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_long)(a) << HALF_BITS) | (b))
  50
  51 /* select a type for digits in base B: use unsigned short if they fit */
  52 #if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff
  53 typedef unsigned short digit;
  54 #else
  55 typedef u_long digit;
  56 #endif
  57
  58 /*
  59  * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
  60  * `fall out' the left (there never will be any such anyway).
  61  * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
  62  */
  63 static void
  64 shl(digit *p, int len, int sh)
  65 {
  66         int i;
  67
  68         for (i = 0; i < len; i++)
  69                 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
  70         p[i] = LHALF(p[i] << sh);
  71 }
  72
  73 /*
  74  * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
  75  *
  76  * We do this in base 2-sup-HALF_BITS, so that all intermediate products
  77  * fit within u_long.  As a consequence, the maximum length dividend and
  78  * divisor are 4 `digits' in this base (they are shorter if they have
  79  * leading zeros).
  80  */
  81 u_quad_t
  82 __qdivrem(uq, vq, arq)
  83         u_quad_t uq, vq, *arq;
  84 {
  85         union uu tmp;
  86         digit *u, *v, *q;
  87         digit v1, v2;
  88         u_long qhat, rhat, t;
  89         int m, n, d, j, i;
  90         digit uspace[5], vspace[5], qspace[5];
  91
  92         /*
  93          * Take care of special cases: divide by zero, and u < v.
  94          */
  95         if (vq == 0) {
  96                 /* divide by zero. */
  97                 static volatile const unsigned int zero = 0;
  98
  99                 tmp.ul[H] = tmp.ul[L] = 1 / zero;
 100                 if (arq)
 101                         *arq = uq;
 102                 return (tmp.q);
 103         }
 104         if (uq < vq) {
 105                 if (arq)
 106                         *arq = uq;
 107                 return (0);
 108         }
 109         u = &uspace[0];
 110         v = &vspace[0];
 111         q = &qspace[0];
 112
 113         /*
 114          * Break dividend and divisor into digits in base B, then
 115          * count leading zeros to determine m and n.  When done, we
 116          * will have:
 117          *      u = (u[1]u[2]...u[m+n]) sub B
 118          *      v = (v[1]v[2]...v[n]) sub B
 119          *      v[1] != 0
 120          *      1 < n <= 4 (if n = 1, we use a different division algorithm)
 121          *      m >= 0 (otherwise u < v, which we already checked)
 122          *      m + n = 4
 123          * and thus
 124          *      m = 4 - n <= 2
 125          */
 126         tmp.uq = uq;
 127         u[0] = 0;
 128         u[1] = HHALF(tmp.ul[H]);
 129         u[2] = LHALF(tmp.ul[H]);
 130         u[3] = HHALF(tmp.ul[L]);
 131         u[4] = LHALF(tmp.ul[L]);
 132         tmp.uq = vq;
 133         v[1] = HHALF(tmp.ul[H]);
 134         v[2] = LHALF(tmp.ul[H]);
 135         v[3] = HHALF(tmp.ul[L]);
 136         v[4] = LHALF(tmp.ul[L]);
 137         for (n = 4; v[1] == 0; v++) {
 138                 if (--n == 1) {
 139                         u_long rbj;     /* r*B+u[j] (not root boy jim) */
 140                         digit q1, q2, q3, q4;
 141
 142                         /*
 143                          * Change of plan, per exercise 16.
 144                          *      r = 0;
 145                          *      for j = 1..4:
 146                          *              q[j] = floor((r*B + u[j]) / v),
 147                          *              r = (r*B + u[j]) % v;
 148                          * We unroll this completely here.
 149                          */
 150                         t = v[2];       /* nonzero, by definition */
 151                         q1 = u[1] / t;
 152                         rbj = COMBINE(u[1] % t, u[2]);
 153                         q2 = rbj / t;
 154                         rbj = COMBINE(rbj % t, u[3]);
 155                         q3 = rbj / t;
 156                         rbj = COMBINE(rbj % t, u[4]);
 157                         q4 = rbj / t;
 158                         if (arq)
 159                                 *arq = rbj % t;
 160                         tmp.ul[H] = COMBINE(q1, q2);
 161                         tmp.ul[L] = COMBINE(q3, q4);
 162                         return (tmp.q);
 163                 }
 164         }
 165
 166         /*
 167          * By adjusting q once we determine m, we can guarantee that
 168          * there is a complete four-digit quotient at &qspace[1] when
 169          * we finally stop.
 170          */
 171         for (m = 4 - n; u[1] == 0; u++)
 172                 m--;
 173         for (i = 4 - m; --i >= 0;)
 174                 q[i] = 0;
 175         q += 4 - m;
 176
 177         /*
 178          * Here we run Program D, translated from MIX to C and acquiring
 179          * a few minor changes.
 180          *
 181          * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
 182          */
 183         d = 0;
 184         for (t = v[1]; t < B / 2; t <<= 1)
 185                 d++;
 186         if (d > 0) {
 187                 shl(&u[0], m + n, d);           /* u <<= d */
 188                 shl(&v[1], n - 1, d);           /* v <<= d */
 189         }
 190         /*
 191          * D2: j = 0.
