external/bsd/pcc/dist/pcc-libs/libpcc/qdivrem.c

   1 /*      $Id: qdivrem.c,v 1.1.1.2 2009/09/04 00:27:36 gmcgarry Exp $     */
   2 /*      $NetBSD: qdivrem.c,v 1.1 2005/12/20 19:28:51 christos Exp $     */
   3
   4 /*-
   5  * Copyright (c) 1992, 1993
   6  *      The Regents of the University of California.  All rights reserved.
   7  *
   8  * This software was developed by the Computer Systems Engineering group
   9  * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
  10  * contributed to Berkeley.
  11  *
  12  * Redistribution and use in source and binary forms, with or without
  13  * modification, are permitted provided that the following conditions
  14  * are met:
  15  * 1. Redistributions of source code must retain the above copyright
  16  *    notice, this list of conditions and the following disclaimer.
  17  * 2. Redistributions in binary form must reproduce the above copyright
  18  *    notice, this list of conditions and the following disclaimer in the
  19  *    documentation and/or other materials provided with the distribution.
  20  * 3. Neither the name of the University nor the names of its contributors
  21  *    may be used to endorse or promote products derived from this software
  22  *    without specific prior written permission.
  23  *
  24  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
  25  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
  26  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
  27  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
  28  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
  29  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
  30  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
  31  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
  32  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
  33  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
  34  * SUCH DAMAGE.
  35  */
  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 "quad.h"
  43
  44 #define B       ((int)1 << HALF_BITS)   /* digit base */
  45
  46 /* Combine two `digits' to make a single two-digit number. */
  47 #define COMBINE(a, b) (((unsigned int)(a) << HALF_BITS) | (b))
  48
  49 /* select a type for digits in base B: use unsigned short if they fit */
  50 #if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
  51 typedef unsigned short digit;
  52 #else
  53 typedef unsigned int digit;
  54 #endif
  55
  56 static void shl(digit *p, int len, int sh);
  57
  58 /*
  59  * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
  60  *
  61  * We do this in base 2-sup-HALF_BITS, so that all intermediate products
  62  * fit within unsigned int.  As a consequence, the maximum length dividend and
  63  * divisor are 4 `digits' in this base (they are shorter if they have
  64  * leading zeros).
  65  */
  66 u_quad_t
  67 __qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
  68 {
  69         union uu tmp;
  70         digit *u, *v, *q;
  71         digit v1, v2;
  72         unsigned int qhat, rhat, t;
  73         int m, n, d, j, i;
  74         digit uspace[5], vspace[5], qspace[5];
  75
  76         /*
  77          * Take care of special cases: divide by zero, and u < v.
  78          */
  79         if (vq == 0) {
  80                 /* divide by zero. */
  81                 static volatile const unsigned int zero = 0;
  82
  83                 tmp.ul[H] = tmp.ul[L] = 1 / zero;
  84                 if (arq)
  85                         *arq = uq;
  86                 return (tmp.q);
  87         }
  88         if (uq < vq) {
  89                 if (arq)
  90                         *arq = uq;
  91                 return (0);
  92         }
  93         u = &uspace[0];
  94         v = &vspace[0];
  95         q = &qspace[0];
  96
  97         /*
  98          * Break dividend and divisor into digits in base B, then
  99          * count leading zeros to determine m and n.  When done, we
 100          * will have:
 101          *      u = (u[1]u[2]...u[m+n]) sub B
 102          *      v = (v[1]v[2]...v[n]) sub B
 103          *      v[1] != 0
 104          *      1 < n <= 4 (if n = 1, we use a different division algorithm)
 105          *      m >= 0 (otherwise u < v, which we already checked)
 106          *      m + n = 4
 107          * and thus
 108          *      m = 4 - n <= 2
 109          */
 110         tmp.uq = uq;
 111         u[0] = 0;
 112         u[1] = (digit)HHALF(tmp.ul[H]);
 113         u[2] = (digit)LHALF(tmp.ul[H]);
 114         u[3] = (digit)HHALF(tmp.ul[L]);
 115         u[4] = (digit)LHALF(tmp.ul[L]);
 116         tmp.uq = vq;
 117         v[1] = (digit)HHALF(tmp.ul[H]);
 118         v[2] = (digit)LHALF(tmp.ul[H]);
 119         v[3] = (digit)HHALF(tmp.ul[L]);
 120         v[4] = (digit)LHALF(tmp.ul[L]);
 121         for (n = 4; v[1] == 0; v++) {
 122                 if (--n == 1) {
 123                         unsigned int rbj;       /* r*B+u[j] (not root boy jim) */
 124                         digit q1, q2, q3, q4;
 125
 126                         /*
 127                          * Change of plan, per exercise 16.
 128                          *      r = 0;
 129                          *      for j = 1..4:
 130                          *              q[j] = floor((r*B + u[j]) / v),
 131                          *              r = (r*B + u[j]) % v;
 132                          * We unroll this completely here.
