Hi,
Thanks! Your solution is faster, but I like to wait implementing it
unless it is really a measurable performance problem, because the
solution is quite complex :-) Maybe I will just switch to BitSet at
some point.
Regards,
Thomas
On Thu, Oct 16, 2008 at 5:25 PM, Alex <[EMAIL PROTECTED]> wrote:
>
> Hello Thomas,
>
> I checked again and you are correct the code is not broken, it just
> iterates bit by bit twice on the first long and on the last long.
> If you'd like to try below is the code that does not iterate bit by
> bit at all. I did not test it with the H2 myself.
>
> public int nextSetBit1(int fromIndex) {
> int result = -1;
>
> int dataIndex = fromIndex >> ADDRESS_BITS;
> int bitIndex = fromIndex - (dataIndex << ADDRESS_BITS);
>
> long l = data[dataIndex];
> l = clearLowBits(l, bitIndex);
> while(l==0) {
> dataIndex++;
> if (dataIndex >= data.length) break;
> l = data[dataIndex];
> }
>
> if (l != 0)
> result = (dataIndex << ADDRESS_BITS) + lowestSetBit(l);
>
>
> return result;
> }
>
>
> private static int lowestSetBit(long data)
> {
> int result = 0;
>
> data &= -data;
>
> if ((data & 0xffffffff00000000L) != 0) result += 32;
> if ((data & 0xffff0000ffff0000L) != 0) result += 16;
> if ((data & 0xff00ff00ff00ff00L) != 0) result += 8;
> if ((data & 0xf0f0f0f0f0f0f0f0L) != 0) result += 4;
> if ((data & 0xccccccccccccccccL) != 0) result += 2;
> if ((data & 0xaaaaaaaaaaaaaaaaL) != 0) result += 1;
>
> return result;
> }
>
> private static long clearLowBits(long data, int cnt)
> {
> long result = data;
>
> if (cnt > 0)
> {
> long mask = 1;
> if (cnt > 1) mask = (0xAAAAAAAAAAAAAAAAL >>> (BITS-cnt)) |
> (0xAAAAAAAAAAAAAAAAL
> >>> (BITS-(cnt-1)));
> result &= ~mask;
> }
>
> return result;
> }
>
>
> >
>
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