arch/sparc/fpu_sqrt.c

123  * zero bit at the top of x.  Doing so means that q is not going to acquire
124  * a 1 bit in the first trip around the loop (since x0 < 2^NBITS).  If the
159  * bit at a time, from the top down, and is not used itself in the loop
161  * in an integer, one word at a time, which saves a bit of work.  Also,
162  * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
174  * Furthermore, we can prove with a bit of work that x never exceeds y by
181  * In fact, we want even one more bit (for a carry, to avoid compares), or
190 	u_int bit, q, tt;
251 	 * set the top bit in q, so we can do that manually and start
252 	 * the loop at the next bit down instead.  We must be sure to
256 	 * save work.  To avoid `(1U << 31) << 1', we also do the top bit
259 	 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
260 	 * t3 = y3, t? |= bit' for the appropriate word.  Since the bit
268 	bit = FP_1;
270 	/* if (x >= (t0 = y0 | bit)) { */	/* always true */
271 		q = bit;
272 		x0 -= bit;
273 		y0 = bit << 1;
276 	while ((bit >>= 1) != 0) {	/* for remaining bits in q0 */
278 		t0 = y0 | bit;		/* t = y + bit */
281 			q |= bit;	/*	q += bit */
282 			y0 |= bit << 1;	/*	y += bit << 1 */
294 	bit = 1 << 31;
296 	t1 = bit;
301 		q = bit;		/*	q += bit */
302 		y0 |= 1;		/*	y += bit << 1 */
305 	while ((bit >>= 1) != 0) {	/* for remaining bits in q1 */
307 		t1 = y1 | bit;
312 			q |= bit;
313 			y1 |= bit << 1;
325 	bit = 1 << 31;
327 	t2 = bit;
333 		q = bit;
337 	while ((bit >>= 1) != 0) {
339 		t2 = y2 | bit;
345 			q |= bit;
346 			y2 |= bit << 1;
358 	bit = 1 << 31;
360 	t3 = bit;
367 		q = bit;
371 	while ((bit >>= 1) != 0) {
373 		t3 = y3 | bit;
380 			q |= bit;
381 			y3 |= bit << 1;