| 1 | /* $NetBSD: qdivrem.c,v 1.12 2005/12/11 12:24:37 christos Exp $ */
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| 2 |
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| 3 | /*-
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| 4 | * Copyright (c) 1992, 1993
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| 5 | * The Regents of the University of California. All rights reserved.
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| 6 | *
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| 7 | * This software was developed by the Computer Systems Engineering group
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| 8 | * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
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| 9 | * contributed to Berkeley.
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| 10 | *
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| 11 | * Redistribution and use in source and binary forms, with or without
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| 12 | * modification, are permitted provided that the following conditions
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| 13 | * are met:
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| 14 | * 1. Redistributions of source code must retain the above copyright
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| 15 | * notice, this list of conditions and the following disclaimer.
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| 16 | * 2. Redistributions in binary form must reproduce the above copyright
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| 17 | * notice, this list of conditions and the following disclaimer in the
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| 18 | * documentation and/or other materials provided with the distribution.
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| 19 | * 3. Neither the name of the University nor the names of its contributors
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| 20 | * may be used to endorse or promote products derived from this software
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| 21 | * without specific prior written permission.
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| 22 | *
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| 23 | * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
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| 24 | * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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| 25 | * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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| 26 | * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
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| 27 | * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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| 28 | * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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| 29 | * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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| 30 | * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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| 31 | * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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| 32 | * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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| 33 | * SUCH DAMAGE.
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| 34 | */
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| 35 |
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| 36 | /*#include <sys/cdefs.h>
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| 37 | #if defined(LIBC_SCCS) && !defined(lint)
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| 38 | #if 0
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| 39 | static char sccsid[] = "@(#)qdivrem.c 8.1 (Berkeley) 6/4/93";
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| 40 | #else
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| 41 | __RCSID("$NetBSD: qdivrem.c,v 1.12 2005/12/11 12:24:37 christos Exp $");
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| 42 | #endif
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| 43 | #endif*/ /* LIBC_SCCS and not lint */
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| 44 |
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| 45 | /*
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| 46 | * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
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| 47 | * section 4.3.1, pp. 257--259.
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| 48 | */
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| 49 |
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| 50 | #include "quad.h"
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| 51 |
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| 52 | #define B ((int)1 << HALF_BITS) /* digit base */
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| 53 |
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| 54 | /* Combine two `digits' to make a single two-digit number. */
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| 55 | #define COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
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| 56 |
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| 57 | /* select a type for digits in base B: use unsigned short if they fit */
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| 58 | #if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
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| 59 | typedef unsigned short digit;
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| 60 | #else
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| 61 | typedef u_int digit;
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| 62 | #endif
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| 63 |
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| 64 | static void shl __P((digit *p, int len, int sh));
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| 65 |
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| 66 | /*
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| 67 | * __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
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| 68 | *
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| 69 | * We do this in base 2-sup-HALF_BITS, so that all intermediate products
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| 70 | * fit within u_int. As a consequence, the maximum length dividend and
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| 71 | * divisor are 4 `digits' in this base (they are shorter if they have
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| 72 | * leading zeros).
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| 73 | */
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| 74 | u_quad_t
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| 75 | __qdivrem(uq, vq, arq)
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| 76 | u_quad_t uq, vq, *arq;
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| 77 | {
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| 78 | union uu tmp;
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| 79 | digit *u, *v, *q;
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| 80 | digit v1, v2;
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| 81 | u_int qhat, rhat, t;
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| 82 | int m, n, d, j, i;
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| 83 | digit uspace[5], vspace[5], qspace[5];
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| 84 |
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| 85 | /*
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| 86 | * Take care of special cases: divide by zero, and u < v.
