This is the second part of the Arbitrary Precision Integer (ABI) workshop. Part one was about the introduction and how to convert integers to ABIs. If you haven't read that one, I suggest you do so now.

This one will be just a short snippet on how to convert our ABI back to a normal integer.

You can find the code in Haxe on github. However, the basic idea is easy to adapt for other languages.

### Lets get to work!

The first thing we should do is to check if an integer variable can even hold the number we want to store, or if an overflow would occur. If that is the case, an appropriate warning should be printed. Depending on the use case, you might even thrown an exception.

Now we start by creating an empty integer (the return value) with value 0.
We start to iterate through the individual bits of our ABI, starting from the end. This means we start the smallest number 1 and end with the biggest power of 2 that is a part of our integer.

Now we just need to check if the bit we are looking at is true. If this is the case, we add

`(1 << i)`

to the return value. Here the arrows indicate the bitshift operation (multiplication by a power of two) and i is the position of the bit we are looking at.

After summing up all the powers of two, we can just return the sum and voila: there is our integer.