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