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  * 3. 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  * $FreeBSD: src/lib/libstand/qdivrem.c,v 1.2 1999/08/28 00:05:33 peter Exp $
  34  * $DragonFly: src/lib/libstand/qdivrem.c,v 1.4 2005/12/11 02:27:26 swildner Exp $
  35  *      From: Id: qdivrem.c,v 1.7 1997/11/07 09:20:40 phk Exp
  36  */
  37
  38 /*
  39  * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
  40  * section 4.3.1, pp. 257--259.
  41  */
  42
  43 #include "quad.h"
  44
  45 #define B       (1 << HALF_BITS)        /* digit base */
  46
  47 /* Combine two `digits' to make a single two-digit number. */
  48 #define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
  49
  50 _Static_assert(sizeof(int) / 2 == sizeof(short),
  51         "Bitwise functions in libstand are broken on this architecture");
  52
  53 /* select a type for digits in base B: use unsigned short if they fit */
  54 typedef unsigned short digit;
  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  */
  61 static 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_int.  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  */
  79 u_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_int 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 (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_int 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_int 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
 275 /*
 276  * Divide two unsigned quads.
 277  */
 278
 279 u_quad_t
 280 __udivdi3(u_quad_t a, u_quad_t b)
 281 {
 282
 283         return (__qdivrem(a, b, (u_quad_t *)0));
 284 }
 285
 286 /*
 287  * Return remainder after dividing two unsigned quads.
 288  */
 289 u_quad_t
 290 __umoddi3(u_quad_t a, u_quad_t b)
 291 {
 292         u_quad_t r;
 293
 294         (void)__qdivrem(a, b, &r);
 295         return (r);
 296 }