qdivrem.c revision 1541
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
38#if defined(LIBC_SCCS) && !defined(lint)
39static char sccsid[] = "@(#)qdivrem.c	8.1 (Berkeley) 6/4/93";
40#endif /* LIBC_SCCS and not lint */
41
42/*
43 * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
44 * section 4.3.1, pp. 257--259.
45 */
46
47#include "quad.h"
48
49#define	B	(1 << HALF_BITS)	/* digit base */
50
51/* Combine two `digits' to make a single two-digit number. */
52#define	COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))
53
54/* select a type for digits in base B: use unsigned short if they fit */
55#if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
56typedef unsigned short digit;
57#else
58typedef u_long digit;
59#endif
60
61/*
62 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
63 * `fall out' the left (there never will be any such anyway).
64 * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
65 */
66static void
67shl(register digit *p, register int len, register int sh)
68{
69	register int i;
70
71	for (i = 0; i < len; i++)
72		p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
73	p[i] = LHALF(p[i] << sh);
74}
75
76/*
77 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
78 *
79 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
80 * fit within u_long.  As a consequence, the maximum length dividend and
81 * divisor are 4 `digits' in this base (they are shorter if they have
82 * leading zeros).
83 */
84u_quad_t
85__qdivrem(uq, vq, arq)
86	u_quad_t uq, vq, *arq;
87{
88	union uu tmp;
89	digit *u, *v, *q;
90	register digit v1, v2;
91	u_long qhat, rhat, t;
92	int m, n, d, j, i;
93	digit uspace[5], vspace[5], qspace[5];
94
95	/*
96	 * Take care of special cases: divide by zero, and u < v.
97	 */
98	if (vq == 0) {
99		/* divide by zero. */
100		static volatile const unsigned int zero = 0;
101
102		tmp.ul[H] = tmp.ul[L] = 1 / zero;
103		if (arq)
104			*arq = uq;
105		return (tmp.q);
106	}
107	if (uq < vq) {
108		if (arq)
109			*arq = uq;
110		return (0);
111	}
112	u = &uspace[0];
113	v = &vspace[0];
114	q = &qspace[0];
115
116	/*
117	 * Break dividend and divisor into digits in base B, then
118	 * count leading zeros to determine m and n.  When done, we
119	 * will have:
120	 *	u = (u[1]u[2]...u[m+n]) sub B
121	 *	v = (v[1]v[2]...v[n]) sub B
122	 *	v[1] != 0
123	 *	1 < n <= 4 (if n = 1, we use a different division algorithm)
124	 *	m >= 0 (otherwise u < v, which we already checked)
125	 *	m + n = 4
126	 * and thus
127	 *	m = 4 - n <= 2
128	 */
129	tmp.uq = uq;
130	u[0] = 0;
131	u[1] = HHALF(tmp.ul[H]);
132	u[2] = LHALF(tmp.ul[H]);
133	u[3] = HHALF(tmp.ul[L]);
134	u[4] = LHALF(tmp.ul[L]);
135	tmp.uq = vq;
136	v[1] = HHALF(tmp.ul[H]);
137	v[2] = LHALF(tmp.ul[H]);
138	v[3] = HHALF(tmp.ul[L]);
139	v[4] = LHALF(tmp.ul[L]);
140	for (n = 4; v[1] == 0; v++) {
141		if (--n == 1) {
142			u_long rbj;	/* r*B+u[j] (not root boy jim) */
143			digit q1, q2, q3, q4;
144
145			/*
146			 * Change of plan, per exercise 16.
147			 *	r = 0;
148			 *	for j = 1..4:
149			 *		q[j] = floor((r*B + u[j]) / v),
150			 *		r = (r*B + u[j]) % v;
151			 * We unroll this completely here.
