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#include <sys/cdefs.h> 37/* 38 * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed), 39 * section 4.3.1, pp. 257--259. 40 */ 41 42#include <libkern/quad.h> 43 44#define B (1 << HALF_BITS) /* digit base */ 45 46/* Combine two `digits' to make a single two-digit number. */ 47#define COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b)) 48 49/* select a type for digits in base B: use unsigned short if they fit */ 50#if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff 51typedef unsigned short digit; 52#else 53typedef u_long digit; 54#endif 55 56/* 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. 60 */ 61static void 62__shl(digit *p, int len, int sh) 63{ 64 int i; 65 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); 69} 70 71/* 72 * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v. 73 * 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 77 * leading zeros). 78 */ 79u_quad_t 80__qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq) 81{ 82 union uu tmp; 83 digit *u, *v, *q; 84 digit v1, v2; 85 u_long qhat, rhat, t; 86 int m, n, d, j, i; 87 digit uspace[5], vspace[5], qspace[5]; 88 89 /* 90 * Take care of special cases: divide by zero, and u < v. 91 */ 92 if (__predict_false(vq == 0)) { 93 /* divide by zero. */ 94 static volatile const unsigned int zero = 0; 95 96 tmp.ul[H] = tmp.ul[L] = 1 / zero; 97 if (arq) 98 *arq = uq; 99 return (tmp.q); 100 } 101 if (uq < vq) { 102 if (arq) 103 *arq = uq; 104 return (0); 105 } 106 u = &uspace[0]; 107 v = &vspace[0]; 108 q = &qspace[0]; 109 110 /* 111 * Break dividend and divisor into digits in base B, then 112 * count leading zeros to determine m and n. When done, we 113 * will have: 114 * u = (u[1]u[2]...u[m+n]) sub B 115 * v = (v[1]v[2]...v[n]) sub B 116 * v[1] != 0 117 * 1 < n <= 4 (if n = 1, we use a different division algorithm) 118 * m >= 0 (otherwise u < v, which we already checked) 119 * m + n = 4 120 * and thus 121 * m = 4 - n <= 2 122 */ 123 tmp.uq = uq; 124 u[0] = 0; 125 u[1] = HHALF(tmp.ul[H]); 126 u[2] = LHALF(tmp.ul[H]); 127 u[3] = HHALF(tmp.ul[L]); 128 u[4] = LHALF(tmp.ul[L]); 129 tmp.uq = vq; 130 v[1] = HHALF(tmp.ul[H]); 131 v[2] = LHALF(tmp.ul[H]); 132 v[3] = HHALF(tmp.ul[L]); 133 v[4] = LHALF(tmp.ul[L]); 134 for (n = 4; v[1] == 0; v++) { 135 if (--n == 1) { 136 u_long rbj; /* r*B+u[j] (not root boy jim) */ 137 digit q1, q2, q3, q4; 138 139 /* 140 * Change of plan, per exercise 16. 141 * r = 0; 142 * for j = 1..4: 143 * q[j] = floor((r*B + u[j]) / v), 144 * r = (r*B + u[j]) % v; 145 * We unroll this completely here. 146 */ 147 t = v[2]; /* nonzero, by definition */ 148 q1 = u[1] / t; 149 rbj = COMBINE(u[1] % t, u[2]); 150 q2 = rbj / t; 151 rbj = COMBINE(rbj % t, u[3]); 152 q3 = rbj / t; 153 rbj = COMBINE(rbj % t, u[4]); 154 q4 = rbj / t; 155 if (arq) 156 *arq = rbj % t; 157 tmp.ul[H] = COMBINE(q1, q2); 158 tmp.ul[L] = COMBINE(q3, q4); 159 return (tmp.q); 160 } 161 } 162 163 /* 164 * By adjusting q once we determine m, we can guarantee that 165 * there is a complete four-digit quotient at &qspace[1] when 166 * we finally stop. 167 */ 168 for (m = 4 - n; u[1] == 0; u++) 169 m--; 170 for (i = 4 - m; --i >= 0;) 171 q[i] = 0; 172 q += 4 - m; 173 174 /* 175 * Here we run Program D, translated from MIX to C and acquiring 176 * a few minor changes. 