/* $OpenBSD: qdivrem.c,v 1.5 2003/06/02 23:28:08 millert Exp $ */ /* $NetBSD: qdivrem.c,v 1.5 1995/10/07 09:26:40 mycroft Exp $ */ /*- * Copyright (c) 1992, 1993 * The Regents of the University of California. All rights reserved. * * This software was developed by the Computer Systems Engineering group * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and * contributed to Berkeley. * * Redistribution and use in source and binary forms, with or without * modification, are permitted provided that the following conditions * are met: * 1. Redistributions of source code must retain the above copyright * notice, this list of conditions and the following disclaimer. * 2. Redistributions in binary form must reproduce the above copyright * notice, this list of conditions and the following disclaimer in the * documentation and/or other materials provided with the distribution. * 3. Neither the name of the University nor the names of its contributors * may be used to endorse or promote products derived from this software * without specific prior written permission. * * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF * SUCH DAMAGE. */ #if defined(LIBC_SCCS) && !defined(lint) #ifdef notdef static char sccsid[] = "@(#)qdivrem.c 8.1 (Berkeley) 6/4/93"; #endif static char rcsid[] = "$OpenBSD: qdivrem.c,v 1.5 2003/06/02 23:28:08 millert Exp $"; #endif /* LIBC_SCCS and not lint */ /* * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed), * section 4.3.1, pp. 257--259. */ #include "quad.h" #define B ((long)1 << HALF_BITS) /* digit base */ /* Combine two `digits' to make a single two-digit number. */ #define COMBINE(a, b) (((u_long)(a) << HALF_BITS) | (b)) /* select a type for digits in base B: use unsigned short if they fit */ #if ULONG_MAX == 0xffffffff && USHRT_MAX >= 0xffff typedef unsigned short digit; #else typedef u_long digit; #endif static void shl(digit *p, int len, int sh); /* * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v. * * We do this in base 2-sup-HALF_BITS, so that all intermediate products * fit within u_long. As a consequence, the maximum length dividend and * divisor are 4 `digits' in this base (they are shorter if they have * leading zeros). */ u_quad_t __qdivrem(uq, vq, arq) u_quad_t uq, vq, *arq; { union uu tmp; digit *u, *v, *q; register digit v1, v2; u_long qhat, rhat, t; int m, n, d, j, i; digit uspace[5], vspace[5], qspace[5]; /* * Take care of special cases: divide by zero, and u < v. */ if (vq == 0) { /* divide by zero. */ static volatile const unsigned int zero = 0; tmp.ul[H] = tmp.ul[L] = 1 / zero; if (arq) *arq = uq; return (tmp.q); } if (uq < vq) { if (arq) *arq = uq; return (0); } u = &uspace[0]; v = &vspace[0]; q = &qspace[0]; /* * Break dividend and divisor into digits in base B, then * count leading zeros to determine m and n. When done, we * will have: * u = (u[1]u[2]...u[m+n]) sub B * v = (v[1]v[2]...v[n]) sub B * v[1] != 0 * 1 < n <= 4 (if n = 1, we use a different division algorithm) * m >= 0 (otherwise u < v, which we already checked) * m + n = 4 * and thus * m = 4 - n <= 2 */ tmp.uq = uq; u[0] = 0; u[1] = HHALF(tmp.ul[H]); u[2] = LHALF(tmp.ul[H]); u[3] = HHALF(tmp.ul[L]); u[4] = LHALF(tmp.ul[L]); tmp.uq = vq; v[1] = HHALF(tmp.ul[H]); v[2] = LHALF(tmp.ul[H]); v[3] = HHALF(tmp.ul[L]); v[4] = LHALF(tmp.ul[L]); for (n = 4; v[1] == 0; v++) { if (--n == 1) { u_long rbj; /* r*B+u[j] (not root boy jim) */ digit q1, q2, q3, q4; /* * Change of plan, per exercise 16. * r = 0; * for j = 1..4: * q[j] = floor((r*B + u[j]) / v), * r = (r*B + u[j]) % v; * We unroll this completely here. */ t = v[2]; /* nonzero, by definition */ q1 = u[1] / t; rbj = COMBINE(u[1] % t, u[2]); q2 = rbj / t; rbj = COMBINE(rbj % t, u[3]); q3 = rbj / t; rbj = COMBINE(rbj % t, u[4]); q4 = rbj / t; if (arq) *arq = rbj % t; tmp.ul[H] = COMBINE(q1, q2); tmp.ul[L] = COMBINE(q3, q4); return (tmp.q); } } /* * By adjusting q once we determine m, we can guarantee that * there is a complete four-digit quotient at &qspace[1] when * we finally stop. */ for (m = 4 - n; u[1] == 0; u++) m--; for (i = 4 - m; --i >= 0;) q[i] = 0; q += 4 - m; /* * Here we run Program D, translated from MIX to C and acquiring * a few minor changes. * * D1: choose multiplier 1 << d to ensure v[1] >= B/2. */ d = 0; for (t = v[1]; t < B / 2; t <<= 1) d++; if (d > 0) { shl(&u[0], m + n, d); /* u <<= d */ shl(&v[1], n - 1, d); /* v <<= d */ } /* * D2: j = 0. */ j = 0; v1 = v[1]; /* for D3 -- note that v[1..n] are constant */ v2 = v[2]; /* for D3 */ do { register digit uj0, uj1, uj2; /* * D3: Calculate qhat (\^q, in TeX notation). * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and * let rhat = (u[j]*B + u[j+1]) mod v[1]. * While rhat < B and v[2]*qhat > rhat*B+u[j+2], * decrement qhat and increase rhat correspondingly. * Note that if rhat >= B, v[2]*qhat < rhat*B. */ uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */ uj1 = u[j + 1]; /* for D3 only */ uj2 = u[j + 2]; /* for D3 only */ if (uj0 == v1) { qhat = B; rhat = uj1; goto qhat_too_big; } else { u_long n = COMBINE(uj0, uj1); qhat = n / v1; rhat = n % v1; } while (v2 * qhat > COMBINE(rhat, uj2)) { qhat_too_big: qhat--; if ((rhat += v1) >= B) break; } /* * D4: Multiply and subtract. * The variable `t' holds any borrows across the loop. * We split this up so that we do not require v[0] = 0, * and to eliminate a final special case. */ for (t = 0, i = n; i > 0; i--) { t = u[i + j] - v[i] * qhat - t; u[i + j] = LHALF(t); t = (B - HHALF(t)) & (B - 1); } t = u[j] - t; u[j] = LHALF(t); /* * D5: test remainder. * There is a borrow if and only if HHALF(t) is nonzero; * in that (rare) case, qhat was too large (by exactly 1). * Fix it by adding v[1..n] to u[j..j+n]. */ if (HHALF(t)) { qhat--; for (t = 0, i = n; i > 0; i--) { /* D6: add back. */ t += u[i + j] + v[i]; u[i + j] = LHALF(t); t = HHALF(t); } u[j] = LHALF(u[j] + t); } q[j] = qhat; } while (++j <= m); /* D7: loop on j. */ /* * If caller wants the remainder, we have to calculate it as * u[m..m+n] >> d (this is at most n digits and thus fits in * u[m+1..m+n], but we may need more source digits). */ if (arq) { if (d) { for (i = m + n; i > m; --i) u[i] = (u[i] >> d) | LHALF(u[i - 1] << (HALF_BITS - d)); u[i] = 0; } tmp.ul[H] = COMBINE(uspace[1], uspace[2]); tmp.ul[L] = COMBINE(uspace[3], uspace[4]); *arq = tmp.q; } tmp.ul[H] = COMBINE(qspace[1], qspace[2]); tmp.ul[L] = COMBINE(qspace[3], qspace[4]); return (tmp.q); } /* * Shift p[0]..p[len] left `sh' bits, ignoring any bits that * `fall out' the left (there never will be any such anyway). * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS. */ static void shl(p, len, sh) register digit *p; register int len; register int sh; { register int i; for (i = 0; i < len; i++) p[i] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh)); p[i] = LHALF(p[i] << sh); }