xref: /freebsd/lib/libc/quad/qdivrem.c (revision 3078531de10dcae44b253a35125c949ff4235284)
1 /*-
2  * SPDX-License-Identifier: BSD-3-Clause
3  *
4  * Copyright (c) 1992, 1993
5  *	The Regents of the University of California.  All rights reserved.
6  *
7  * This software was developed by the Computer Systems Engineering group
8  * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
9  * contributed to Berkeley.
10  *
11  * Redistribution and use in source and binary forms, with or without
12  * modification, are permitted provided that the following conditions
13  * are met:
14  * 1. Redistributions of source code must retain the above copyright
15  *    notice, this list of conditions and the following disclaimer.
16  * 2. Redistributions in binary form must reproduce the above copyright
17  *    notice, this list of conditions and the following disclaimer in the
18  *    documentation and/or other materials provided with the distribution.
19  * 3. Neither the name of the University nor the names of its contributors
20  *    may be used to endorse or promote products derived from this software
21  *    without specific prior written permission.
22  *
23  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
24  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
25  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
26  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
27  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
28  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
29  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
30  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
31  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
32  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
33  * SUCH DAMAGE.
34  */
35 
36 #if defined(LIBC_SCCS) && !defined(lint)
37 static char sccsid[] = "@(#)qdivrem.c	8.1 (Berkeley) 6/4/93";
38 #endif /* LIBC_SCCS and not lint */
39 #include <sys/cdefs.h>
40 __FBSDID("$FreeBSD$");
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	(1L << 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
56 typedef unsigned short digit;
57 #else
58 typedef 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  */
66 static void
67 shl(digit *p, int len, int sh)
68 {
69 	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  */
84 u_quad_t
85 __qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
86 {
87 	union uu tmp;
88 	digit *u, *v, *q;
89 	digit v1, v2;
90 	u_long qhat, rhat, t;
91 	int m, n, d, j, i;
92 	digit uspace[5], vspace[5], qspace[5];
93 
94 	/*
95 	 * Take care of special cases: divide by zero, and u < v.
96 	 */
97 	if (vq == 0) {
98 		/* divide by zero. */
99 		static volatile const unsigned int zero = 0;
100 
101 		tmp.ul[H] = tmp.ul[L] = 1 / zero;
102 		if (arq)
103 			*arq = uq;
104 		return (tmp.q);
105 	}
106 	if (uq < vq) {
107 		if (arq)
108 			*arq = uq;
109 		return (0);
110 	}
111 	u = &uspace[0];
112 	v = &vspace[0];
113 	q = &qspace[0];
114 
115 	/*
116 	 * Break dividend and divisor into digits in base B, then
117 	 * count leading zeros to determine m and n.  When done, we
118 	 * will have:
119 	 *	u = (u[1]u[2]...u[m+n]) sub B
120 	 *	v = (v[1]v[2]...v[n]) sub B
121 	 *	v[1] != 0
122 	 *	1 < n <= 4 (if n = 1, we use a different division algorithm)
123 	 *	m >= 0 (otherwise u < v, which we already checked)
124 	 *	m + n = 4
125 	 * and thus
126 	 *	m = 4 - n <= 2
127 	 */
128 	tmp.uq = uq;
129 	u[0] = 0;
130 	u[1] = HHALF(tmp.ul[H]);
131 	u[2] = LHALF(tmp.ul[H]);
132 	u[3] = HHALF(tmp.ul[L]);
133 	u[4] = LHALF(tmp.ul[L]);
134 	tmp.uq = vq;
135 	v[1] = HHALF(tmp.ul[H]);
136 	v[2] = LHALF(tmp.ul[H]);
137 	v[3] = HHALF(tmp.ul[L]);
138 	v[4] = LHALF(tmp.ul[L]);
139 	for (n = 4; v[1] == 0; v++) {
140 		if (--n == 1) {
141 			u_long rbj;	/* r*B+u[j] (not root boy jim) */
142 			digit q1, q2, q3, q4;
143 
144 			/*
145 			 * Change of plan, per exercise 16.
146 			 *	r = 0;
147 			 *	for j = 1..4:
148 			 *		q[j] = floor((r*B + u[j]) / v),
149 			 *		r = (r*B + u[j]) % v;
150 			 * We unroll this completely here.
