here for an explanation of the 255.255.255.255 broadcast address: serverfault.com/questions/338593/… – joeqwerty Jan 8 at 23:09
feedback

Your Answer

 
or
required, but never shown

Not the answer you're looking for? Browse other questions tagged or ask your own question.