usr/src/boot/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 #include <sys/stddef.h>
  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_int)(a) << HALF_BITS) | (b))
  50
  51 _Static_assert(sizeof(int) / 2 == sizeof(short),
  52         "Bitwise functions in libstand are broken on this architecture\n");
  53
  54 /* select a type for digits in base B: use unsigned short if they fit */
  55 typedef unsigned short digit;
  56
  57 /*
  58  * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
  59  * `fall out' the left (there never will be any such anyway).
  60  * We may assume len >= 0.  NOTE THAT THIS WRITES len+1 DIGITS.
  61  */
  62 static void
  63 shl(digit *p, int len, int sh)
  64 {
  65         int i;
  66
  67         for (i = 0; i < len; i++)
  68                 p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
  69         p[i] = LHALF(p[i] << sh);
  70 }
  71
  72 /*
  73  * __udivmoddi4(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
  74  *
  75  * We do this in base 2-sup-HALF_BITS, so that all intermediate products
  76  * fit within u_int.  As a consequence, the maximum length dividend and
  77  * divisor are 4 `digits' in this base (they are shorter if they have
  78  * leading zeros).
  79  */
  80 u_quad_t
  81 __udivmoddi4(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
  82 {
  83         union uu tmp;
  84         digit *u, *v, *q;
  85         digit v1, v2;
  86         u_int qhat, rhat, t;
  87         int m, n, d, j, i;
  88         digit uspace[5], vspace[5], qspace[5];
  89
  90         /*
  91          * Take care of special cases: divide by zero, and u < v.
  92          */
  93         if (vq == 0) {
  94                 /* divide by zero. */
  95                 static volatile const unsigned int zero = 0;
  96
  97                 tmp.ul[H] = tmp.ul[L] = 1 / zero;
  98                 if (arq)
  99                         *arq = uq;
 100                 return (tmp.q);
 101         }
 102         if (uq < vq) {
 103                 if (arq)
 104                         *arq = uq;
 105                 return (0);
 106         }
 107         u = &uspace[0];
 108         v = &vspace[0];
 109         q = &qspace[0];
 110
 111         /*
 112          * Break dividend and divisor into digits in base B, then
 113          * count leading zeros to determine m and n.  When done, we
 114          * will have:
 115          *      u = (u[1]u[2]...u[m+n]) sub B
 116          *      v = (v[1]v[2]...v[n]) sub B
 117          *      v[1] != 0
 118          *      1 < n <= 4 (if n = 1, we use a different division algorithm)
 119          *      m >= 0 (otherwise u < v, which we already checked)
 120          *      m + n = 4
 121          * and thus
 122          *      m = 4 - n <= 2
 123          */
 124         tmp.uq = uq;
 125         u[0] = 0;
 126         u[1] = HHALF(tmp.ul[H]);
 127         u[2] = LHALF(tmp.ul[H]);
 128         u[3] = HHALF(tmp.ul[L]);
 129         u[4] = LHALF(tmp.ul[L]);
 130         tmp.uq = vq;
 131         v[1] = HHALF(tmp.ul[H]);
 132         v[2] = LHALF(tmp.ul[H]);
 133         v[3] = HHALF(tmp.ul[L]);
 134         v[4] = LHALF(tmp.ul[L]);
 135         for (n = 4; v[1] == 0; v++) {
 136                 if (--n == 1) {
 137                         u_int rbj;      /* r*B+u[j] (not root boy jim) */
 138                         digit q1, q2, q3, q4;
 139
 140                         /*
 141                          * Change of plan, per exercise 16.
 142                          *      r = 0;
 143                          *      for j = 1..4:
 144                          *              q[j] = floor((r*B + u[j]) / v),
 145                          *              r = (r*B + u[j]) % v;
 146                          * We unroll this completely here.
 147                          */
 148                         t = v[2];       /* nonzero, by definition */
 149                         q1 = u[1] / t;
 150                         rbj = COMBINE(u[1] % t, u[2]);
 151                         q2 = rbj / t;
 152                         rbj = COMBINE(rbj % t, u[3]);
 153                         q3 = rbj / t;
 154                         rbj = COMBINE(rbj % t, u[4]);
 155                         q4 = rbj / t;
 156                         if (arq)
 157                                 *arq = rbj % t;
 158                         tmp.ul[H] = COMBINE(q1, q2);
 159                         tmp.ul[L] = COMBINE(q3, q4);
 160                         return (tmp.q);
 161                 }
 162         }
 163
 164         /*
 165          * By adjusting q once we determine m, we can guarantee that
 166          * there is a complete four-digit quotient at &qspace[1] when
 167          * we finally stop.
 168          */
 169         for (m = 4 - n; u[1] == 0; u++)
 170                 m--;
 171         for (i = 4 - m; --i >= 0;)
 172                 q[i] = 0;
 173         q += 4 - m;
 174
 175         /*
 176          * Here we run Program D, translated from MIX to C and acquiring
 177          * a few minor changes.
 178          *
 179          * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
 180          */
 181         d = 0;
 182         for (t = v[1]; t < B / 2; t <<= 1)
 183                 d++;
 184         if (d > 0) {
 185                 shl(&u[0], m + n, d);           /* u <<= d */
 186                 shl(&v[1], n - 1, d);           /* v <<= d */
 187         }
 188         /*
 189          * D2: j = 0.
