[vJASS] 2^n system

[Starquizer]

The system that is like that:
1 = 1
2 = 2
3 = 1 + 2
4 = 4
5 = 4 + 1
6 = 4 + 2
7 = 4 + 2 + 1
8 = 8
9 = 8 + 1
10 = 8 + 2
...
and so on, I don't know what this system is called but all I know is that it is 2ⁿ where n is a +ve integer.

 

[ap0calypse]

Shouldn't 9 and 10 be '8+1' and '8+2' respectively?

Anyway, it's not that much of a system.
You just have to use a loop (high to low), see if (yourValue) > PowerOf2(loop).
If so, then (yourValue) = (yourValue - PowerOf2(loop)) and you know 1 of the powers.
Continue until (yourValue) == 0, or until the loop hits 0.

Edit: by the way, BonusMod uses this.
You can have a look at that if you want.

 

[Nestharus]

It's just binary...

you represent every possible value using binary digits.

A given digit position is n and then that digit can either be 0 or 1.

 

[Dr Super Good]

The system is called a Binary Number.

 

[Starquizer]

The system is called a Binary Number.

I guessed it but wasn't sure, thanks guys for the help.