1 /* $NetBSD: qdivrem.c,v 1.1 2005/12/20 19:28:51 christos 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 1.1 2005/12/20 19:28:51 christos 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(uq, vq, arq) 76 u_quad_t uq, vq, *arq; 77 { 78 union uu tmp; 79 digit *u, *v, *q; 80 digit v1, v2; 81 u_int qhat, rhat, t; 82 int m, n, d, j, i; 83 digit uspace[5], vspace[5], qspace[5]; 84 85 /* 86 * Take care of special cases: divide by zero, and u < v. 87 */ 88 if (vq == 0) { 89 /* divide by zero. */ 90 static volatile const unsigned int zero = 0; 91 92 tmp.ul[H] = tmp.ul[L] = 1 / zero; 93 if (arq) 94 *arq = uq; 95 return (tmp.q); 96 } 97 if (uq < vq) { 98 if (arq) 99 *arq = uq; 100 return (0); 101 } 102 u = &uspace[0]; 103 v = &vspace[0]; 104 q = &qspace[0]; 105 106 /* 107 * Break dividend and divisor into digits in base B, then 108 * count leading zeros to determine m and n. When done, we 109 * will have: 110 * u = (u[1]u[2]...u[m+n]) sub B 111 * v = (v[1]v[2]...v[n]) sub B 112 * v[1] != 0 113 * 1 < n <= 4 (if n = 1, we use a different division algorithm) 114 * m >= 0 (otherwise u < v, which we already checked) 115 * m + n = 4 116 * and thus 117 * m = 4 - n <= 2 118 */ 119 tmp.uq = uq; 120 u[0] = 0; 121 u[1] = (digit)HHALF(tmp.ul[H]); 122 u[2] = (digit)LHALF(tmp.ul[H]); 123 u[3] = (digit)HHALF(tmp.ul[L]); 124 u[4] = (digit)LHALF(tmp.ul[L]); 125 tmp.uq = vq; 126 v[1] = (digit)HHALF(tmp.ul[H]); 127 v[2] = (digit)LHALF(tmp.ul[H]); 128 v[3] = (digit)HHALF(tmp.ul[L]); 129 v[4] = (digit)LHALF(tmp.ul[L]); 130 for (n = 4; v[1] == 0; v++) { 131 if (--n == 1) { 132 u_int rbj; /* r*B+u[j] (not root boy jim) */ 133 digit q1, q2, q3, q4; 134 135 /* 136 * Change of plan, per exercise 16. 137 * r = 0; 138 * for j = 1..4: 139 * q[j] = floor((r*B + u[j]) / v), 140 * r = (r*B + u[j]) % v; 141 * We unroll this completely here. 142 */ 143 t = v[2]; /* nonzero, by definition */ 144 q1 = (digit)(u[1] / t); 145 rbj = COMBINE(u[1] % t, u[2]); 146 q2 = (digit)(rbj / t); 147 rbj = COMBINE(rbj % t, u[3]); 148 q3 = (digit)(rbj / t); 149 rbj = COMBINE(rbj % t, u[4]); 150 q4 = (digit)(rbj / t); 151 if (arq) 152 *arq = rbj % t; 153 tmp.ul[H] = COMBINE(q1, q2); 154 tmp.ul[L] = COMBINE(q3, q4); 155 return (tmp.q); 156 } 157 } 158 159 /* 160 * By adjusting q once we determine m, we can guarantee that 161 * there is a complete four-digit quotient at &qspace[1] when 162 * we finally stop. 163 */ 164 for (m = 4 - n; u[1] == 0; u++) 165 m--; 166 for (i = 4 - m; --i >= 0;) 167 q[i] = 0; 168 q += 4 - m; 169 170 /* 171 * Here we run Program D, translated from MIX to C and acquiring 172 * a few minor changes. 173 * 174 * D1: choose multiplier 1 << d to ensure v[1] >= B/2. 175 */ 176 d = 0; 177 for (t = v[1]; t < B / 2; t <<= 1) 178 d++; 179 if (d > 0) { 180 shl(&u[0], m + n, d); /* u <<= d */ 181 shl(&v[1], n - 1, d); /* v <<= d */ 182 } 183 /* 184 * D2: j = 0. 