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 * From: Id: qdivrem.c,v 1.7 1997/11/07 09:20:40 phk Exp
34 */
35
36 #include <sys/cdefs.h>
37 #include <sys/stddef.h>
38
39 /*
40 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
41 * section 4.3.1, pp. 257--259.
42 */
43
44 #include "quad.h"
45
46 #define B (1 << HALF_BITS) /* digit base */
47
48 /* Combine two `digits' to make a single two-digit number. */
49 #define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
50
51 _Static_assert(sizeof(int) / 2 == sizeof(short),
52 "Bitwise functions in libstand are broken on this architecture\n");
53
54 /* select a type for digits in base B: use unsigned short if they fit */
55 typedef unsigned short digit;
56
57 /*
58 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
59 * `fall out' the left (there never will be any such anyway).
60 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
61 */
62 static void
shl(digit * p,int len,int sh)63 shl(digit *p, int len, int sh)
64 {
65 int i;
66
67 for (i = 0; i < len; i++)
68 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
69 p[i] = LHALF(p[i] << sh);
70 }
71
72 /*
73 * __udivmoddi4(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
74 *
75 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
76 * fit within u_int. As a consequence, the maximum length dividend and
77 * divisor are 4 `digits' in this base (they are shorter if they have
78 * leading zeros).
79 */
80 u_quad_t
__udivmoddi4(u_quad_t uq,u_quad_t vq,u_quad_t * arq)81 __udivmoddi4(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
82 {
83 union uu tmp;
84 digit *u, *v, *q;
85 digit v1, v2;
86 u_int qhat, rhat, t;
87 int m, n, d, j, i;
88 digit uspace[5], vspace[5], qspace[5];
89
90 /*
91 * Take care of special cases: divide by zero, and u < v.
92 */
93 if (vq == 0) {
94 /* divide by zero. */
95 static volatile const unsigned int zero = 0;
96
97 tmp.ul[H] = tmp.ul[L] = 1 / zero;
98 if (arq)
99 *arq = uq;
100 return (tmp.q);
101 }
102 if (uq < vq) {
103 if (arq)
104 *arq = uq;
105 return (0);
106 }
107 u = &uspace[0];
108 v = &vspace[0];
109 q = &qspace[0];
110
111 /*
112 * Break dividend and divisor into digits in base B, then
113 * count leading zeros to determine m and n. When done, we
114 * will have:
115 * u = (u[1]u[2]...u[m+n]) sub B
116 * v = (v[1]v[2]...v[n]) sub B
117 * v[1] != 0
118 * 1 < n <= 4 (if n = 1, we use a different division algorithm)
119 * m >= 0 (otherwise u < v, which we already checked)
120 * m + n = 4
121 * and thus
122 * m = 4 - n <= 2
123 */
124 tmp.uq = uq;
125 u[0] = 0;
126 u[1] = HHALF(tmp.ul[H]);
127 u[2] = LHALF(tmp.ul[H]);
128 u[3] = HHALF(tmp.ul[L]);
129 u[4] = LHALF(tmp.ul[L]);
130 tmp.uq = vq;
131 v[1] = HHALF(tmp.ul[H]);
132 v[2] = LHALF(tmp.ul[H]);
133 v[3] = HHALF(tmp.ul[L]);
134 v[4] = LHALF(tmp.ul[L]);
135 for (n = 4; v[1] == 0; v++) {
136 if (--n == 1) {
137 u_int rbj; /* r*B+u[j] (not root boy jim) */
138 digit q1, q2, q3, q4;
139
140 /*
141 * Change of plan, per exercise 16.
142 * r = 0;
143 * for j = 1..4:
144 * q[j] = floor((r*B + u[j]) / v),
145 * r = (r*B + u[j]) % v;
146 * We unroll this completely here.
147 */
148 t = v[2]; /* nonzero, by definition */
149 q1 = u[1] / t;
150 rbj = COMBINE(u[1] % t, u[2]);
151 q2 = rbj / t;
152 rbj = COMBINE(rbj % t, u[3]);
153 q3 = rbj / t;
154 rbj = COMBINE(rbj % t, u[4]);
155 q4 = rbj / t;
156 if (arq)
157 *arq = rbj % t;
158 tmp.ul[H] = COMBINE(q1, q2);
159 tmp.ul[L] = COMBINE(q3, q4);
160 return (tmp.q);
161 }
162 }
163
164 /*
165 * By adjusting q once we determine m, we can guarantee that
166 * there is a complete four-digit quotient at &qspace[1] when
167 * we finally stop.
168 */
169 for (m = 4 - n; u[1] == 0; u++)
170 m--;
171 for (i = 4 - m; --i >= 0;)
172 q[i] = 0;
173 q += 4 - m;
174
175 /*
176 * Here we run Program D, translated from MIX to C and acquiring
177 * a few minor changes.
178 *
179 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
180 */
181 d = 0;
182 for (t = v[1]; t < B / 2; t <<= 1)
183 d++;
184 if (d > 0) {
185 shl(&u[0], m + n, d); /* u <<= d */
186 shl(&v[1], n - 1, d); /* v <<= d */
187 }
188 /*
189 * D2: j = 0.
