Question

void convert(int bTen) {
    System.out.println("Base 10 = " + bTen);
    int bTwo = 0;
    int leftOver = bTen;
    while (leftOver > 0) {
        int i = 0;
        int remains = 0;
        while (remains >= 0) {
            remains = leftOver - (int)Math.pow(2, i);
            i++;
        }
        bTwo += Math.pow(10, i - 2);
        leftOver = leftOver - (int)Math.pow(2, i - 2);
    }
    System.out.println("Base 2 = " + bTwo);
}

I was wondering why the above code can convert numbers in base-10 to base-2. I know how to write a program to convert base-2 to base-10 but I can’t seem to understand how to do the reverse.

Answer 1

First, I think this conversion is unreasonable and shouldn’t be done such way.

Base-10 or Base-2 is simply a textual representation of same number. However your logic is changing a number (A) to another number (B), for which if you are reading B in base-10 it will look the same as base-2 of A.

Anyway, the idea of the logic of your quoted code is something like this:

loop until input is not zero (i.e. loop until all binary digits processed)
  find the position (i) of the most significant bit that is 1
  set position of i of result being 1
  subtract 2^i from input
end loop