stand/lib/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  *      From: Id: qdivrem.c,v 1.7 1997/11/07 09:20:40 phk Exp
  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 #include <sys/endian.h> /* _QUAD_HIGHWORD */
  44
  45 #define B       (1 << HALF_BITS)        /* digit base */
  46
  47 /*
  48  * Define high and low longwords.
  49  */
  50 #define H       _QUAD_HIGHWORD
  51 #define L       _QUAD_LOWWORD
  52
  53 /* Combine two `digits' to make a single two-digit number. */
  54 #define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
  55
  56 _Static_assert(sizeof(int) / 2 == sizeof(short),
  57         "Bitwise functions in libstand are broken on this architecture");
  58
  59 /* select a type for digits in base B: use unsigned short if they fit */
  60 typedef unsigned short digit;
  61
  62 /*
  63  * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
  64  * `fall out' the left (there never will be any such anyway).
  65  * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
  66  */
  67 static void
  68 shl(digit *p, int len, int sh)
  69 {
  70         int i;
  71
  72         for (i = 0; i < len; i++)
  73                 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
  74         p[i] = LHALF(p[i] << sh);
  75 }
  76
  77 /*
  78  * __udivmoddi4(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
  79  *
  80  * We do this in base 2-sup-HALF_BITS, so that all intermediate products
  81  * fit within u_int.  As a consequence, the maximum length dividend and
  82  * divisor are 4 `digits' in this base (they are shorter if they have
  83  * leading zeros).
  84  */
  85 u_quad_t
  86 __udivmoddi4(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
  87 {
  88         union uu tmp;
  89         digit *u, *v, *q;
  90         digit v1, v2;
  91         u_int qhat, rhat, t;
  92         int m, n, d, j, i;
  93         digit uspace[5], vspace[5], qspace[5];
  94
  95         /*
  96          * Take care of special cases: divide by zero, and u < v.
  97          */
  98         if (vq == 0) {
  99                 /* divide by zero. */
 100                 static volatile const unsigned int zero = 0;
 101
 102                 tmp.ul[H] = tmp.ul[L] = 1 / zero;
 103                 if (arq)
 104                         *arq = uq;
 105                 return (tmp.q);
 106         }
 107         if (uq < vq) {
 108                 if (arq)
 109                         *arq = uq;
 110                 return (0);
 111         }
 112         u = &uspace[0];
 113         v = &vspace[0];
 114         q = &qspace[0];
 115
 116         /*
 117          * Break dividend and divisor into digits in base B, then
 118          * count leading zeros to determine m and n.  When done, we
 119          * will have:
 120          *      u = (u[1]u[2]...u[m+n]) sub B
 121          *      v = (v[1]v[2]...v[n]) sub B
 122          *      v[1] != 0
 123          *      1 < n <= 4 (if n = 1, we use a different division algorithm)
 124          *      m >= 0 (otherwise u < v, which we already checked)
 125          *      m + n = 4
 126          * and thus
 127          *      m = 4 - n <= 2
 128          */
 129         tmp.uq = uq;
 130         u[0] = 0;
 131         u[1] = HHALF(tmp.ul[H]);
 132         u[2] = LHALF(tmp.ul[H]);
 133         u[3] = HHALF(tmp.ul[L]);
 134         u[4] = LHALF(tmp.ul[L]);
 135         tmp.uq = vq;
 136         v[1] = HHALF(tmp.ul[H]);
 137         v[2] = LHALF(tmp.ul[H]);
 138         v[3] = HHALF(tmp.ul[L]);
 139         v[4] = LHALF(tmp.ul[L]);
 140         for (n = 4; v[1] == 0; v++) {
 141                 if (--n == 1) {
 142                         u_int rbj;      /* r*B+u[j] (not root boy jim) */
 143                         digit q1, q2, q3, q4;
 144
 145                         /*
 146                          * Change of plan, per exercise 16.
 147                          *      r = 0;
 148                          *      for j = 1..4:
 149                          *              q[j] = floor((r*B + u[j]) / v),
 150                          *              r = (r*B + u[j]) % v;
 151                          * We unroll this completely here.
 152                          */
 153                         t = v[2];       /* nonzero, by definition */
 154                         q1 = u[1] / t;
 155                         rbj = COMBINE(u[1] % t, u[2]);
 156                         q2 = rbj / t;
 157                         rbj = COMBINE(rbj % t, u[3]);
 158                         q3 = rbj / t;
 159                         rbj = COMBINE(rbj % t, u[4]);
 160                         q4 = rbj / t;
 161                         if (arq)
 162                                 *arq = rbj % t;
 163                         tmp.ul[H] = COMBINE(q1, q2);
 164                         tmp.ul[L] = COMBINE(q3, q4);
 165                         return (tmp.q);
 166                 }
 167         }
 168
 169         /*
 170          * By adjusting q once we determine m, we can guarantee that
 171          * there is a complete four-digit quotient at &qspace[1] when
 172          * we finally stop.
