Bitwise AND & usable host counts

Network, broadcast, and hosts

Lesson 4 of 20 · 11 min · Module 1: Addresses are numbers

Two addresses are on the same network when their network bits match. A bitwise AND with the mask removes the host bits and reveals the network address.

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By the end, you’ll be able to…

Find the network address with AND
Find the broadcast address with OR and NOT
Count usable host addresses

The idea

Convert the address to one integer: (a << 24) | (b << 16) | (c << 8) | d. The network is ip & mask. The broadcast address sets every host bit to 1: network | (~mask & 0xFFFFFFFF). For prefixes up to /30, usable hosts are 2 ** (32 - prefix) - 2: the network and broadcast addresses are reserved.

ip = (192 << 24) | (168 << 16) | (1 << 8) | 77
mask = (0xFFFFFFFF << 8) & 0xFFFFFFFF
print(ip & mask)

01AND keeps what the mask keeps

Where the mask has a 1, the network bit stays; where it has a 0, the result is 0. 192.168.1.77 AND 255.255.255.0 is 192.168.1.0.

02Broadcast fills the host bits

~mask flips every bit, giving ones in the host part. OR-ing them onto the network address gives the last address in the subnet: 192.168.1.255 for a /24.

print(subnet_info("10.20.30.130", 26)["broadcast"])

Output

10.20.30.191

03Two addresses are reserved

A /26 has 2^6 = 64 addresses. The first identifies the network and the last is the broadcast address, which leaves 62 for devices.

Your turn

Write subnet_info(ip, prefix) for prefixes 1 to 30. Return a dictionary with "network" and "broadcast" as dotted strings and "hosts" as the number of usable host addresses. Calculate it yourself: do not import ipaddress in this lesson.

What the checks look for

Describes 192.168.1.77/24
Describes 10.20.30.130/26
Describes a /30 point-to-point link and a /20
Does not use the ipaddress module

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Take this with you

Network = address AND mask; broadcast = network OR inverted mask.

Go a little further

Add "first" and "last" keys for the first and last usable host addresses.