#define LT16(n) n, n, n, n, n, n, n, n, n, n, n, n, n, n, n, n
static const unsigned char LogTable256[256] =
{
	0, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4,
	LT16(5), LT16(6), LT16(6), LT16(7), LT16(7), LT16(7), LT16(7),
	LT16(8), LT16(8), LT16(8), LT16(8), LT16(8), LT16(8), LT16(8), LT16(8)
};

static inline int
AllocSetFreeIndex(Size size)
{
	int			idx;
	unsigned int t,
				tsize;

	if (size > (1 << ALLOC_MINBITS))
	{
		tsize = (size - 1) >> ALLOC_MINBITS;

		/*
		 * At this point we need to obtain log2(tsize)+1, ie, the number of
		 * not-all-zero bits at the right.  We used to do this with a
		 * shift-and-count loop, but this function is enough of a hotspot to
		 * justify micro-optimization effort.  The best approach seems to be
		 * to use a lookup table.  Note that this code assumes that
		 * ALLOCSET_NUM_FREELISTS <= 17, since we only cope with two bytes of
		 * the tsize value.
		 */
		t = tsize >> 8;
		idx = t ? LogTable256[t] + 8 : LogTable256[tsize];

		Assert(idx < ALLOCSET_NUM_FREELISTS);
	}
	else
		idx = 0;

	return idx;
}