Check if two CIDR addresses intersect?

Question

Given two CIDR addresses say 192.168.2.0/14 and 192.168.2.0/32

How do I check if two ip addresses overlap in \"python2.6\"??

I have gone through netaddr and

Answer 1

If don't have netaddr at hand for testing, but I think you could check if the first and last address of the first network are both contained in the second:

net_1 = IPNetwork("192.168.2.0/14")
net_2 = IPNetwork("192.168.2.0/32")
if net_1.first in net_2 and net_1.last in net_2:
    # do something

BTW, IPNetwork line 1102 defines a __contains__ method. But I'm not certain the line 1127 isn't broken ? You should test and report a bug if it is.

Answer 2

I wrote this simple command line tool, based on the netaddr lib.

pip install ipconflict

Example:

ipconflict 10.0.0.0/22 10.0.1.0/24

Output:

conflict found: 10.0.1.0/24 <-> 10.0.1.0/22

Answer 3

Using ipaddr:

>>> import ipaddr
>>> n1 = ipaddr.IPNetwork('192.168.1.0/24')
>>> n2 = ipaddr.IPNetwork('192.168.2.0/24')
>>> n3 = ipaddr.IPNetwork('192.168.2.0/25')
>>> n1.overlaps(n2)
False
>>> n1.overlaps(n3)
False
>>> n2.overlaps(n3)
True
>>> n2.overlaps(n1)
False

Answer 4

I'll assume you actually want both CIDRs to represent ranges, even though in your example, 192.168.2.0/32 represents only one address. Also note that in 192.168.2.0/14, the .2. is meaningless, because the 14-bit prefix doesn't reach the third octet.

Anyway, there are a several ways to do this. You could notice that for them to overlap, one must always be a subset of the other:

def cidrsOverlap(cidr0, cidr1):
    return cidr0 in cidr1 or cidr1 in cidr0

Or you could notice that for the ranges to overlap, the first range's lowest address must be less than or equal to the second range's highest address, and vice versa. Thus:

def cidrsOverlap(cidr0, cidr1):
    return cidr0.first <= cidr1.last and cidr1.first <= cidr0.last

print cidrsOverlap(IPNetwork('192.168.2.0/24'), IPNetwork('192.168.3.0/24'))
# prints False

print cidrsOverlap(IPNetwork('192.168.2.0/23'), IPNetwork('192.168.3.0/24'))
# prints True