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 rcsid[] = "$OpenBSD: qdivrem.c,v 1.8 2005/02/13 03:37:14 jsg Exp $";
36 #endif /* LIBC_SCCS and not lint */
37
38 /*
39 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
40 * section 4.3.1, pp. 257--259.
41 */
42
43 #include "quad.h"
44
45 #define B ((int)1 << HALF_BITS) /* digit base */
46
47 /* Combine two `digits' to make a single two-digit number. */
48 #define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
49
50 /* select a type for digits in base B: use unsigned short if they fit */
51 #if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
52 typedef unsigned short digit;
53 #else
54 typedef u_int digit;
55 #endif
56
57 static void shl(digit *p, int len, int sh);
58
59 /*
60 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
61 *
62 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
63 * fit within u_int. As a consequence, the maximum length dividend and
64 * divisor are 4 `digits' in this base (they are shorter if they have
65 * leading zeros).
66 */
67 u_quad_t
68 __qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
69 {
70 union uu tmp;
71 digit *u, *v, *q;
72 digit v1, v2;
73 u_int qhat, rhat, t;
74 int m, n, d, j, i;
75 digit uspace[5], vspace[5], qspace[5];
76
77 /*
78 * Take care of special cases: divide by zero, and u < v.
79 */
80 if (vq == 0) {
81 /* divide by zero. */
82 static volatile const unsigned int zero = 0;
83
84 tmp.ul[H] = tmp.ul[L] = 1 / zero;
85 if (arq)
86 *arq = uq;
87 return (tmp.q);
88 }
89 if (uq < vq) {
90 if (arq)
91 *arq = uq;
92 return (0);
93 }
94 u = &uspace[0];
95 v = &vspace[0];
96 q = &qspace[0];
97
98 /*
99 * Break dividend and divisor into digits in base B, then
100 * count leading zeros to determine m and n. When done, we
101 * will have:
102 * u = (u[1]u[2]...u[m+n]) sub B
103 * v = (v[1]v[2]...v[n]) sub B
104 * v[1] != 0
105 * 1 < n <= 4 (if n = 1, we use a different division algorithm)
106 * m >= 0 (otherwise u < v, which we already checked)
107 * m + n = 4
108 * and thus
109 * m = 4 - n <= 2
110 */
111 tmp.uq = uq;
112 u[0] = 0;
113 u[1] = (digit)HHALF(tmp.ul[H]);
114 u[2] = (digit)LHALF(tmp.ul[H]);
115 u[3] = (digit)HHALF(tmp.ul[L]);
116 u[4] = (digit)LHALF(tmp.ul[L]);
117 tmp.uq = vq;
118 v[1] = (digit)HHALF(tmp.ul[H]);
119 v[2] = (digit)LHALF(tmp.ul[H]);
120 v[3] = (digit)HHALF(tmp.ul[L]);
121 v[4] = (digit)LHALF(tmp.ul[L]);
122 for (n = 4; v[1] == 0; v++) {
123 if (--n == 1) {
124 u_int rbj; /* r*B+u[j] (not root boy jim) */
125 digit q1, q2, q3, q4;
126
127 /*
128 * Change of plan, per exercise 16.
129 * r = 0;
130 * for j = 1..4:
131 * q[j] = floor((r*B + u[j]) / v),
132 * r = (r*B + u[j]) % v;
133 * We unroll this completely here.
134 */
135 t = v[2]; /* nonzero, by definition */
136 q1 = (digit)(u[1] / t);
137 rbj = COMBINE(u[1] % t, u[2]);
138 q2 = (digit)(rbj / t);
139 rbj = COMBINE(rbj % t, u[3]);
140 q3 = (digit)(rbj / t);
141 rbj = COMBINE(rbj % t, u[4]);
142 q4 = (digit)(rbj / t);
143 if (arq)
144 *arq = rbj % t;
145 tmp.ul[H] = COMBINE(q1, q2);
146 tmp.ul[L] = COMBINE(q3, q4);
147 return (tmp.q);
148 }
149 }
150
151 /*
152 * By adjusting q once we determine m, we can guarantee that
153 * there is a complete four-digit quotient at &qspace[1] when
154 * we finally stop.