 133                          */
 134                         t = v[2];       /* nonzero, by definition */
 135                         q1 = (digit)(u[1] / t);
 136                         rbj = COMBINE(u[1] % t, u[2]);
 137                         q2 = (digit)(rbj / t);
 138                         rbj = COMBINE(rbj % t, u[3]);
 139                         q3 = (digit)(rbj / t);
 140                         rbj = COMBINE(rbj % t, u[4]);
 141                         q4 = (digit)(rbj / t);
 142                         if (arq)
 143                                 *arq = rbj % t;
 144                         tmp.ul[H] = COMBINE(q1, q2);
 145                         tmp.ul[L] = COMBINE(q3, q4);
 146                         return (tmp.q);
 147                 }
 148         }
 149
 150         /*
 151          * By adjusting q once we determine m, we can guarantee that
 152          * there is a complete four-digit quotient at &qspace[1] when
 153          * we finally stop.
 154          */
 155         for (m = 4 - n; u[1] == 0; u++)
 156                 m--;
 157         for (i = 4 - m; --i >= 0;)
 158                 q[i] = 0;
 159         q += 4 - m;
 160
 161         /*
 162          * Here we run Program D, translated from MIX to C and acquiring
 163          * a few minor changes.
 164          *
 165          * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
 166          */
 167         d = 0;
 168         for (t = v[1]; t < B / 2; t <<= 1)
 169                 d++;
 170         if (d > 0) {
 171                 shl(&u[0], m + n, d);           /* u <<= d */
 172                 shl(&v[1], n - 1, d);           /* v <<= d */
 173         }
 174         /*
 175          * D2: j = 0.
 176          */
 177         j = 0;
 178         v1 = v[1];      /* for D3 -- note that v[1..n] are constant */
 179         v2 = v[2];      /* for D3 */
 180         do {
 181                 digit uj0, uj1, uj2;
 182
 183                 /*
 184                  * D3: Calculate qhat (\^q, in TeX notation).
 185                  * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
 186                  * let rhat = (u[j]*B + u[j+1]) mod v[1].
 187                  * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
 188                  * decrement qhat and increase rhat correspondingly.
 189                  * Note that if rhat >= B, v[2]*qhat < rhat*B.
 190                  */
 191                 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
 192                 uj1 = u[j + 1]; /* for D3 only */
 193                 uj2 = u[j + 2]; /* for D3 only */
 194                 if (uj0 == v1) {
 195                         qhat = B;
 196                         rhat = uj1;
 197                         goto qhat_too_big;
 198                 } else {
 199                         unsigned int nn = COMBINE(uj0, uj1);
 200                         qhat = nn / v1;
 201                         rhat = nn % v1;
 202                 }
 203                 while (v2 * qhat > COMBINE(rhat, uj2)) {
 204         qhat_too_big:
 205                         qhat--;
 206                         if ((rhat += v1) >= B)
 207                                 break;
 208                 }
 209                 /*
 210                  * D4: Multiply and subtract.
 211                  * The variable `t' holds any borrows across the loop.
 212                  * We split this up so that we do not require v[0] = 0,
 213                  * and to eliminate a final special case.
 214                  */
 215                 for (t = 0, i = n; i > 0; i--) {
 216                         t = u[i + j] - v[i] * qhat - t;
 217                         u[i + j] = (digit)LHALF(t);
 218                         t = (B - HHALF(t)) & (B - 1);
 219                 }
 220                 t = u[j] - t;
 221                 u[j] = (digit)LHALF(t);
 222                 /*
 223                  * D5: test remainder.
 224                  * There is a borrow if and only if HHALF(t) is nonzero;
 225                  * in that (rare) case, qhat was too large (by exactly 1).
 226                  * Fix it by adding v[1..n] to u[j..j+n].
 227                  */
 228                 if (HHALF(t)) {
 229                         qhat--;
 230                         for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
 231                                 t += u[i + j] + v[i];
 232                                 u[i + j] = (digit)LHALF(t);
 233                                 t = HHALF(t);
 234                         }
 235                         u[j] = (digit)LHALF(u[j] + t);
 236                 }
 237                 q[j] = (digit)qhat;
 238         } while (++j <= m);             /* D7: loop on j. */
 239
 240         /*
 241          * If caller wants the remainder, we have to calculate it as
 242          * u[m..m+n] >> d (this is at most n digits and thus fits in
 243          * u[m+1..m+n], but we may need more source digits).
 244          */
 245         if (arq) {
 246                 if (d) {
 247                         for (i = m + n; i > m; --i)
 248                                 u[i] = (digit)(((unsigned int)u[i] >> d) |
 249                                     LHALF((unsigned int)u[i - 1] << (HALF_BITS - d)));
 250                         u[i] = 0;
 251                 }
 252                 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
 253                 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
 254                 *arq = tmp.q;
 255         }
 256
 257         tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
 258         tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
 259         return (tmp.q);
 260 }
 261
 262 /*
 263  * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
 264  * `fall out' the left (there never will be any such anyway).
 265  * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
 266  */
 267 static void
 268 shl(digit *p, int len, int sh)
 269 {
 270         int i;
 271
 272         for (i = 0; i < len; i++)
 273                 p[i] = (digit)(LHALF((unsigned int)p[i] << sh) |
 274                     ((unsigned int)p[i + 1] >> (HALF_BITS - sh)));
 275         p[i] = (digit)(LHALF((unsigned int)p[i] << sh));
 276 }