2 * Copyright (c) 1992, 1993
3 * The Regents of the University of California. All rights reserved.
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.
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10 * modification, are permitted provided that the following conditions
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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
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17 * 4. 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.
21 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
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23 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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33 * From: Id: qdivrem.c,v 1.7 1997/11/07 09:20:40 phk Exp
36 #include <sys/cdefs.h>
37 __FBSDID("$FreeBSD$");
40 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
41 * section 4.3.1, pp. 257--259.
46 #define B (1 << HALF_BITS) /* digit base */
48 /* Combine two `digits' to make a single two-digit number. */
49 #define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
51 _Static_assert(sizeof(int) / 2 == sizeof(short),
52 "Bitwise functions in libstand are broken on this architecture\n");
54 /* select a type for digits in base B: use unsigned short if they fit */
55 typedef unsigned short digit;
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.
63 shl(digit *p, int len, int sh)
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);
73 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
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
81 __qdivrem(uq, vq, arq)
82 u_quad_t uq, vq, *arq;
89 digit uspace[5], vspace[5], qspace[5];
92 * Take care of special cases: divide by zero, and u < v.
96 static volatile const unsigned int zero = 0;
98 tmp.ul[H] = tmp.ul[L] = 1 / zero;
113 * Break dividend and divisor into digits in base B, then
114 * count leading zeros to determine m and n. When done, we
116 * u = (u[1]u[2]...u[m+n]) sub B
117 * v = (v[1]v[2]...v[n]) sub B
119 * 1 < n <= 4 (if n = 1, we use a different division algorithm)
120 * m >= 0 (otherwise u < v, which we already checked)
127 u[1] = HHALF(tmp.ul[H]);
128 u[2] = LHALF(tmp.ul[H]);
129 u[3] = HHALF(tmp.ul[L]);
130 u[4] = LHALF(tmp.ul[L]);
132 v[1] = HHALF(tmp.ul[H]);
133 v[2] = LHALF(tmp.ul[H]);
134 v[3] = HHALF(tmp.ul[L]);
135 v[4] = LHALF(tmp.ul[L]);
136 for (n = 4; v[1] == 0; v++) {
138 u_int rbj; /* r*B+u[j] (not root boy jim) */
139 digit q1, q2, q3, q4;
142 * Change of plan, per exercise 16.
145 * q[j] = floor((r*B + u[j]) / v),
146 * r = (r*B + u[j]) % v;
147 * We unroll this completely here.
149 t = v[2]; /* nonzero, by definition */
151 rbj = COMBINE(u[1] % t, u[2]);
153 rbj = COMBINE(rbj % t, u[3]);
155 rbj = COMBINE(rbj % t, u[4]);
159 tmp.ul[H] = COMBINE(q1, q2);
160 tmp.ul[L] = COMBINE(q3, q4);
166 * By adjusting q once we determine m, we can guarantee that
167 * there is a complete four-digit quotient at &qspace[1] when
170 for (m = 4 - n; u[1] == 0; u++)
172 for (i = 4 - m; --i >= 0;)
177 * Here we run Program D, translated from MIX to C and acquiring
178 * a few minor changes.
180 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
183 for (t = v[1]; t < B / 2; t <<= 1)
186 shl(&u[0], m + n, d); /* u <<= d */
187 shl(&v[1], n - 1, d); /* v <<= d */
193 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
194 v2 = v[2]; /* for D3 */
199 * D3: Calculate qhat (\^q, in TeX notation).
200 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
201 * let rhat = (u[j]*B + u[j+1]) mod v[1].
202 * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
203 * decrement qhat and increase rhat correspondingly.
204 * Note that if rhat >= B, v[2]*qhat < rhat*B.
206 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
207 uj1 = u[j + 1]; /* for D3 only */
208 uj2 = u[j + 2]; /* for D3 only */
214 u_int nn = COMBINE(uj0, uj1);
218 while (v2 * qhat > COMBINE(rhat, uj2)) {
221 if ((rhat += v1) >= B)
225 * D4: Multiply and subtract.
226 * The variable `t' holds any borrows across the loop.
227 * We split this up so that we do not require v[0] = 0,
228 * and to eliminate a final special case.
230 for (t = 0, i = n; i > 0; i--) {
231 t = u[i + j] - v[i] * qhat - t;
233 t = (B - HHALF(t)) & (B - 1);
238 * D5: test remainder.
239 * There is a borrow if and only if HHALF(t) is nonzero;
240 * in that (rare) case, qhat was too large (by exactly 1).
241 * Fix it by adding v[1..n] to u[j..j+n].
245 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
246 t += u[i + j] + v[i];
250 u[j] = LHALF(u[j] + t);
253 } while (++j <= m); /* D7: loop on j. */
256 * If caller wants the remainder, we have to calculate it as
257 * u[m..m+n] >> d (this is at most n digits and thus fits in
258 * u[m+1..m+n], but we may need more source digits).
262 for (i = m + n; i > m; --i)
264 LHALF(u[i - 1] << (HALF_BITS - d));
267 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
268 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
272 tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
273 tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
278 * Divide two unsigned quads.
286 return (__qdivrem(a, b, (u_quad_t *)0));
290 * Return remainder after dividing two unsigned quads.
298 (void)__qdivrem(a, b, &r);
303 * Divide two signed quads.
304 * ??? if -1/2 should produce -1 on this machine, this code is wrong
314 ua = -(u_quad_t)a, neg = 1;
318 ub = -(u_quad_t)b, neg ^= 1;
321 uq = __qdivrem(ua, ub, (u_quad_t *)0);
322 return (neg ? -uq : uq);
326 * Return remainder after dividing two signed quads.
329 * If -1/2 should produce -1 on this machine, this code is wrong.
339 ua = -(u_quad_t)a, neg = 1;
346 (void)__qdivrem(ua, ub, &ur);
347 return (neg ? -ur : ur);