xref: /freebsd/lib/libc/quad/qdivrem.c (revision 47606b869eb149ebb7135d6594c3b9b9f05b9aed)
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. 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 
34 #if defined(LIBC_SCCS) && !defined(lint)
35 static char sccsid[] = "@(#)qdivrem.c	8.1 (Berkeley) 6/4/93";
36 #endif /* LIBC_SCCS and not lint */
37 #include <sys/cdefs.h>
38 __FBSDID("$FreeBSD$");
39 
40 /*
41  * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
42  * section 4.3.1, pp. 257--259.
43  */
44 
45 #include "quad.h"
46 
47 #define	B	(1 << HALF_BITS)	/* digit base */
48 
49 /* Combine two `digits' to make a single two-digit number. */
50 #define	COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))
51 
52 /* select a type for digits in base B: use unsigned short if they fit */
53 #if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
54 typedef unsigned short digit;
55 #else
56 typedef u_long digit;
57 #endif
58 
59 /*
60  * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
61  * `fall out' the left (there never will be any such anyway).
62  * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
63  */
64 static void
65 shl(digit *p, int len, int sh)
66 {
67 	int i;
68 
69 	for (i = 0; i < len; i++)
70 		p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
71 	p[i] = LHALF(p[i] << sh);
72 }
73 
74 /*
75  * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
76  *
77  * We do this in base 2-sup-HALF_BITS, so that all intermediate products
78  * fit within u_long.  As a consequence, the maximum length dividend and
79  * divisor are 4 `digits' in this base (they are shorter if they have
80  * leading zeros).
81  */
82 u_quad_t
83 __qdivrem(uq, vq, arq)
84 	u_quad_t uq, vq, *arq;
85 {
86 	union uu tmp;
87 	digit *u, *v, *q;
88 	digit v1, v2;
89 	u_long qhat, rhat, t;
90 	int m, n, d, j, i;
91 	digit uspace[5], vspace[5], qspace[5];
92 
93 	/*
94 	 * Take care of special cases: divide by zero, and u < v.
95 	 */
96 	if (vq == 0) {
97 		/* divide by zero. */
98 		static volatile const unsigned int zero = 0;
99 
100 		tmp.ul[H] = tmp.ul[L] = 1 / zero;
101 		if (arq)
102 			*arq = uq;
103 		return (tmp.q);
104 	}
105 	if (uq < vq) {
106 		if (arq)
107 			*arq = uq;
108 		return (0);
109 	}
110 	u = &uspace[0];
111 	v = &vspace[0];
112 	q = &qspace[0];
113 
114 	/*
115 	 * Break dividend and divisor into digits in base B, then
116 	 * count leading zeros to determine m and n.  When done, we
117 	 * will have:
118 	 *	u = (u[1]u[2]...u[m+n]) sub B
119 	 *	v = (v[1]v[2]...v[n]) sub B
120 	 *	v[1] != 0
121 	 *	1 < n <= 4 (if n = 1, we use a different division algorithm)
122 	 *	m >= 0 (otherwise u < v, which we already checked)
123 	 *	m + n = 4
124 	 * and thus
125 	 *	m = 4 - n <= 2
126 	 */
127 	tmp.uq = uq;
128 	u[0] = 0;
129 	u[1] = HHALF(tmp.ul[H]);
130 	u[2] = LHALF(tmp.ul[H]);
131 	u[3] = HHALF(tmp.ul[L]);
132 	u[4] = LHALF(tmp.ul[L]);
133 	tmp.uq = vq;
134 	v[1] = HHALF(tmp.ul[H]);
135 	v[2] = LHALF(tmp.ul[H]);
136 	v[3] = HHALF(tmp.ul[L]);
137 	v[4] = LHALF(tmp.ul[L]);
138 	for (n = 4; v[1] == 0; v++) {
139 		if (--n == 1) {
140 			u_long rbj;	/* r*B+u[j] (not root boy jim) */
141 			digit q1, q2, q3, q4;
142 
143 			/*
144 			 * Change of plan, per exercise 16.
145 			 *	r = 0;
146 			 *	for j = 1..4:
147 			 *		q[j] = floor((r*B + u[j]) / v),
148 			 *		r = (r*B + u[j]) % v;
149 			 * We unroll this completely here.