 192          */
 193         j = 0;
 194         v1 = v[1];      /* for D3 -- note that v[1..n] are constant */
 195         v2 = v[2];      /* for D3 */
 196         do {
 197                 digit uj0, uj1, uj2;
 198
 199                 /*
 200                  * D3: Calculate qhat (\^q, in TeX notation).
 201                  * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
 202                  * let rhat = (u[j]*B + u[j+1]) mod v[1].
 203                  * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
 204                  * decrement qhat and increase rhat correspondingly.
 205                  * Note that if rhat >= B, v[2]*qhat < rhat*B.
 206                  */
 207                 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
 208                 uj1 = u[j + 1]; /* for D3 only */
 209                 uj2 = u[j + 2]; /* for D3 only */
 210                 if (uj0 == v1) {
 211                         qhat = B;
 212                         rhat = uj1;
 213                         goto qhat_too_big;
 214                 } else {
 215                         u_long nn = COMBINE(uj0, uj1);
 216                         qhat = nn / v1;
 217                         rhat = nn % v1;
 218                 }
 219                 while (v2 * qhat > COMBINE(rhat, uj2)) {
 220         qhat_too_big:
 221                         qhat--;
 222                         if ((rhat += v1) >= B)
 223                                 break;
 224                 }
 225                 /*
 226                  * D4: Multiply and subtract.
 227                  * The variable `t' holds any borrows across the loop.
 228                  * We split this up so that we do not require v[0] = 0,
 229                  * and to eliminate a final special case.
 230                  */
 231                 for (t = 0, i = n; i > 0; i--) {
 232                         t = u[i + j] - v[i] * qhat - t;
 233                         u[i + j] = LHALF(t);
 234                         t = (B - HHALF(t)) & (B - 1);
 235                 }
 236                 t = u[j] - t;
 237                 u[j] = LHALF(t);
 238                 /*
 239                  * D5: test remainder.
 240                  * There is a borrow if and only if HHALF(t) is nonzero;
 241                  * in that (rare) case, qhat was too large (by exactly 1).
 242                  * Fix it by adding v[1..n] to u[j..j+n].
 243                  */
 244                 if (HHALF(t)) {
 245                         qhat--;
 246                         for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
 247                                 t += u[i + j] + v[i];
 248                                 u[i + j] = LHALF(t);
 249                                 t = HHALF(t);
 250                         }
 251                         u[j] = LHALF(u[j] + t);
 252                 }
 253                 q[j] = qhat;
 254         } while (++j <= m);             /* D7: loop on j. */
 255
 256         /*
 257          * If caller wants the remainder, we have to calculate it as
 258          * u[m..m+n] >> d (this is at most n digits and thus fits in
 259          * u[m+1..m+n], but we may need more source digits).
 260          */
 261         if (arq) {
 262                 if (d) {
 263                         for (i = m + n; i > m; --i)
 264                                 u[i] = (u[i] >> d) |
 265                                     LHALF(u[i - 1] << (HALF_BITS - d));
 266                         u[i] = 0;
 267                 }
 268                 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
 269                 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
 270                 *arq = tmp.q;
 271         }
 272
 273         tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
 274         tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
 275         return (tmp.q);
 276 }
 277
 278 /*
 279  * Divide two unsigned quads.
 280  */
 281
 282 u_quad_t
 283 __udivdi3(a, b)
 284         u_quad_t a, b;
 285 {
 286
 287         return (__qdivrem(a, b, (u_quad_t *)0));
 288 }
 289
 290 /*
 291  * Return remainder after dividing two unsigned quads.
 292  */
 293 u_quad_t
 294 __umoddi3(a, b)
 295         u_quad_t a, b;
 296 {
 297         u_quad_t r;
 298
 299         (void)__qdivrem(a, b, &r);
 300         return (r);
 301 }
 302
 303 /*
 304  * Divide two signed quads.
 305  * ??? if -1/2 should produce -1 on this machine, this code is wrong
 306  */
 307 quad_t
 308 __divdi3(a, b)
 309         quad_t a, b;
 310 {
 311         u_quad_t ua, ub, uq;
 312         int neg;
 313
 314         if (a < 0)
 315                 ua = -(u_quad_t)a, neg = 1;
 316         else
 317                 ua = a, neg = 0;
 318         if (b < 0)
 319                 ub = -(u_quad_t)b, neg ^= 1;
 320         else
 321                 ub = b;
 322         uq = __qdivrem(ua, ub, (u_quad_t *)0);
 323         return (neg ? -uq : uq);
 324 }
 325
 326 /*
 327  * Return remainder after dividing two signed quads.
 328  *
 329  * XXX
 330  * If -1/2 should produce -1 on this machine, this code is wrong.
 331  */
 332 quad_t
 333 __moddi3(a, b)
 334         quad_t a, b;
 335 {
 336         u_quad_t ua, ub, ur;
 337         int neg;
 338
 339         if (a < 0)
 340                 ua = -(u_quad_t)a, neg = 1;
 341         else
 342                 ua = a, neg = 0;
 343         if (b < 0)
 344                 ub = -(u_quad_t)b;
 345         else
 346                 ub = b;
 347         (void)__qdivrem(ua, ub, &ur);
 348         return (neg ? -ur : ur);
 349 }