usr/src/boot/lib/libstand/qdivrem.c

   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 }