 190          */
 191         j = 0;
 192         v1 = v[1];      /* for D3 -- note that v[1..n] are constant */
 193         v2 = v[2];      /* for D3 */
 194         do {
 195                 digit uj0, uj1, uj2;
 196
 197                 /*
 198                  * D3: Calculate qhat (\^q, in TeX notation).
 199                  * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
 200                  * let rhat = (u[j]*B + u[j+1]) mod v[1].
 201                  * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
 202                  * decrement qhat and increase rhat correspondingly.
 203                  * Note that if rhat >= B, v[2]*qhat < rhat*B.
 204                  */
 205                 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
 206                 uj1 = u[j + 1]; /* for D3 only */
 207                 uj2 = u[j + 2]; /* for D3 only */
 208                 if (uj0 == v1) {
 209                         qhat = B;
 210                         rhat = uj1;
 211                         goto qhat_too_big;
 212                 } else {
 213                         u_int nn = COMBINE(uj0, uj1);
 214                         qhat = nn / v1;
 215                         rhat = nn % v1;
 216                 }
 217                 while (v2 * qhat > COMBINE(rhat, uj2)) {
 218         qhat_too_big:
 219                         qhat--;
 220                         if ((rhat += v1) >= B)
 221                                 break;
 222                 }
 223                 /*
 224                  * D4: Multiply and subtract.
 225                  * The variable `t' holds any borrows across the loop.
 226                  * We split this up so that we do not require v[0] = 0,
 227                  * and to eliminate a final special case.
 228                  */
 229                 for (t = 0, i = n; i > 0; i--) {
 230                         t = u[i + j] - v[i] * qhat - t;
 231                         u[i + j] = LHALF(t);
 232                         t = (B - HHALF(t)) & (B - 1);
 233                 }
 234                 t = u[j] - t;
 235                 u[j] = LHALF(t);
 236                 /*
 237                  * D5: test remainder.
 238                  * There is a borrow if and only if HHALF(t) is nonzero;
 239                  * in that (rare) case, qhat was too large (by exactly 1).
 240                  * Fix it by adding v[1..n] to u[j..j+n].
 241                  */
 242                 if (HHALF(t)) {
 243                         qhat--;
 244                         for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
 245                                 t += u[i + j] + v[i];
 246                                 u[i + j] = LHALF(t);
 247                                 t = HHALF(t);
 248                         }
 249                         u[j] = LHALF(u[j] + t);
 250                 }
 251                 q[j] = qhat;
 252         } while (++j <= m);             /* D7: loop on j. */
 253
 254         /*
 255          * If caller wants the remainder, we have to calculate it as
 256          * u[m..m+n] >> d (this is at most n digits and thus fits in
 257          * u[m+1..m+n], but we may need more source digits).
 258          */
 259         if (arq) {
 260                 if (d) {
 261                         for (i = m + n; i > m; --i)
 262                                 u[i] = (u[i] >> d) |
 263                                     LHALF(u[i - 1] << (HALF_BITS - d));
 264                         u[i] = 0;
 265                 }
 266                 tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
 267                 tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
 268                 *arq = tmp.q;
 269         }
 270
 271         tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
 272         tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
 273         return (tmp.q);
 274 }
 275
 276 /*
 277  * Divide two unsigned quads.
 278  */
 279
 280 u_quad_t
 281 __udivdi3(u_quad_t a, u_quad_t b)
 282 {
 283
 284         return (__udivmoddi4(a, b, NULL));
 285 }
 286
 287 /*
 288  * Return remainder after dividing two unsigned quads.
 289  */
 290 u_quad_t
 291 __umoddi3(u_quad_t a, u_quad_t b)
 292 {
 293         u_quad_t r;
 294
 295         (void)__udivmoddi4(a, b, &r);
 296         return (r);
 297 }
 298
 299 /*
 300  * Divide two signed quads.
 301  * ??? if -1/2 should produce -1 on this machine, this code is wrong
 302  */
 303 quad_t
 304 __divdi3(quad_t a, quad_t b)
 305 {
 306         u_quad_t ua, ub, uq;
 307         int neg;
 308
 309         if (a < 0)
 310                 ua = -(u_quad_t)a, neg = 1;
 311         else
 312                 ua = a, neg = 0;
 313         if (b < 0)
 314                 ub = -(u_quad_t)b, neg ^= 1;
 315         else
 316                 ub = b;
 317         uq = __udivmoddi4(ua, ub, NULL);
 318         return (neg ? -uq : uq);
 319 }
 320
 321 /*
 322  * Return remainder after dividing two signed quads.
 323  *
 324  * XXX
 325  * If -1/2 should produce -1 on this machine, this code is wrong.
 326  */
 327 quad_t
 328 __moddi3(quad_t a, quad_t b)
 329 {
 330         u_quad_t ua, ub, ur;
 331         int neg;
 332
 333         if (a < 0)
 334                 ua = -(u_quad_t)a, neg = 1;
 335         else
 336                 ua = a, neg = 0;
 337         if (b < 0)
 338                 ub = -(u_quad_t)b;
 339         else
 340                 ub = b;
 341         (void)__udivmoddi4(ua, ub, &ur);
 342         return (neg ? -ur : ur);
 343 }
 344
 345 quad_t
 346 __divmoddi4(quad_t a, quad_t b, quad_t *r)
 347 {
 348         quad_t d;
 349
 350         d = __divdi3(a, b);
 351         *r = a - (b * d);
 352
 353         return (d);
 354 }