185 */ 186 j = 0; 187 v1 = v[1]; /* for D3 -- note that v[1..n] are constant */ 188 v2 = v[2]; /* for D3 */ 189 do { 190 digit uj0, uj1, uj2; 191 192 /* 193 * D3: Calculate qhat (\^q, in TeX notation). 194 * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and 195 * let rhat = (u[j]*B + u[j+1]) mod v[1]. 196 * While rhat < B and v[2]*qhat > rhat*B+u[j+2], 197 * decrement qhat and increase rhat correspondingly. 198 * Note that if rhat >= B, v[2]*qhat < rhat*B. 199 */ 200 uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */ 201 uj1 = u[j + 1]; /* for D3 only */ 202 uj2 = u[j + 2]; /* for D3 only */ 203 if (uj0 == v1) { 204 qhat = B; 205 rhat = uj1; 206 goto qhat_too_big; 207 } else { 208 u_int nn = COMBINE(uj0, uj1); 209 qhat = nn / v1; 210 rhat = nn % v1; 211 } 212 while (v2 * qhat > COMBINE(rhat, uj2)) { 213 qhat_too_big: 214 qhat--; 215 if ((rhat += v1) >= B) 216 break; 217 } 218 /* 219 * D4: Multiply and subtract. 220 * The variable `t' holds any borrows across the loop. 221 * We split this up so that we do not require v[0] = 0, 222 * and to eliminate a final special case. 223 */ 224 for (t = 0, i = n; i > 0; i--) { 225 t = u[i + j] - v[i] * qhat - t; 226 u[i + j] = (digit)LHALF(t); 227 t = (B - HHALF(t)) & (B - 1); 228 } 229 t = u[j] - t; 230 u[j] = (digit)LHALF(t); 231 /* 232 * D5: test remainder. 233 * There is a borrow if and only if HHALF(t) is nonzero; 234 * in that (rare) case, qhat was too large (by exactly 1). 235 * Fix it by adding v[1..n] to u[j..j+n]. 236 */ 237 if (HHALF(t)) { 238 qhat--; 239 for (t = 0, i = n; i > 0; i--) { /* D6: add back. */ 240 t += u[i + j] + v[i]; 241 u[i + j] = (digit)LHALF(t); 242 t = HHALF(t); 243 } 244 u[j] = (digit)LHALF(u[j] + t); 245 } 246 q[j] = (digit)qhat; 247 } while (++j <= m); /* D7: loop on j. */ 248 249 /* 250 * If caller wants the remainder, we have to calculate it as 251 * u[m..m+n] >> d (this is at most n digits and thus fits in 252 * u[m+1..m+n], but we may need more source digits). 253 */ 254 if (arq) { 255 if (d) { 256 for (i = m + n; i > m; --i) 257 u[i] = (digit)(((u_int)u[i] >> d) | 258 LHALF((u_int)u[i - 1] << (HALF_BITS - d))); 259 u[i] = 0; 260 } 261 tmp.ul[H] = COMBINE(uspace[1], uspace[2]); 262 tmp.ul[L] = COMBINE(uspace[3], uspace[4]); 263 *arq = tmp.q; 264 } 265 266 tmp.ul[H] = COMBINE(qspace[1], qspace[2]); 267 tmp.ul[L] = COMBINE(qspace[3], qspace[4]); 268 return (tmp.q); 269 } 270 271 /* 272 * Shift p[0]..p[len] left `sh' bits, ignoring any bits that 273 * `fall out' the left (there never will be any such anyway). 274 * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS. 275 */ 276 static void 277 shl(digit *p, int len, int sh) 278 { 279 int i; 280 281 for (i = 0; i < len; i++) 282 p[i] = (digit)(LHALF((u_int)p[i] << sh) | 283 ((u_int)p[i + 1] >> (HALF_BITS - sh))); 284 p[i] = (digit)(LHALF((u_int)p[i] << sh)); 285 } 286