On 22 Mar 2021, at 13:07, Faisal Mahmood firstname.lastname@example.org wrote:
Thanks Jakub, I wasn't aware of the netaddr library but just gave it a play and it does seem very similar and I think it's very useful and completely valid.
I think the subtle difference is also that my implementation allows you to specify the next prefix as well, so it won't just find the next prefix of the same size. This is a common use case when you are trying to find free address spaces, you will need to look for networks of different sizes depending on what you are doing.
Currently, you could say that you were given a network of 10.200.20.0/24 and asked to split this network into a bunch of /26 networks, you can do this easily using the subnets method:
[IPv4Network('10.200.20.0/26'), IPv4Network('10.200.20.64/26'), IPv4Network('10.200.20.128/26'), IPv4Network('10.200.20.192/26')]
That is very simple and effective, but not a very realistic example of how you would split networks up. Given how limited the IPv4 address space is, normally you may have to use that /24 block for multiple things, so can't just simply split it up into /26's, you may need to instead get two /30's, one /27 and one /25. Currently, I don't think there is any straightforward way to do this without a 'next_network' method that I have implemented.
Example, given a network of 10.200.20.0/24, to get two /30's out of it, one /27 and one /25, I would do the following with my method:
first_network = IPv4Network("10.200.20.0/30")
# first_network = IPv4Network("10.200.20.0/30")
Then get the next one (note not specifying prefix just gives me another /30 - i.e. same prefix size):
second_network = first_network.next_network()
# second_network = IPv4Network("10.200.20.4/30")
Then I would need to get the /27, so do this:
third_network = second_network.next_network(new_prefixlen=27)
# third_network = IPv4Network("10.200.20.32/27)
Finally the /25:
fourth_network = third_network.next_network(new_prefixlen=25)
# fourth_network = IPv4Network("10.200.20.128/25)
When you are dealing with the same prefix size for each new network, I think it's just a simple case of adding 1 to the broadcast address each time, but when you have different prefix sizes it's a bit more complicated.
On Sat, 20 Mar 2021 at 22:04, Jakub Stasiak email@example.com wrote:
I don’t know if this is gonna support Faisal’s cause (“It’s used in a third party library in the wild, that means stdlib could use it!”) or the exact opposite (“it’s provided by a third party library, so may as well use third party library and stdlib doesn’t necessarily need this”) but the quite popular netaddr library that I’m a maintainer of does have IPNetwork.next() and IPNetwork.previous() methods. The main difference is that in netaddr:
- next() and previous() can produce networks arbitrary number of steps away, not just the first closest network forwards or backwards
- going from a network to its supernetwork or subnetwork is done through a separate set of supernet() and subnet() methods
I can’t say how many library users actually consume this particular section of the API, of course, but I expect this API to be used and it seems rather useful.
 https://netaddr.readthedocs.io/en/latest/api.html#netaddr.IPNetwork.next  https://netaddr.readthedocs.io/en/latest/api.html#netaddr.IPNetwork.previous  https://netaddr.readthedocs.io/en/latest/api.html#netaddr.IPNetwork.subnet  https://netaddr.readthedocs.io/en/latest/api.html#netaddr.IPNetwork.supernet
Thank you for the explanation – now I understand better how that prefix parameter is useful.
It’s not *that* difficult to emulate this using netaddr, there’s an extra switch-to-supernet or switch-to-subnet step but that’s about it:
from netaddr import IPNetwork first_network = IPNetwork("10.200.20.0/30") second_network = first_network.next() second_network
third_network = second_network.supernet(27).next() third_network
fourth_network = third_network.supernet(25).next() fourth_network
That said, if merging this into the Python stdlib doesn’t work out I’m open to improving netaddr’s interface to better serve this use case.
Best wishes, Jakub