150 			 */
151 			t = v[2];	/* nonzero, by definition */
152 			q1 = u[1] / t;
153 			rbj = COMBINE(u[1] % t, u[2]);
154 			q2 = rbj / t;
155 			rbj = COMBINE(rbj % t, u[3]);
156 			q3 = rbj / t;
157 			rbj = COMBINE(rbj % t, u[4]);
158 			q4 = rbj / t;
159 			if (arq)
160 				*arq = rbj % t;
161 			tmp.ul[H] = COMBINE(q1, q2);
162 			tmp.ul[L] = COMBINE(q3, q4);
163 			return (tmp.q);
164 		}
165 	}
166 
167 	/*
168 	 * By adjusting q once we determine m, we can guarantee that
169 	 * there is a complete four-digit quotient at &qspace[1] when
170 	 * we finally stop.
171 	 */
172 	for (m = 4 - n; u[1] == 0; u++)
173 		m--;
174 	for (i = 4 - m; --i >= 0;)
175 		q[i] = 0;
176 	q += 4 - m;
177 
178 	/*
179 	 * Here we run Program D, translated from MIX to C and acquiring
180 	 * a few minor changes.
181 	 *
182 	 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
183 	 */
184 	d = 0;
185 	for (t = v[1]; t < B / 2; t <<= 1)
186 		d++;
187 	if (d > 0) {
188 		shl(&u[0], m + n, d);		/* u <<= d */
189 		shl(&v[1], n - 1, d);		/* v <<= d */
190 	}
191 	/*
192 	 * D2: j = 0.
193 	 */
194 	j = 0;
195 	v1 = v[1];	/* for D3 -- note that v[1..n] are constant */
196 	v2 = v[2];	/* for D3 */
197 	do {
198 		digit uj0, uj1, uj2;
199 
200 		/*
201 		 * D3: Calculate qhat (\^q, in TeX notation).
202 		 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
203 		 * let rhat = (u[j]*B + u[j+1]) mod v[1].
204 		 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
205 		 * decrement qhat and increase rhat correspondingly.
206 		 * Note that if rhat >= B, v[2]*qhat < rhat*B.
207 		 */
208 		uj0 = u[j + 0];	/* for D3 only -- note that u[j+...] change */
209 		uj1 = u[j + 1];	/* for D3 only */
210 		uj2 = u[j + 2];	/* for D3 only */
211 		if (uj0 == v1) {
212 			qhat = B;
213 			rhat = uj1;
214 			goto qhat_too_big;
215 		} else {
216 			u_long n = COMBINE(uj0, uj1);
217 			qhat = n / v1;
218 			rhat = n % v1;
219 		}
220 		while (v2 * qhat > COMBINE(rhat, uj2)) {
221 	qhat_too_big:
222 			qhat--;
223 			if ((rhat += v1) >= B)
224 				break;
225 		}
226 		/*
227 		 * D4: Multiply and subtract.
228 		 * The variable `t' holds any borrows across the loop.
229 		 * We split this up so that we do not require v[0] = 0,
230 		 * and to eliminate a final special case.
231 		 */
232 		for (t = 0, i = n; i > 0; i--) {
233 			t = u[i + j] - v[i] * qhat - t;
234 			u[i + j] = LHALF(t);
235 			t = (B - HHALF(t)) & (B - 1);
236 		}
237 		t = u[j] - t;
238 		u[j] = LHALF(t);
239 		/*
240 		 * D5: test remainder.
241 		 * There is a borrow if and only if HHALF(t) is nonzero;
242 		 * in that (rare) case, qhat was too large (by exactly 1).
243 		 * Fix it by adding v[1..n] to u[j..j+n].
244 		 */
245 		if (HHALF(t)) {
246 			qhat--;
247 			for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
248 				t += u[i + j] + v[i];
249 				u[i + j] = LHALF(t);
250 				t = HHALF(t);
251 			}
252 			u[j] = LHALF(u[j] + t);
253 		}
254 		q[j] = qhat;
255 	} while (++j <= m);		/* D7: loop on j. */
256 
257 	/*
258 	 * If caller wants the remainder, we have to calculate it as
259 	 * u[m..m+n] >> d (this is at most n digits and thus fits in
260 	 * u[m+1..m+n], but we may need more source digits).
261 	 */
262 	if (arq) {
263 		if (d) {
264 			for (i = m + n; i > m; --i)
265 				u[i] = (u[i] >> d) |
266 				    LHALF(u[i - 1] << (HALF_BITS - d));
267 			u[i] = 0;
268 		}
269 		tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
270 		tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
271 		*arq = tmp.q;
272 	}
273 
274 	tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
275 	tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
276 	return (tmp.q);
277 }
278