IP Address CIDR Bitmask Conversion

Question

In looking for a programming function to validate an IP address within a subnet, I had to calculate the subnet mask from a given CIDR bitmask value ie. 192.168.0.0/24, the value 24 in this example.

The first step is to convert the bitmask value (24) into an unsigned 32bit integer. I found this on the internet (specific to our 24 bitmask example): mask = (( 2^24 ) - 1) * ( 2^(32-24) ). This gives me an integer value of 4,294,967,040 which is exactly the binary number I am looking for where the first 24 bits are 1, while the remaining bits are 0 11111111.11111111.11111111.00000000. Obviously, this works for any bitmask value 0-32.

I would like to understand why this works: mask = (( 2^bitmask ) - 1) * (2^( 32-bitmask ))

Answer 1

The binary expansion of $2^n-1$ consists of $n$ ones. This is because the binary expansion of $2^n$ consists of a one followed by $n$ zeroes. This is similar to $10^3-1 = 999$ in decimal.

If we want these $n$ bits to be the left $n$ bits out of $m$ bits, then we need to add $m-n$ zeroes to the right. This is the same as multiplying everything by $2^{n-m}$. Compare $100 \cdots 54 = 5400$. This operation is usually known as left shift, and many languages support it natively. For example, in C you would use the following code to generate the mask:

((1 << 24) - 1) << (32-24)