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 * From: Id: qdivrem.c,v 1.7 1997/11/07 09:20:40 phk Exp 34 */ 35 36 #include <sys/cdefs.h> 37 __FBSDID("$FreeBSD$"); 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 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 * __qdivrem(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 81 __qdivrem(uq, vq, arq) 82 u_quad_t uq, vq, *arq; 83 { 84 union uu tmp; 85 digit *u, *v, *q; 86 digit v1, v2; 87 u_int qhat, rhat, t; 88 int m, n, d, j, i; 89 digit uspace[5], vspace[5], qspace[5]; 90 91 /* 92 * Take care of special cases: divide by zero, and u < v. 93 */ 94 if (vq == 0) { 95 /* divide by zero. */ 96 static volatile const unsigned int zero = 0; 97 98 tmp.ul[H] = tmp.ul[L] = 1 / zero; 99 if (arq) 100 *arq = uq; 101 return (tmp.q); 102 } 103 if (uq < vq) { 104 if (arq) 105 *arq = uq; 106 return (0); 107 } 108 u = &uspace[0]; 109 v = &vspace[0]; 110 q = &qspace[0]; 111 112 /* 113 * Break dividend and divisor into digits in base B, then 114 * count leading zeros to determine m and n. When done, we 115 * will have: 116 * u = (u[1]u[2]...u[m+n]) sub B 117 * v = (v[1]v[2]...v[n]) sub B 118 * v[1] != 0 119 * 1 < n <= 4 (if n = 1, we use a different division algorithm) 120 * m >= 0 (otherwise u < v, which we already checked) 121 * m + n = 4 122 * and thus 123 * m = 4 - n <= 2 124 */ 125 tmp.uq = uq; 126 u[0] = 0; 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]); 131 tmp.uq = vq; 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++) { 137 if (--n == 1) { 138 u_int rbj; /* r*B+u[j] (not root boy jim) */ 139 digit q1, q2, q3, q4; 140 141 /* 142 * Change of plan, per exercise 16. 143 * r = 0; 144 * for j = 1..4: 145 * q[j] = floor((r*B + u[j]) / v), 146 * r = (r*B + u[j]) % v; 147 * We unroll this completely here. 148 */ 149 t = v[2]; /* nonzero, by definition */ 150 q1 = u[1] / t; 151 rbj = COMBINE(u[1] % t, u[2]); 152 q2 = rbj / t; 153 rbj = COMBINE(rbj % t, u[3]); 154 q3 = rbj / t; 155 rbj = COMBINE(rbj % t, u[4]); 156 q4 = rbj / t; 157 if (arq) 158 *arq = rbj % t; 159 tmp.ul[H] = COMBINE(q1, q2); 160 tmp.ul[L] = COMBINE(q3, q4); 161 return (tmp.q); 162 } 163 } 164 165 /* 166 * By adjusting q once we determine m, we can guarantee that 167 * there is a complete four-digit quotient at &qspace[1] when 168 * we finally stop. 169 */ 170 for (m = 4 - n; u[1] == 0; u++) 171 m--; 172 for (i = 4 - m; --i >= 0;) 173 q[i] = 0; 174 q += 4 - m; 175 176 /* 177 * Here we run Program D, translated from MIX to C and acquiring 178 * a few minor changes. 179 * 180 * D1: choose multiplier 1 << d to ensure v[1] >= B/2. 181 */ 182 d = 0; 183 for (t = v[1]; t < B / 2; t <<= 1) 184 d++; 185 if (d > 0) { 186 shl(&u[0], m + n, d); /* u <<= d */ 187 shl(&v[1], n - 1, d); /* v <<= d */ 188 } 189 /* 190 * D2: j = 0. 191 */ 192 j = 0; 193 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */ 194 v2 = v[2]; /* for D3 */ 195 do { 196 digit uj0, uj1, uj2; 197 198 /* 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. 205 */ 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 */ 209 if (uj0 == v1) { 210 qhat = B; 211 rhat = uj1; 212 goto qhat_too_big; 213 } else { 214 u_int nn = COMBINE(uj0, uj1); 215 qhat = nn / v1; 216 rhat = nn % v1; 217 } 218 while (v2 * qhat > COMBINE(rhat, uj2)) { 219 qhat_too_big: 220 qhat--; 221 if ((rhat += v1) >= B) 222 break; 223 } 224 /* 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. 229 */ 230 for (t = 0, i = n; i > 0; i--) { 231 t = u[i + j] - v[i] * qhat - t; 232 u[i + j] = LHALF(t); 233 t = (B - HHALF(t)) & (B - 1); 234 } 235 t = u[j] - t; 236 u[j] = LHALF(t); 237 /* 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]. 242 */ 243 if (HHALF(t)) { 244 qhat--; 245 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */ 246 t += u[i + j] + v[i]; 247 u[i + j] = LHALF(t); 248 t = HHALF(t); 249 } 250 u[j] = LHALF(u[j] + t); 251 } 252 q[j] = qhat; 253 } while (++j <= m); /* D7: loop on j. */ 254 255 /* 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). 259 */ 260 if (arq) { 261 if (d) { 262 for (i = m + n; i > m; --i) 263 u[i] = (u[i] >> d) | 264 LHALF(u[i - 1] << (HALF_BITS - d)); 265 u[i] = 0; 266 } 267 tmp.ul[H] = COMBINE(uspace[1], uspace[2]); 268 tmp.ul[L] = COMBINE(uspace[3], uspace[4]); 269 *arq = tmp.q; 270 } 271 272 tmp.ul[H] = COMBINE(qspace[1], qspace[2]); 273 tmp.ul[L] = COMBINE(qspace[3], qspace[4]); 274 return (tmp.q); 275 } 276 277 /* 278 * Divide two unsigned quads. 279 */ 280 281 u_quad_t 282 __udivdi3(a, b) 283 u_quad_t a, b; 284 { 285 286 return (__qdivrem(a, b, (u_quad_t *)0)); 287 } 288 289 /* 290 * Return remainder after dividing two unsigned quads. 291 */ 292 u_quad_t 293 __umoddi3(a, b) 294 u_quad_t a, b; 295 { 296 u_quad_t r; 297 298 (void)__qdivrem(a, b, &r); 299 return (r); 300 } 301 302 /* 303 * Divide two signed quads. 304 * ??? if -1/2 should produce -1 on this machine, this code is wrong 305 */ 306 quad_t 307 __divdi3(a, b) 308 quad_t a, b; 309 { 310 u_quad_t ua, ub, uq; 311 int neg; 312 313 if (a < 0) 314 ua = -(u_quad_t)a, neg = 1; 315 else 316 ua = a, neg = 0; 317 if (b < 0) 318 ub = -(u_quad_t)b, neg ^= 1; 319 else 320 ub = b; 321 uq = __qdivrem(ua, ub, (u_quad_t *)0); 322 return (neg ? -uq : uq); 323 } 324 325 /* 326 * Return remainder after dividing two signed quads. 327 * 328 * XXX 329 * If -1/2 should produce -1 on this machine, this code is wrong. 330 */ 331 quad_t 332 __moddi3(a, b) 333 quad_t a, b; 334 { 335 u_quad_t ua, ub, ur; 336 int neg; 337 338 if (a < 0) 339 ua = -(u_quad_t)a, neg = 1; 340 else 341 ua = a, neg = 0; 342 if (b < 0) 343 ub = -(u_quad_t)b; 344 else 345 ub = b; 346 (void)__qdivrem(ua, ub, &ur); 347 return (neg ? -ur : ur); 348 } 349