177 * 178 * D1: choose multiplier 1 << d to ensure v[1] >= B/2. 179 */ 180 d = 0; 181 for (t = v[1]; t < B / 2; t <<= 1) 182 d++; 183 if (d > 0) { 184 __shl(&u[0], m + n, d); /* u <<= d */ 185 __shl(&v[1], n - 1, d); /* v <<= d */ 186 } 187 /* 188 * D2: j = 0. 189 */ 190 j = 0; 191 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */ 192 v2 = v[2]; /* for D3 */ 193 do { 194 digit uj0, uj1, uj2; 195 196 /* 197 * D3: Calculate qhat (\^q, in TeX notation). 198 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and 199 * let rhat = (u[j]*B + u[j+1]) mod v[1]. 200 * While rhat < B and v[2]*qhat > rhat*B+u[j+2], 201 * decrement qhat and increase rhat correspondingly. 202 * Note that if rhat >= B, v[2]*qhat < rhat*B. 203 */ 204 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */ 205 uj1 = u[j + 1]; /* for D3 only */ 206 uj2 = u[j + 2]; /* for D3 only */ 207 if (uj0 == v1) { 208 qhat = B; 209 rhat = uj1; 210 goto qhat_too_big; 211 } else { 212 u_long nn = COMBINE(uj0, uj1); 213 qhat = nn / v1; 214 rhat = nn % v1; 215 } 216 while (v2 * qhat > COMBINE(rhat, uj2)) { 217 qhat_too_big: 218 qhat--; 219 if ((rhat += v1) >= B) 220 break; 221 } 222 /* 223 * D4: Multiply and subtract. 224 * The variable `t' holds any borrows across the loop. 225 * We split this up so that we do not require v[0] = 0, 226 * and to eliminate a final special case. 227 */ 228 for (t = 0, i = n; i > 0; i--) { 229 t = u[i + j] - v[i] * qhat - t; 230 u[i + j] = LHALF(t); 231 t = (B - HHALF(t)) & (B - 1); 232 } 233 t = u[j] - t; 234 u[j] = LHALF(t); 235 /* 236 * D5: test remainder. 237 * There is a borrow if and only if HHALF(t) is nonzero; 238 * in that (rare) case, qhat was too large (by exactly 1). 239 * Fix it by adding v[1..n] to u[j..j+n]. 240 */ 241 if (HHALF(t)) { 242 qhat--; 243 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */ 244 t += u[i + j] + v[i]; 245 u[i + j] = LHALF(t); 246 t = HHALF(t); 247 } 248 u[j] = LHALF(u[j] + t); 249 } 250 q[j] = qhat; 251 } while (++j <= m); /* D7: loop on j. */ 252 253 /* 254 * If caller wants the remainder, we have to calculate it as 255 * u[m..m+n] >> d (this is at most n digits and thus fits in 256 * u[m+1..m+n], but we may need more source digits). 257 */ 258 if (arq) { 259 if (d) { 260 for (i = m + n; i > m; --i) 261 u[i] = (u[i] >> d) | 262 LHALF(u[i - 1] << (HALF_BITS - d)); 263 u[i] = 0; 264 } 265 tmp.ul[H] = COMBINE(uspace[1], uspace[2]); 266 tmp.ul[L] = COMBINE(uspace[3], uspace[4]); 267 *arq = tmp.q; 268 } 269 270 tmp.ul[H] = COMBINE(qspace[1], qspace[2]); 271 tmp.ul[L] = COMBINE(qspace[3], qspace[4]); 272 return (tmp.q); 273} 274