190 */
191 j = 0;
192 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
193 v2 = v[2]; /* for D3 */
194 do {
195 digit uj0, uj1, uj2;
196
197 /*
198 * D3: Calculate qhat (\^q, in TeX notation).
199 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
200 * let rhat = (u[j]*B + u[j+1]) mod v[1].
201 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
202 * decrement qhat and increase rhat correspondingly.
203 * Note that if rhat >= B, v[2]*qhat < rhat*B.
204 */
205 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
206 uj1 = u[j + 1]; /* for D3 only */
207 uj2 = u[j + 2]; /* for D3 only */
208 if (uj0 == v1) {
209 qhat = B;
210 rhat = uj1;
211 goto qhat_too_big;
212 } else {
213 u_int nn = COMBINE(uj0, uj1);
214 qhat = nn / v1;
215 rhat = nn % v1;
216 }
217 while (v2 * qhat > COMBINE(rhat, uj2)) {
218 qhat_too_big:
219 qhat--;
220 if ((rhat += v1) >= B)
221 break;
222 }
223 /*
224 * D4: Multiply and subtract.
225 * The variable `t' holds any borrows across the loop.
226 * We split this up so that we do not require v[0] = 0,
227 * and to eliminate a final special case.
228 */
229 for (t = 0, i = n; i > 0; i--) {
230 t = u[i + j] - v[i] * qhat - t;
231 u[i + j] = LHALF(t);
232 t = (B - HHALF(t)) & (B - 1);
233 }
234 t = u[j] - t;
235 u[j] = LHALF(t);
236 /*
237 * D5: test remainder.
238 * There is a borrow if and only if HHALF(t) is nonzero;
239 * in that (rare) case, qhat was too large (by exactly 1).
240 * Fix it by adding v[1..n] to u[j..j+n].
241 */
242 if (HHALF(t)) {
243 qhat--;
244 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
245 t += u[i + j] + v[i];
246 u[i + j] = LHALF(t);
247 t = HHALF(t);
248 }
249 u[j] = LHALF(u[j] + t);
250 }
251 q[j] = qhat;
252 } while (++j <= m); /* D7: loop on j. */
253
254 /*
255 * If caller wants the remainder, we have to calculate it as
256 * u[m..m+n] >> d (this is at most n digits and thus fits in
257 * u[m+1..m+n], but we may need more source digits).
258 */
259 if (arq) {
260 if (d) {
261 for (i = m + n; i > m; --i)
262 u[i] = (u[i] >> d) |
263 LHALF(u[i - 1] << (HALF_BITS - d));
264 u[i] = 0;
265 }
266 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
267 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
268 *arq = tmp.q;
269 }
270
271 tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
272 tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
273 return (tmp.q);
274 }
275
276 /*
277 * Divide two unsigned quads.
278 */
279
280 u_quad_t
__udivdi3(u_quad_t a,u_quad_t b)281 __udivdi3(u_quad_t a, u_quad_t b)
282 {
283
284 return (__udivmoddi4(a, b, NULL));
285 }
286
287 /*
288 * Return remainder after dividing two unsigned quads.
289 */
290 u_quad_t
__umoddi3(u_quad_t a,u_quad_t b)291 __umoddi3(u_quad_t a, u_quad_t b)
292 {
293 u_quad_t r;
294
295 (void)__udivmoddi4(a, b, &r);
296 return (r);
297 }
298
299 /*
300 * Divide two signed quads.
301 * ??? if -1/2 should produce -1 on this machine, this code is wrong
302 */
303 quad_t
__divdi3(quad_t a,quad_t b)304 __divdi3(quad_t a, quad_t b)
305 {
306 u_quad_t ua, ub, uq;
307 int neg;
308
309 if (a < 0)
310 ua = -(u_quad_t)a, neg = 1;
311 else
312 ua = a, neg = 0;
313 if (b < 0)
314 ub = -(u_quad_t)b, neg ^= 1;
315 else
316 ub = b;
317 uq = __udivmoddi4(ua, ub, NULL);
318 return (neg ? -uq : uq);
319 }
320
321 /*
322 * Return remainder after dividing two signed quads.
323 *
324 * XXX
325 * If -1/2 should produce -1 on this machine, this code is wrong.
326 */
327 quad_t
__moddi3(quad_t a,quad_t b)328 __moddi3(quad_t a, quad_t b)
329 {
330 u_quad_t ua, ub, ur;
331 int neg;
332
333 if (a < 0)
334 ua = -(u_quad_t)a, neg = 1;
335 else
336 ua = a, neg = 0;
337 if (b < 0)
338 ub = -(u_quad_t)b;
339 else
340 ub = b;
341 (void)__udivmoddi4(ua, ub, &ur);
342 return (neg ? -ur : ur);
343 }
344
345 quad_t
__divmoddi4(quad_t a,quad_t b,quad_t * r)346 __divmoddi4(quad_t a, quad_t b, quad_t *r)
347 {
348 quad_t d;
349
350 d = __divdi3(a, b);
351 if (r != NULL)
352 *r = a - (b * d);
353
354 return (d);
355 }
356