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