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I understand that for classful networks in EIGRP we do not need a subnet mask. Why is this? When I do a show ip protocols the routes appear without a subnet mask which looks somewhat odd.

If I have a 192.168.0.0/16 network and just introduce the network into the EIGRP routing protocol as 192.168.0.0, how will EIGRP know that it is 192.168.0.0/16 and not say 192.168.0.0/24?

Or does EIGRP understand the network address from the host addresses assigned to the router's DC interfaces?

Thanks.

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    What is the point of this question? Classful networking has not been used by anyone with any knowledge in over 20 years. This is like asking how to write a program to run efficiently on an 8086 processor.
    – YLearn
    Commented Apr 22, 2015 at 12:19
  • @YLearn Does Stuart Smith's answer also answer your comment?
    – Todd Wilcox
    Commented Apr 23, 2015 at 13:44
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    @ToddWilcox, no. My point is that no one should be considering "classful networks" in today's day and age and shouldn't use classful defaults. Using classful networking, there is no way to have 192.168.0.0/16 as a valid network.
    – YLearn
    Commented Apr 23, 2015 at 22:18

1 Answer 1

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By default EIGRP uses a classful network, meaning that it assumes that an address in the class A range with have an 8-bit subnet mask. There is no automatic process of checking other than the assumption that if you do not specify the subnet mask in wildcard mask format ( host mask ) then you want to use the default subnet mask for the IP range you've specified.

http://study-ccna.com/eigrp-configuration

As @Todd Wilcox pointed out below, each address range has a class, originally specified in RFC791 ( https://www.ietf.org/rfc/rfc791.txt ), excluding special use addresses ( https://www.rfc-editor.org/rfc/rfc3330 ).

Basically:

Class      Sig Bits      Network Mask  Range

A          0             8             0.0.0.0   - 127.255.255.255

B         10             16            128.0.0.0 - 191.255.255.255

C        110             24            192.0.0.0 - 223.255.255.255

D       1110             undefined     224.0.0.0 - 239.255.255.255

E       1111             undefined     240.0.0.0 - 255.255.255.255

So converting the first octet ( first group of decimal digits in an IP address: 192.168.1.0 ) into binary gives us 11000000, matching the MSB ( most significant bits ) of a C class address, implying a default subnet mask of /24 bits or 255.255.255.0.

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