qdivrem.c revision 96525
1/*-
2 * Copyright (c) 1992, 1993
3 *	The Regents of the University of California.  All rights reserved.
4 *
5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
7 * contributed to Berkeley.
8 *
9 * Redistribution and use in source and binary forms, with or without
10 * modification, are permitted provided that the following conditions
11 * are met:
12 * 1. Redistributions of source code must retain the above copyright
13 *    notice, this list of conditions and the following disclaimer.
14 * 2. Redistributions in binary form must reproduce the above copyright
15 *    notice, this list of conditions and the following disclaimer in the
16 *    documentation and/or other materials provided with the distribution.
17 * 3. All advertising materials mentioning features or use of this software
18 *    must display the following acknowledgement:
19 *	This product includes software developed by the University of
20 *	California, Berkeley and its contributors.
21 * 4. Neither the name of the University nor the names of its contributors
22 *    may be used to endorse or promote products derived from this software
23 *    without specific prior written permission.
24 *
25 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
26 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
27 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
28 * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
29 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
30 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
31 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
32 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
33 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
34 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
35 * SUCH DAMAGE.
36 *
37 * 	From: Id: qdivrem.c,v 1.7 1997/11/07 09:20:40 phk Exp
38 */
39
40#include <sys/cdefs.h>
41__FBSDID("$FreeBSD: head/lib/libstand/qdivrem.c 96525 2002-05-13 13:31:20Z phk $");
42
43/*
44 * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
45 * section 4.3.1, pp. 257--259.
46 */
47
48#include "quad.h"
49
50#define	B	(1 << HALF_BITS)	/* digit base */
51
52/* Combine two `digits' to make a single two-digit number. */
53#define	COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))
54
55/* select a type for digits in base B: use unsigned short if they fit */
56#if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
57typedef unsigned short digit;
58#else
59typedef u_long digit;
60#endif
61
62/*
63 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
64 * `fall out' the left (there never will be any such anyway).
65 * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
66 */
67static void
68shl(digit *p, int len, int sh)
69{
70	int i;
71
72	for (i = 0; i < len; i++)
73		p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
74	p[i] = LHALF(p[i] << sh);
75}
76
77/*
78 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
79 *
80 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
81 * fit within u_long.  As a consequence, the maximum length dividend and
82 * divisor are 4 `digits' in this base (they are shorter if they have
83 * leading zeros).
84 */
85u_quad_t
86__qdivrem(uq, vq, arq)
87	u_quad_t uq, vq, *arq;
88{
89	union uu tmp;
90	digit *u, *v, *q;
91	digit v1, v2;
92	u_long qhat, rhat, t;
93	int m, n, d, j, i;
94	digit uspace[5], vspace[5], qspace[5];
95
96	/*
97	 * Take care of special cases: divide by zero, and u < v.
98	 */
99	if (vq == 0) {
100		/* divide by zero. */
101		static volatile const unsigned int zero = 0;
102
103		tmp.ul[H] = tmp.ul[L] = 1 / zero;
104		if (arq)
105			*arq = uq;
106		return (tmp.q);
107	}
108	if (uq < vq) {
109		if (arq)
110			*arq = uq;
111		return (0);
112	}
113	u = &uspace[0];
114	v = &vspace[0];
115	q = &qspace[0];
116
117	/*
118	 * Break dividend and divisor into digits in base B, then
119	 * count leading zeros to determine m and n.  When done, we
120	 * will have:
121	 *	u = (u[1]u[2]...u[m+n]) sub B
122	 *	v = (v[1]v[2]...v[n]) sub B
123	 *	v[1] != 0
124	 *	1 < n <= 4 (if n = 1, we use a different division algorithm)
125	 *	m >= 0 (otherwise u < v, which we already checked)
126	 *	m + n = 4
127	 * and thus
128	 *	m = 4 - n <= 2
129	 */
130	tmp.uq = uq;
131	u[0] = 0;
132	u[1] = HHALF(tmp.ul[H]);
133	u[2] = LHALF(tmp.ul[H]);
134	u[3] = HHALF(tmp.ul[L]);
135	u[4] = LHALF(tmp.ul[L]);
136	tmp.uq = vq;
137	v[1] = HHALF(tmp.ul[H]);
138	v[2] = LHALF(tmp.ul[H]);
139	v[3] = HHALF(tmp.ul[L]);
140	v[4] = LHALF(tmp.ul[L]);
141	for (n = 4; v[1] == 0; v++) {
142		if (--n == 1) {
143			u_long rbj;	/* r*B+u[j] (not root boy jim) */
144			digit q1, q2, q3, q4;
145
146			/*
147			 * Change of plan, per exercise 16.
148			 *	r = 0;
149			 *	for j = 1..4:
150			 *		q[j] = floor((r*B + u[j]) / v),
151			 *		r = (r*B + u[j]) % v;
152			 * We unroll this completely here.
153			 */
154			t = v[2];	/* nonzero, by definition */
155			q1 = u[1] / t;
156			rbj = COMBINE(u[1] % t, u[2]);
157			q2 = rbj / t;
158			rbj = COMBINE(rbj % t, u[3]);
159			q3 = rbj / t;
160			rbj = COMBINE(rbj % t, u[4]);
161			q4 = rbj / t;
162			if (arq)
163				*arq = rbj % t;
164			tmp.ul[H] = COMBINE(q1, q2);
165			tmp.ul[L] = COMBINE(q3, q4);
166			return (tmp.q);
167		}
168	}
169
170	/*
171	 * By adjusting q once we determine m, we can guarantee that
172	 * there is a complete four-digit quotient at &qspace[1] when
173	 * we finally stop.
