common/lib/libc/quad/qdivrem.c

   1 /*      $NetBSD: qdivrem.c,v 1.2 2009/03/15 22:31:12 cegger Exp $       */
   2
   3 /*-
   4  * Copyright (c) 1992, 1993
   5  *      The Regents of the University of California.  All rights reserved.
   6  *
   7  * This software was developed by the Computer Systems Engineering group
   8  * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
   9  * contributed to Berkeley.
  10  *
  11  * Redistribution and use in source and binary forms, with or without
  12  * modification, are permitted provided that the following conditions
  13  * are met:
  14  * 1. Redistributions of source code must retain the above copyright
  15  *    notice, this list of conditions and the following disclaimer.
  16  * 2. Redistributions in binary form must reproduce the above copyright
  17  *    notice, this list of conditions and the following disclaimer in the
  18  *    documentation and/or other materials provided with the distribution.
  19  * 3. Neither the name of the University nor the names of its contributors
  20  *    may be used to endorse or promote products derived from this software
  21  *    without specific prior written permission.
  22  *
  23  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
  24  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
  25  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
  26  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
  27  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
  28  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
  29  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
  30  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
  31  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
  32  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
  33  * SUCH DAMAGE.
  34  */
  35
  36 #include <sys/cdefs.h>
  37 #if defined(LIBC_SCCS) && !defined(lint)
  38 #if 0
  39 static char sccsid[] = "@(#)qdivrem.c   8.1 (Berkeley) 6/4/93";
  40 #else
  41 __RCSID("$NetBSD: qdivrem.c,v 1.2 2009/03/15 22:31:12 cegger Exp $");
  42 #endif
  43 #endif /* LIBC_SCCS and not lint */
  44
  45 /*
  46  * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
  47  * section 4.3.1, pp. 257--259.
  48  */
  49
  50 #include "quad.h"
  51
  52 #define B       ((int)1 << HALF_BITS)   /* digit base */
  53
  54 /* Combine two `digits' to make a single two-digit number. */
  55 #define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
  56
  57 /* select a type for digits in base B: use unsigned short if they fit */
  58 #if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
  59 typedef unsigned short digit;
  60 #else
  61 typedef u_int digit;
  62 #endif
  63
  64 static void shl __P((digit *p, int len, int sh));
  65
  66 /*
  67  * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
  68  *
  69  * We do this in base 2-sup-HALF_BITS, so that all intermediate products
  70  * fit within u_int.  As a consequence, the maximum length dividend and
  71  * divisor are 4 `digits' in this base (they are shorter if they have
  72  * leading zeros).
  73  */
  74 u_quad_t
  75 __qdivrem(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
  76 {
  77         union uu tmp;
  78         digit *u, *v, *q;
  79         digit v1, v2;
  80         u_int qhat, rhat, t;
  81         int m, n, d, j, i;
  82         digit uspace[5], vspace[5], qspace[5];
  83
  84         /*
  85          * Take care of special cases: divide by zero, and u < v.
  86          */
  87         if (vq == 0) {
  88                 /* divide by zero. */
  89                 static volatile const unsigned int zero = 0;
  90
  91                 tmp.ul[H] = tmp.ul[L] = 1 / zero;
  92                 if (arq)
  93                         *arq = uq;
  94                 return (tmp.q);
  95         }
  96         if (uq < vq) {
  97                 if (arq)
  98                         *arq = uq;
  99                 return (0);
 100         }
 101         u = &uspace[0];
 102         v = &vspace[0];
 103         q = &qspace[0];
 104
 105         /*
 106          * Break dividend and divisor into digits in base B, then
 107          * count leading zeros to determine m and n.  When done, we
 108          * will have:
 109          *      u = (u[1]u[2]...u[m+n]) sub B
 110          *      v = (v[1]v[2]...v[n]) sub B
 111          *      v[1] != 0
 112          *      1 < n <= 4 (if n = 1, we use a different division algorithm)
 113          *      m >= 0 (otherwise u < v, which we already checked)
 114          *      m + n = 4
 115          * and thus
 116          *      m = 4 - n <= 2
 117          */
 118         tmp.uq = uq;
 119         u[0] = 0;
 120         u[1] = (digit)HHALF(tmp.ul[H]);
 121         u[2] = (digit)LHALF(tmp.ul[H]);
 122         u[3] = (digit)HHALF(tmp.ul[L]);
 123         u[4] = (digit)LHALF(tmp.ul[L]);
 124         tmp.uq = vq;
 125         v[1] = (digit)HHALF(tmp.ul[H]);
 126         v[2] = (digit)LHALF(tmp.ul[H]);
 127         v[3] = (digit)HHALF(tmp.ul[L]);
 128         v[4] = (digit)LHALF(tmp.ul[L]);
 129         for (n = 4; v[1] == 0; v++) {
 130                 if (--n == 1) {
 131                         u_int rbj;      /* r*B+u[j] (not root boy jim) */
 132                         digit q1, q2, q3, q4;
 133
 134                         /*
 135                          * Change of plan, per exercise 16.
 136                          *      r = 0;
 137                          *      for j = 1..4:
 138                          *              q[j] = floor((r*B + u[j]) / v),
 139                          *              r = (r*B + u[j]) % v;
 140                          * We unroll this completely here.
