Topic.
float f =Float.parseFloat(binaryString)
int binary= Float.floatToIntBits( f)
//isn't working as I had hoped.
Basically I need to do bitwise ops on a binary String. Obviously not possible.
Thanks again.
That's what I was using and it wasn't working. So let binaryString = "101". Then:
int bin = Integer.parseInt( binaryString, 2)
returns bin as the number 101 or the number 5?
I've been working with parseInt for three weeks and I still get so confused. I know this is very basic sorry but I appreciate it.
> the API says that would say bin=5.
> I need bin=101.
Do you really want 101? Why?
What I mean is you start with the string "101" and the parse method returns...
five, V, the number of fingers on one hand (however you want to write that
number). Why do you want a number that's 20 and a bit times as big?
If you do want a hundred and something returned, parse it base 10. (But
stop calling it a binary string, because that's not how you're using it).
You mentioned bitwise operations. Try:int bin1 = Integer.parseInt("101", 2); // bin1 is 5
int bin2 = Integer.parseInt("110", 2); // bin2 is 6
// prints 100 - Is this not what you want?
System.out.println(Integer.toBinaryString(bin1 & bin2));
Yes, it is confusing.