 173          */
 174         for (m = 4 - n; u[1] == 0; u++)
 175                 m--;
 176         for (i = 4 - m; --i >= 0;)
 177                 q[i] = 0;
 178         q += 4 - m;
 179
 180         /*
 181          * Here we run Program D, translated from MIX to C and acquiring
 182          * a few minor changes.
 183          *
 184          * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
 185          */
 186         d = 0;
 187         for (t = v[1]; t < B / 2; t <<= 1)
 188                 d++;
 189         if (d > 0) {
 190                 shl(&u[0], m + n, d);           /* u <<= d */
 191                 shl(&v[1], n - 1, d);           /* v <<= d */
 192         }
 193         /*
 194          * D2: j = 0.
 195          */
 196         j = 0;
 197         v1 = v[1];      /* for D3 -- note that v[1..n] are constant */
 198         v2 = v[2];      /* for D3 */
 199         do {
 200                 digit uj0, uj1, uj2;
 201
 202                 /*
 203                  * D3: Calculate qhat (\^q, in TeX notation).
 204                  * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
 205                  * let rhat = (u[j]*B + u[j+1]) mod v[1].
 206                  * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
 207                  * decrement qhat and increase rhat correspondingly.
 208                  * Note that if rhat >= B, v[2]*qhat < rhat*B.
 209                  */
 210                 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
 211                 uj1 = u[j + 1]; /* for D3 only */
 212                 uj2 = u[j + 2]; /* for D3 only */
 213                 if (uj0 == v1) {
 214                         qhat = B;
 215                         rhat = uj1;
 216                         goto qhat_too_big;
 217                 } else {
 218                         u_int nn = COMBINE(uj0, uj1);
 219                         qhat = nn / v1;
 220                         rhat = nn % v1;
 221                 }
 222                 while (v2 * qhat > COMBINE(rhat, uj2)) {
 223         qhat_too_big:
 224                         qhat--;
 225                         if ((rhat += v1) >= B)
 226                                 break;
 227                 }
 228                 /*
 229                  * D4: Multiply and subtract.
 230                  * The variable `t' holds any borrows across the loop.
 231                  * We split this up so that we do not require v[0] = 0,
 232                  * and to eliminate a final special case.
 233                  */
 234                 for (t = 0, i = n; i > 0; i--) {
 235                         t = u[i + j] - v[i] * qhat - t;
 236                         u[i + j] = LHALF(t);
 237                         t = (B - HHALF(t)) & (B - 1);
 238                 }
 239                 t = u[j] - t;
 240                 u[j] = LHALF(t);
 241                 /*
 242                  * D5: test remainder.
 243                  * There is a borrow if and only if HHALF(t) is nonzero;
 244                  * in that (rare) case, qhat was too large (by exactly 1).
 245                  * Fix it by adding v[1..n] to u[j..j+n].
 246                  */
 247                 if (HHALF(t)) {
 248                         qhat--;
 249                         for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
 250                                 t += u[i + j] + v[i];
 251                                 u[i + j] = LHALF(t);
 252                                 t = HHALF(t);
 253                         }
 254                         u[j] = LHALF(u[j] + t);
 255                 }
 256                 q[j] = qhat;
 257         } while (++j <= m);             /* D7: loop on j. */
 258
 259         /*
 260          * If caller wants the remainder, we have to calculate it as
 261          * u[m..m+n] >> d (this is at most n digits and thus fits in
 262          * u[m+1..m+n], but we may need more source digits).
 263          */
 264         if (arq) {
 265                 if (d) {
 266                         for (i = m + n; i > m; --i)
 267                                 u[i] = (u[i] >> d) |
 268                                     LHALF(u[i - 1] << (HALF_BITS - d));
 269                         u[i] = 0;
 270                 }
 271                 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
 272                 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
 273                 *arq = tmp.q;
 274         }
 275
 276         tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
 277         tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
 278         return (tmp.q);
 279 }
 280
 281 /*
 282  * Divide two unsigned quads.
 283  */
 284
 285 u_quad_t
 286 __udivdi3(u_quad_t a, u_quad_t b)
 287 {
 288
 289         return (__udivmoddi4(a, b, (u_quad_t *)0));
 290 }
 291
 292 /*
 293  * Return remainder after dividing two unsigned quads.
 294  */
 295 u_quad_t
 296 __umoddi3(u_quad_t a, u_quad_t b)
 297 {
 298         u_quad_t r;
 299
 300         (void)__udivmoddi4(a, b, &r);
 301         return (r);
 302 }