Re: Bitwise operations on a byte array
Chris wrote:
Daniel Pitts wrote:
Chris wrote:
I'd like to do something like this:
byte [] array0 = ...
byte [] array1 = ...
array0 |= array1;
The idea being that all the bytes in array0 get ORd with the ones in
array1. Is there an efficient way to do this?
This is just too ugly and unnecessarily slow:
for (int i = 0; i < array0.length; i++) {
array0[i] |= array1[i];
}
Have you considered using a BitSet instead?
Internally, a BitSet is just an array of ints or longs, and faces the
same problem.
Not exactly the same problem...
BitSet's have the operations defined already, so you don't have a "too
ugly" implementation.
Also, they use integers, for a speed increase of a factor around 4
(could be more).
On most modern processors (as someone else pointed out), you would have
to implement it as a loop through every element anyway. There is no
instruction that I know of that says "For every bit in the range a
through b, or it with the corrisponding bit in the range c through
(c+b-a)"
BitSet is definitly the way to go if you find yourself doing those
operations often.
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