 141                          */
 142                         t = v[2];       /* nonzero, by definition */
 143                         q1 = (digit)(u[1] / t);
 144                         rbj = COMBINE(u[1] % t, u[2]);
 145                         q2 = (digit)(rbj / t);
 146                         rbj = COMBINE(rbj % t, u[3]);
 147                         q3 = (digit)(rbj / t);
 148                         rbj = COMBINE(rbj % t, u[4]);
 149                         q4 = (digit)(rbj / t);
 150                         if (arq)
 151                                 *arq = rbj % t;
 152                         tmp.ul[H] = COMBINE(q1, q2);
 153                         tmp.ul[L] = COMBINE(q3, q4);
 154                         return (tmp.q);
 155                 }
 156         }
 157
 158         /*
 159          * By adjusting q once we determine m, we can guarantee that
 160          * there is a complete four-digit quotient at &qspace[1] when
 161          * we finally stop.
 162          */
 163         for (m = 4 - n; u[1] == 0; u++)
 164                 m--;
 165         for (i = 4 - m; --i >= 0;)
 166                 q[i] = 0;
 167         q += 4 - m;
 168
 169         /*
 170          * Here we run Program D, translated from MIX to C and acquiring
 171          * a few minor changes.
 172          *
 173          * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
 174          */
 175         d = 0;
 176         for (t = v[1]; t < B / 2; t <<= 1)
 177                 d++;
 178         if (d > 0) {
 179                 shl(&u[0], m + n, d);           /* u <<= d */
 180                 shl(&v[1], n - 1, d);           /* v <<= d */
 181         }
 182         /*
 183          * D2: j = 0.
 184          */
 185         j = 0;
 186         v1 = v[1];      /* for D3 -- note that v[1..n] are constant */
 187         v2 = v[2];      /* for D3 */
 188         do {
 189                 digit uj0, uj1, uj2;
 190
 191                 /*
 192                  * D3: Calculate qhat (\^q, in TeX notation).
 193                  * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
 194                  * let rhat = (u[j]*B + u[j+1]) mod v[1].
 195                  * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
 196                  * decrement qhat and increase rhat correspondingly.
 197                  * Note that if rhat >= B, v[2]*qhat < rhat*B.
 198                  */
 199                 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
 200                 uj1 = u[j + 1]; /* for D3 only */
 201                 uj2 = u[j + 2]; /* for D3 only */
 202                 if (uj0 == v1) {
 203                         qhat = B;
 204                         rhat = uj1;
 205                         goto qhat_too_big;
 206                 } else {
 207                         u_int nn = COMBINE(uj0, uj1);
 208                         qhat = nn / v1;
 209                         rhat = nn % v1;
 210                 }
 211                 while (v2 * qhat > COMBINE(rhat, uj2)) {
 212         qhat_too_big:
 213                         qhat--;
 214                         if ((rhat += v1) >= B)
 215                                 break;
 216                 }
 217                 /*
 218                  * D4: Multiply and subtract.
 219                  * The variable `t' holds any borrows across the loop.
 220                  * We split this up so that we do not require v[0] = 0,
 221                  * and to eliminate a final special case.
 222                  */
 223                 for (t = 0, i = n; i > 0; i--) {
 224                         t = u[i + j] - v[i] * qhat - t;
 225                         u[i + j] = (digit)LHALF(t);
 226                         t = (B - HHALF(t)) & (B - 1);
 227                 }
 228                 t = u[j] - t;
 229                 u[j] = (digit)LHALF(t);
 230                 /*
 231                  * D5: test remainder.
 232                  * There is a borrow if and only if HHALF(t) is nonzero;
 233                  * in that (rare) case, qhat was too large (by exactly 1).
 234                  * Fix it by adding v[1..n] to u[j..j+n].
 235                  */
 236                 if (HHALF(t)) {
 237                         qhat--;
 238                         for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
 239                                 t += u[i + j] + v[i];
 240                                 u[i + j] = (digit)LHALF(t);
 241                                 t = HHALF(t);
 242                         }
 243                         u[j] = (digit)LHALF(u[j] + t);
 244                 }
 245                 q[j] = (digit)qhat;
 246         } while (++j <= m);             /* D7: loop on j. */
 247
 248         /*
 249          * If caller wants the remainder, we have to calculate it as
 250          * u[m..m+n] >> d (this is at most n digits and thus fits in
 251          * u[m+1..m+n], but we may need more source digits).
 252          */
 253         if (arq) {
 254                 if (d) {
 255                         for (i = m + n; i > m; --i)
 256                                 u[i] = (digit)(((u_int)u[i] >> d) |
 257                                     LHALF((u_int)u[i - 1] << (HALF_BITS - d)));
 258                         u[i] = 0;
 259                 }
 260                 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
 261                 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
 262                 *arq = tmp.q;
 263         }
 264
 265         tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
 266         tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
 267         return (tmp.q);
 268 }
 269
 270 /*
 271  * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
 272  * `fall out' the left (there never will be any such anyway).
 273  * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
 274  */
 275 static void
 276 shl(digit *p, int len, int sh)
 277 {
 278         int i;
 279
 280         for (i = 0; i < len; i++)
 281                 p[i] = (digit)(LHALF((u_int)p[i] << sh) |
 282                     ((u_int)p[i + 1] >> (HALF_BITS - sh)));
 283         p[i] = (digit)(LHALF((u_int)p[i] << sh));
 284 }