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 * 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. 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#include <sys/cdefs.h> 35__FBSDID("$FreeBSD: stable/11/sys/libkern/qdivrem.c 319284 2017-05-31 05:45:06Z delphij $"); 36 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(uq, vq, arq) 81 u_quad_t uq, vq, *arq; 82{ 83 union uu tmp; 84 digit *u, *v, *q; 85 digit v1, v2; 86 u_long 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_long 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_long 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