174	 */
175	for (m = 4 - n; u[1] == 0; u++)
176		m--;
177	for (i = 4 - m; --i >= 0;)
178		q[i] = 0;
179	q += 4 - m;
180
181	/*
182	 * Here we run Program D, translated from MIX to C and acquiring
183	 * a few minor changes.
184	 *
185	 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
186	 */
187	d = 0;
188	for (t = v[1]; t < B / 2; t <<= 1)
189		d++;
190	if (d > 0) {
191		shl(&u[0], m + n, d);		/* u <<= d */
192		shl(&v[1], n - 1, d);		/* v <<= d */
193	}
194	/*
195	 * D2: j = 0.
196	 */
197	j = 0;
198	v1 = v[1];	/* for D3 -- note that v[1..n] are constant */
199	v2 = v[2];	/* for D3 */
200	do {
201		digit uj0, uj1, uj2;
202
203		/*
204		 * D3: Calculate qhat (\^q, in TeX notation).
205		 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
206		 * let rhat = (u[j]*B + u[j+1]) mod v[1].
207		 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
208		 * decrement qhat and increase rhat correspondingly.
209		 * Note that if rhat >= B, v[2]*qhat < rhat*B.
210		 */
211		uj0 = u[j + 0];	/* for D3 only -- note that u[j+...] change */
212		uj1 = u[j + 1];	/* for D3 only */
213		uj2 = u[j + 2];	/* for D3 only */
214		if (uj0 == v1) {
215			qhat = B;
216			rhat = uj1;
217			goto qhat_too_big;
218		} else {
219			u_long nn = COMBINE(uj0, uj1);
220			qhat = nn / v1;
221			rhat = nn % v1;
222		}
223		while (v2 * qhat > COMBINE(rhat, uj2)) {
224	qhat_too_big:
225			qhat--;
226			if ((rhat += v1) >= B)
227				break;
228		}
229		/*
230		 * D4: Multiply and subtract.
231		 * The variable `t' holds any borrows across the loop.
232		 * We split this up so that we do not require v[0] = 0,
233		 * and to eliminate a final special case.
234		 */
235		for (t = 0, i = n; i > 0; i--) {
236			t = u[i + j] - v[i] * qhat - t;
237			u[i + j] = LHALF(t);
238			t = (B - HHALF(t)) & (B - 1);
239		}
240		t = u[j] - t;
241		u[j] = LHALF(t);
242		/*
243		 * D5: test remainder.
244		 * There is a borrow if and only if HHALF(t) is nonzero;
245		 * in that (rare) case, qhat was too large (by exactly 1).
246		 * Fix it by adding v[1..n] to u[j..j+n].
247		 */
248		if (HHALF(t)) {
249			qhat--;
250			for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
251				t += u[i + j] + v[i];
252				u[i + j] = LHALF(t);
253				t = HHALF(t);
254			}
255			u[j] = LHALF(u[j] + t);
256		}
257		q[j] = qhat;
258	} while (++j <= m);		/* D7: loop on j. */
259
260	/*
261	 * If caller wants the remainder, we have to calculate it as
262	 * u[m..m+n] >> d (this is at most n digits and thus fits in
263	 * u[m+1..m+n], but we may need more source digits).
264	 */
265	if (arq) {
266		if (d) {
267			for (i = m + n; i > m; --i)
268				u[i] = (u[i] >> d) |
269				    LHALF(u[i - 1] << (HALF_BITS - d));
270			u[i] = 0;
271		}
272		tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
273		tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
274		*arq = tmp.q;
275	}
276
277	tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
278	tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
279	return (tmp.q);
280}
281
282/*
283 * Divide two unsigned quads.
284 */
285
286u_quad_t
287__udivdi3(a, b)
288	u_quad_t a, b;
289{
290
291	return (__qdivrem(a, b, (u_quad_t *)0));
292}
293
294/*
295 * Return remainder after dividing two unsigned quads.
296 */
297u_quad_t
298__umoddi3(a, b)
299	u_quad_t a, b;
300{
301	u_quad_t r;
302
303	(void)__qdivrem(a, b, &r);
304	return (r);
305}
306
307/*
308 * Divide two signed quads.
309 * ??? if -1/2 should produce -1 on this machine, this code is wrong
310 */
311quad_t
312__divdi3(a, b)
313        quad_t a, b;
314{
315	u_quad_t ua, ub, uq;
316	int neg;
317
318	if (a < 0)
319		ua = -(u_quad_t)a, neg = 1;
320	else
321		ua = a, neg = 0;
322	if (b < 0)
323		ub = -(u_quad_t)b, neg ^= 1;
324	else
325		ub = b;
326	uq = __qdivrem(ua, ub, (u_quad_t *)0);
327	return (neg ? -uq : uq);
328}
329
330/*
331 * Return remainder after dividing two signed quads.
332 *
333 * XXX
334 * If -1/2 should produce -1 on this machine, this code is wrong.
335 */
336quad_t
337__moddi3(a, b)
338        quad_t a, b;
339{
340	u_quad_t ua, ub, ur;
341	int neg;
342
343	if (a < 0)
344		ua = -(u_quad_t)a, neg = 1;
345	else
346		ua = a, neg = 0;
347	if (b < 0)
348		ub = -(u_quad_t)b;
349	else
350		ub = b;
351	(void)__qdivrem(ua, ub, &ur);
352	return (neg ? -ur : ur);
353}
354