Mayeul <mayeul.marguet@free.fr> wrote in
news:4a708393$0$9973$426a74cc@news.free.fr:
RVic wrote:
<snip>
Anyway, if all you want to do is to check for even parity, you need
neither. All you need to do is count the number of 1's in a byte and
check this number is even.
(On a side note, I know of no direct way to do that in Java.
boolean even = true;
for(int i = 0; i < 8; i++) {
if((theByte & (1 << i)) != 0) {
even = !even;
}
}
Maybe?)
Or a table lookup could be used. There are only 256 values in the table.
The table could be computed offline and stored in the source code, or the
table could be computed at initialization (perhaps by using the counting
routine that you showed above). Also, I suspect that there may be faster
ways to count. Depending upon how the jvm is implemented, having (1 << i) in
the loop may be inefficent. It may be more efficient to do
if ((theByte & 1) != 0)
as the test, and
theByte >>= 1 ;
for the shift at the end of each iteration. Only one shift is needed per
iteration, so there are no questions about whether the processor has a barrel
shifter.
exercise for the reader. (Extending it to float and double is left