151 			 */
152 			t = v[2];	/* nonzero, by definition */
153 			q1 = u[1] / t;
154 			rbj = COMBINE(u[1] % t, u[2]);
155 			q2 = rbj / t;
156 			rbj = COMBINE(rbj % t, u[3]);
157 			q3 = rbj / t;
158 			rbj = COMBINE(rbj % t, u[4]);
159 			q4 = rbj / t;
160 			if (arq)
161 				*arq = rbj % t;
162 			tmp.ul[H] = COMBINE(q1, q2);
163 			tmp.ul[L] = COMBINE(q3, q4);
164 			return (tmp.q);
165 		}
166 	}
167 
168 	/*
169 	 * By adjusting q once we determine m, we can guarantee that
170 	 * there is a complete four-digit quotient at &qspace[1] when
171 	 * we finally stop.
172 	 */
173 	for (m = 4 - n; u[1] == 0; u++)
174 		m--;
175 	for (i = 4 - m; --i >= 0;)
176 		q[i] = 0;
177 	q += 4 - m;
178 
179 	/*
180 	 * Here we run Program D, translated from MIX to C and acquiring
181 	 * a few minor changes.
182 	 *
183 	 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
184 	 */
185 	d = 0;
186 	for (t = v[1]; t < B / 2; t <<= 1)
187 		d++;
188 	if (d > 0) {
189 		shl(&u[0], m + n, d);		/* u <<= d */
190 		shl(&v[1], n - 1, d);		/* v <<= d */
191 	}
192 	/*
193 	 * D2: j = 0.
194 	 */
195 	j = 0;
196 	v1 = v[1];	/* for D3 -- note that v[1..n] are constant */
197 	v2 = v[2];	/* for D3 */
198 	do {
199 		digit uj0, uj1, uj2;
200 
201 		/*
202 		 * D3: Calculate qhat (\^q, in TeX notation).
203 		 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
204 		 * let rhat = (u[j]*B + u[j+1]) mod v[1].
205 		 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
206 		 * decrement qhat and increase rhat correspondingly.
207 		 * Note that if rhat >= B, v[2]*qhat < rhat*B.
208 		 */
209 		uj0 = u[j + 0];	/* for D3 only -- note that u[j+...] change */
210 		uj1 = u[j + 1];	/* for D3 only */
211 		uj2 = u[j + 2];	/* for D3 only */
212 		if (uj0 == v1) {
213 			qhat = B;
214 			rhat = uj1;
215 			goto qhat_too_big;
216 		} else {
217 			u_long n = COMBINE(uj0, uj1);
218 			qhat = n / v1;
219 			rhat = n % v1;
220 		}
221 		while (v2 * qhat > COMBINE(rhat, uj2)) {
222 	qhat_too_big:
223 			qhat--;
224 			if ((rhat += v1) >= B)
225 				break;
226 		}
227 		/*
228 		 * D4: Multiply and subtract.
229 		 * The variable `t' holds any borrows across the loop.
230 		 * We split this up so that we do not require v[0] = 0,
231 		 * and to eliminate a final special case.
232 		 */
233 		for (t = 0, i = n; i > 0; i--) {
234 			t = u[i + j] - v[i] * qhat - t;
235 			u[i + j] = LHALF(t);
236 			t = (B - HHALF(t)) & (B - 1);
237 		}
238 		t = u[j] - t;
239 		u[j] = LHALF(t);
240 		/*
241 		 * D5: test remainder.
242 		 * There is a borrow if and only if HHALF(t) is nonzero;
243 		 * in that (rare) case, qhat was too large (by exactly 1).
244 		 * Fix it by adding v[1..n] to u[j..j+n].
245 		 */
246 		if (HHALF(t)) {
247 			qhat--;
248 			for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
249 				t += u[i + j] + v[i];
250 				u[i + j] = LHALF(t);
251 				t = HHALF(t);
252 			}
253 			u[j] = LHALF(u[j] + t);
254 		}
255 		q[j] = qhat;
256 	} while (++j <= m);		/* D7: loop on j. */
257 
258 	/*
259 	 * If caller wants the remainder, we have to calculate it as
260 	 * u[m..m+n] >> d (this is at most n digits and thus fits in
261 	 * u[m+1..m+n], but we may need more source digits).
262 	 */
263 	if (arq) {
264 		if (d) {
265 			for (i = m + n; i > m; --i)
266 				u[i] = (u[i] >> d) |
267 				    LHALF(u[i - 1] << (HALF_BITS - d));
268 			u[i] = 0;
269 		}
270 		tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
271 		tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
272 		*arq = tmp.q;
273 	}
274 
275 	tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
276 	tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
277 	return (tmp.q);
278 }
279