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126  * zero bit at the top of x.  Doing so means that q is not going to acquire
127  * a 1 bit in the first trip around the loop (since x0 < 2^NBITS).  If the
162  * bit at a time, from the top down, and is not used itself in the loop
164  * in an integer, one word at a time, which saves a bit of work.  Also,
165  * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
177  * Furthermore, we can prove with a bit of work that x never exceeds y by
184  * In fact, we want even one more bit (for a carry, to avoid compares), or
192 	uint32_t bit, q, tt;
253 	 * set the top bit in q, so we can do that manually and start
254 	 * the loop at the next bit down instead.  We must be sure to
258 	 * save work.  To avoid `(1 << 31) << 1', we also do the top bit
261 	 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
262 	 * t2 = y2, t? |= bit' for the appropriate word.  Since the bit
270 	bit = FP_1;
272 	/* if (x >= (t0 = y0 | bit)) { */	/* always true */
273 		q = bit;
274 		x0 -= bit;
275 		y0 = bit << 1;
278 	while ((bit >>= 1) != 0) {	/* for remaining bits in q0 */
280 		t0 = y0 | bit;		/* t = y + bit */
283 			q |= bit;	/*	q += bit */
284 			y0 |= bit << 1;	/*	y += bit << 1 */
296 	bit = 1 << 31;
298 	t1 = bit;
303 		q = bit;		/*	q += bit */
304 		y0 |= 1;		/*	y += bit << 1 */
307 	while ((bit >>= 1) != 0) {	/* for remaining bits in q1 */
309 		t1 = y1 | bit;
314 			q |= bit;
315 			y1 |= bit << 1;
327 	bit = 1 << 31;
329 	t2 = bit;
335 		q |= bit;
339 	while ((bit >>= 1) != 0) {
341 		t2 = y2 | bit;
347 			q |= bit;
348 			y2 |= bit << 1;