🚀 go-pugleaf

What about using this algorithm to find the lowest bit set in a word

	if (p & (p-1)) return(-1);	/* returns -1 if > 1 bit set */
	return(bitcount(p ^ (p-1)));	/* use XOR, returns 32 if no bit set */

where bitcount counts the number of bits in a word?  bitcount exists
in the hardware of several cdc machines and elsewhere.  On other
machines, it can be implemented in 16 arithmetic ops or much less if
the computer has a fast % (remainder) operation.  For instance, on the
68020 % is slow enough that the arithmetic bitcount() subroutine is
faster.  In fact, arithmetic bitcount can run faster than 8-bit
character lookups (assuming a memory lookup is 2 cycles, whereas an
arithmetic op is one cycle).  Table lookups only win with 16-bit
tables:

bitcount(n)
long unsigned n;
{
	long unsigned tmp;

        tmp = n - ((n >> 1) & 033333333333) - ((n >> 2) & 011111111111);
        tmp = ((tmp + (tmp >> 3)) & 030707070707);
        tmp =  (tmp + (tmp >> 6));
        tmp = (tmp + (tmp >> 12) + (tmp >> 24)) & 077;
                        /* 6+3+3+5 = 16 ops */
--