152			 */
153			t = v[2];	/* nonzero, by definition */
154			q1 = u[1] / t;
155			rbj = COMBINE(u[1] % t, u[2]);
156			q2 = rbj / t;
157			rbj = COMBINE(rbj % t, u[3]);
158			q3 = rbj / t;
159			rbj = COMBINE(rbj % t, u[4]);
160			q4 = rbj / t;
161			if (arq)
162				*arq = rbj % t;
163			tmp.ul[H] = COMBINE(q1, q2);
164			tmp.ul[L] = COMBINE(q3, q4);
165			return (tmp.q);
166		}
167	}
168
169	/*
170	 * By adjusting q once we determine m, we can guarantee that
171	 * there is a complete four-digit quotient at &qspace[1] when
172	 * we finally stop.
173	 */
174	for (m = 4 - n; u[1] == 0; u++)
175		m--;
176	for (i = 4 - m; --i >= 0;)
177		q[i] = 0;
178	q += 4 - m;
179
180	/*
181	 * Here we run Program D, translated from MIX to C and acquiring
182	 * a few minor changes.
183	 *
184	 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
185	 */
186	d = 0;
187	for (t = v[1]; t < B / 2; t <<= 1)
188		d++;
189	if (d > 0) {
190		shl(&u[0], m + n, d);		/* u <<= d */
191		shl(&v[1], n - 1, d);		/* v <<= d */
192	}
193	/*
194	 * D2: j = 0.
195	 */
196	j = 0;
197	v1 = v[1];	/* for D3 -- note that v[1..n] are constant */
198	v2 = v[2];	/* for D3 */
199	do {
200		register digit uj0, uj1, uj2;
201
202		/*
203		 * D3: Calculate qhat (\^q, in TeX notation).
204		 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
205		 * let rhat = (u[j]*B + u[j+1]) mod v[1].
206		 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
207		 * decrement qhat and increase rhat correspondingly.
208		 * Note that if rhat >= B, v[2]*qhat < rhat*B.
209		 */
210		uj0 = u[j + 0];	/* for D3 only -- note that u[j+...] change */
211		uj1 = u[j + 1];	/* for D3 only */
212		uj2 = u[j + 2];	/* for D3 only */
213		if (uj0 == v1) {
214			qhat = B;
215			rhat = uj1;
216			goto qhat_too_big;
217		} else {
218			u_long n = COMBINE(uj0, uj1);
219			qhat = n / v1;
220			rhat = n % v1;
221		}
222		while (v2 * qhat > COMBINE(rhat, uj2)) {
223	qhat_too_big:
224			qhat--;
225			if ((rhat += v1) >= B)
226				break;
227		}
228		/*
229		 * D4: Multiply and subtract.
230		 * The variable `t' holds any borrows across the loop.
231		 * We split this up so that we do not require v[0] = 0,
232		 * and to eliminate a final special case.
233		 */
234		for (t = 0, i = n; i > 0; i--) {
235			t = u[i + j] - v[i] * qhat - t;
236			u[i + j] = LHALF(t);
237			t = (B - HHALF(t)) & (B - 1);
238		}
239		t = u[j] - t;
240		u[j] = LHALF(t);
241		/*
242		 * D5: test remainder.
243		 * There is a borrow if and only if HHALF(t) is nonzero;
244		 * in that (rare) case, qhat was too large (by exactly 1).
245		 * Fix it by adding v[1..n] to u[j..j+n].
246		 */
247		if (HHALF(t)) {
248			qhat--;
249			for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
250				t += u[i + j] + v[i];
251				u[i + j] = LHALF(t);
252				t = HHALF(t);
253			}
254			u[j] = LHALF(u[j] + t);
255		}
256		q[j] = qhat;
257	} while (++j <= m);		/* D7: loop on j. */
258
259	/*
260	 * If caller wants the remainder, we have to calculate it as
261	 * u[m..m+n] >> d (this is at most n digits and thus fits in
262	 * u[m+1..m+n], but we may need more source digits).
263	 */
264	if (arq) {
265		if (d) {
266			for (i = m + n; i > m; --i)
267				u[i] = (u[i] >> d) |
268				    LHALF(u[i - 1] << (HALF_BITS - d));
269			u[i] = 0;
270		}
271		tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
272		tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
273		*arq = tmp.q;
274	}
275
276	tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
277	tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
278	return (tmp.q);
279}
280