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.
9 * Redistribution and use in source and binary forms, with or without
10 * modification, are permitted provided that the following conditions
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 * 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
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
34 #include <sys/cdefs.h>
35 __FBSDID("$FreeBSD$");
38 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
39 * section 4.3.1, pp. 257--259.
42 #include <libkern/quad.h>
44 #define B (1 << HALF_BITS) /* digit base */
46 /* Combine two `digits' to make a single two-digit number. */
47 #define COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b))
49 /* select a type for digits in base B: use unsigned short if they fit */
50 #if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
51 typedef unsigned short digit;
57 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
58 * `fall out' the left (there never will be any such anyway).
59 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
62 __shl(register digit *p, register int len, register int sh)
66 for (i = 0; i < len; i++)
67 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
68 p[i] = LHALF(p[i] << sh);
72 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
74 * We do this in base 2-sup-HALF_BITS, so that all intermediate products
75 * fit within u_long. As a consequence, the maximum length dividend and
76 * divisor are 4 `digits' in this base (they are shorter if they have
80 __qdivrem(uq, vq, arq)
81 u_quad_t uq, vq, *arq;
85 register digit v1, v2;
88 digit uspace[5], vspace[5], qspace[5];
91 * Take care of special cases: divide by zero, and u < v.
95 static volatile const unsigned int zero = 0;
97 tmp.ul[H] = tmp.ul[L] = 1 / zero;
112 * Break dividend and divisor into digits in base B, then
113 * count leading zeros to determine m and n. When done, we
115 * u = (u[1]u[2]...u[m+n]) sub B
116 * v = (v[1]v[2]...v[n]) sub B
118 * 1 < n <= 4 (if n = 1, we use a different division algorithm)
119 * m >= 0 (otherwise u < v, which we already checked)
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]);
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++) {
137 u_long rbj; /* r*B+u[j] (not root boy jim) */
138 digit q1, q2, q3, q4;
141 * Change of plan, per exercise 16.
144 * q[j] = floor((r*B + u[j]) / v),
145 * r = (r*B + u[j]) % v;
146 * We unroll this completely here.
148 t = v[2]; /* nonzero, by definition */
150 rbj = COMBINE(u[1] % t, u[2]);
152 rbj = COMBINE(rbj % t, u[3]);
154 rbj = COMBINE(rbj % t, u[4]);
158 tmp.ul[H] = COMBINE(q1, q2);
159 tmp.ul[L] = COMBINE(q3, q4);
165 * By adjusting q once we determine m, we can guarantee that
166 * there is a complete four-digit quotient at &qspace[1] when
169 for (m = 4 - n; u[1] == 0; u++)
171 for (i = 4 - m; --i >= 0;)
176 * Here we run Program D, translated from MIX to C and acquiring
177 * a few minor changes.
179 * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
182 for (t = v[1]; t < B / 2; t <<= 1)
185 __shl(&u[0], m + n, d); /* u <<= d */
186 __shl(&v[1], n - 1, d); /* v <<= d */
192 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
193 v2 = v[2]; /* for D3 */
195 register digit uj0, uj1, uj2;
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.
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 */
213 u_long nn = COMBINE(uj0, uj1);
217 while (v2 * qhat > COMBINE(rhat, uj2)) {
220 if ((rhat += v1) >= B)
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.
229 for (t = 0, i = n; i > 0; i--) {
230 t = u[i + j] - v[i] * qhat - t;
232 t = (B - HHALF(t)) & (B - 1);
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].
244 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
245 t += u[i + j] + v[i];
249 u[j] = LHALF(u[j] + t);
252 } while (++j <= m); /* D7: loop on j. */
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).
261 for (i = m + n; i > m; --i)
263 LHALF(u[i - 1] << (HALF_BITS - d));
266 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
267 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
271 tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
272 tmp.ul[L] = COMBINE(qspace[3], qspace[4]);