xref: /freebsd/lib/libc/quad/qdivrem.c (revision 389e4940069316fe667ffa263fa7d6390d0a960f)
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	(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
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(uq, vq, arq)
86 	u_quad_t uq, vq, *arq;
87 {
88 	union uu tmp;
89 	digit *u, *v, *q;
90 	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 		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