Re: Bitwise operations on a byte array

Patricia Shanahan <>
Fri, 05 Jan 2007 05:30:19 GMT
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.

I'm not sure what you mean by "the same problem". The byte-based method
may be unnecessarily slow, because of the numeric promotions. That does
not apply to BitSet.

The logic for doing a bitwise or in the processor has a finite width,
probably 64 bits. Something, somewhere, must chop the data into 64 bit
chunks and issue a series of commands, one for each chunk. What's so bad
about BitSet doing it?


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