Home | History | Annotate | Download | only in fpu

Lines Matching refs:bit

130  * zero bit at the top of x.  Doing so means that q is not going to acquire
131  * a 1 bit in the first trip around the loop (since x0 < 2^NBITS).  If the
166  * bit at a time, from the top down, and is not used itself in the loop
168  * in an integer, one word at a time, which saves a bit of work.  Also,
169  * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
181  * Furthermore, we can prove with a bit of work that x never exceeds y by
188  * In fact, we want even one more bit (for a carry, to avoid compares), or
196 	u_int bit, q, tt;
273 	 * set the top bit in q, so we can do that manually and start
274 	 * the loop at the next bit down instead.  We must be sure to
278 	 * save work.  To avoid `(1 << 31) << 1', we also do the top bit
281 	 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
282 	 * t3 = y3, t? |= bit' for the appropriate word.  Since the bit
290 	bit = FP_1;
292 	/* if (x >= (t0 = y0 | bit)) { */	/* always true */
293 		q = bit;
294 		x0 -= bit;
295 		y0 = bit << 1;
298 	while ((bit >>= 1) != 0) {	/* for remaining bits in q0 */
300 		t0 = y0 | bit;		/* t = y + bit */
303 			q |= bit;	/*	q += bit */
304 			y0 |= bit << 1;	/*	y += bit << 1 */
316 	bit = 1 << 31;
318 	t1 = bit;
323 		q = bit;		/*	q += bit */
324 		y0 |= 1;		/*	y += bit << 1 */
327 	while ((bit >>= 1) != 0) {	/* for remaining bits in q1 */
329 		t1 = y1 | bit;
334 			q |= bit;
335 			y1 |= bit << 1;
347 	bit = 1 << 31;
349 	t2 = bit;
355 		q = bit;
359 	while ((bit >>= 1) != 0) {
361 		t2 = y2 | bit;
367 			q |= bit;
368 			y2 |= bit << 1;
380 	bit = 1 << 31;
382 	t3 = bit;
389 		q = bit;
393 	while ((bit >>= 1) != 0) {
395 		t3 = y3 | bit;
402 			q |= bit;
403 			y3 |= bit << 1;