155 */
156 for (m = 4 - n; u[1] == 0; u++)
157 m--;
158 for (i = 4 - m; --i >= 0;)
159 q[i] = 0;
160 q += 4 - m;
161
162 /*
163 * Here we run Program D, translated from MIX to C and acquiring
164 * a few minor changes.
165 *
166 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
167 */
168 d = 0;
169 for (t = v[1]; t < B / 2; t <<= 1)
170 d++;
171 if (d > 0) {
172 shl(&u[0], m + n, d); /* u <<= d */
173 shl(&v[1], n - 1, d); /* v <<= d */
174 }
175 /*
176 * D2: j = 0.
177 */
178 j = 0;
179 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
180 v2 = v[2]; /* for D3 */
181 do {
182 digit uj0, uj1, uj2;
183
184 /*
185 * D3: Calculate qhat (\^q, in TeX notation).
186 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
187 * let rhat = (u[j]*B + u[j+1]) mod v[1].
188 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
189 * decrement qhat and increase rhat correspondingly.
190 * Note that if rhat >= B, v[2]*qhat < rhat*B.
191 */
192 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
193 uj1 = u[j + 1]; /* for D3 only */
194 uj2 = u[j + 2]; /* for D3 only */
195 if (uj0 == v1) {
196 qhat = B;
197 rhat = uj1;
198 goto qhat_too_big;
199 } else {
200 u_int nn = COMBINE(uj0, uj1);
201 qhat = nn / v1;
202 rhat = nn % v1;
203 }
204 while (v2 * qhat > COMBINE(rhat, uj2)) {
205 qhat_too_big:
206 qhat--;
207 if ((rhat += v1) >= B)
208 break;
209 }
210 /*
211 * D4: Multiply and subtract.
212 * The variable `t' holds any borrows across the loop.
213 * We split this up so that we do not require v[0] = 0,
214 * and to eliminate a final special case.
215 */
216 for (t = 0, i = n; i > 0; i--) {
217 t = u[i + j] - v[i] * qhat - t;
218 u[i + j] = (digit)LHALF(t);
219 t = (B - HHALF(t)) & (B - 1);
220 }
221 t = u[j] - t;
222 u[j] = (digit)LHALF(t);
223 /*
224 * D5: test remainder.
225 * There is a borrow if and only if HHALF(t) is nonzero;
226 * in that (rare) case, qhat was too large (by exactly 1).
227 * Fix it by adding v[1..n] to u[j..j+n].
228 */
229 if (HHALF(t)) {
230 qhat--;
231 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
232 t += u[i + j] + v[i];
233 u[i + j] = (digit)LHALF(t);
234 t = HHALF(t);
235 }
236 u[j] = (digit)LHALF(u[j] + t);
237 }
238 q[j] = (digit)qhat;
239 } while (++j <= m); /* D7: loop on j. */
240
241 /*
242 * If caller wants the remainder, we have to calculate it as
243 * u[m..m+n] >> d (this is at most n digits and thus fits in
244 * u[m+1..m+n], but we may need more source digits).
245 */
246 if (arq) {
247 if (d) {
248 for (i = m + n; i > m; --i)
249 u[i] = (digit)(((u_int)u[i] >> d) |
250 LHALF((u_int)u[i - 1] << (HALF_BITS - d)));
251 u[i] = 0;
252 }
253 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
254 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
255 *arq = tmp.q;
256 }
257
258 tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
259 tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
260 return (tmp.q);
261 }
262
263 /*
264 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
265 * `fall out' the left (there never will be any such anyway).
266 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
267 */
268 static void
269 shl(digit *p, int len, int sh)
270 {
271 int i;
272
273 for (i = 0; i < len; i++)
274 p[i] = (digit)(LHALF((u_int)p[i] << sh) |
275 ((u_int)p[i + 1] >> (HALF_BITS - sh)));
276 p[i] = (digit)(LHALF((u_int)p[i] << sh));
277 }