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| 87 | */
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| 88 | if (vq == 0) {
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| 89 | /* divide by zero. */
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| 90 | static volatile const unsigned int zero = 0;
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| 91 |
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| 92 | tmp.ul[H] = tmp.ul[L] = 1 / zero;
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| 93 | if (arq)
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| 94 | *arq = uq;
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| 95 | return (tmp.q);
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| 96 | }
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| 97 | if (uq < vq) {
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| 98 | if (arq)
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| 99 | *arq = uq;
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| 100 | return (0);
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| 101 | }
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| 102 | u = &uspace[0];
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| 103 | v = &vspace[0];
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| 104 | q = &qspace[0];
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| 105 |
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| 106 | /*
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| 107 | * Break dividend and divisor into digits in base B, then
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| 108 | * count leading zeros to determine m and n. When done, we
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| 109 | * will have:
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| 110 | * u = (u[1]u[2]...u[m+n]) sub B
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| 111 | * v = (v[1]v[2]...v[n]) sub B
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| 112 | * v[1] != 0
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| 113 | * 1 < n <= 4 (if n = 1, we use a different division algorithm)
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| 114 | * m >= 0 (otherwise u < v, which we already checked)
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| 115 | * m + n = 4
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| 116 | * and thus
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| 117 | * m = 4 - n <= 2
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| 118 | */
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| 119 | tmp.uq = uq;
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| 120 | u[0] = 0;
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| 121 | u[1] = (digit)HHALF(tmp.ul[H]);
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| 122 | u[2] = (digit)LHALF(tmp.ul[H]);
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| 123 | u[3] = (digit)HHALF(tmp.ul[L]);
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| 124 | u[4] = (digit)LHALF(tmp.ul[L]);
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| 125 | tmp.uq = vq;
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| 126 | v[1] = (digit)HHALF(tmp.ul[H]);
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| 127 | v[2] = (digit)LHALF(tmp.ul[H]);
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| 128 | v[3] = (digit)HHALF(tmp.ul[L]);
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| 129 | v[4] = (digit)LHALF(tmp.ul[L]);
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| 130 | for (n = 4; v[1] == 0; v++) {
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| 131 | if (--n == 1) {
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| 132 | u_int rbj; /* r*B+u[j] (not root boy jim) */
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| 133 | digit q1, q2, q3, q4;
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| 134 |
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| 135 | /*
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| 136 | * Change of plan, per exercise 16.
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| 137 | * r = 0;
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| 138 | * for j = 1..4:
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| 139 | * q[j] = floor((r*B + u[j]) / v),
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| 140 | * r = (r*B + u[j]) % v;
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| 141 | * We unroll this completely here.
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| 142 | */
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| 143 | t = v[2]; /* nonzero, by definition */
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| 144 | q1 = (digit)(u[1] / t);
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| 145 | rbj = COMBINE(u[1] % t, u[2]);
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| 146 | q2 = (digit)(rbj / t);
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| 147 | rbj = COMBINE(rbj % t, u[3]);
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| 148 | q3 = (digit)(rbj / t);
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| 149 | rbj = COMBINE(rbj % t, u[4]);
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| 150 | q4 = (digit)(rbj / t);
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| 151 | if (arq)
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| 152 | *arq = rbj % t;
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| 153 | tmp.ul[H] = COMBINE(q1, q2);
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| 154 | tmp.ul[L] = COMBINE(q3, q4);
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| 155 | return (tmp.q);
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| 156 | }
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| 157 | }
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| 158 |
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| 159 | /*
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| 160 | * By adjusting q once we determine m, we can guarantee that
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| 161 | * there is a complete four-digit quotient at &qspace[1] when
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| 162 | * we finally stop.
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| 163 | */
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| 164 | for (m = 4 - n; u[1] == 0; u++)
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| 165 | m--;
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| 166 | for (i = 4 - m; --i >= 0;)
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| 167 | q[i] = 0;
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| 168 | q += 4 - m;
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| 169 |
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| 170 | /*
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| 171 | * Here we run Program D, translated from MIX to C and acquiring
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| 172 | * a few minor changes.
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| 173 | *
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| 174 | * D1: choose multiplier 1 << d to ensure v[1] >= B/2.
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| 175 | */
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| 176 | d = 0;
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| 177 | for (t = v[1]; t < B / 2; t <<= 1)
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| 178 | d++;
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| 179 | if (d > 0) {
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| 180 | shl(&u[0], m + n, d); /* u <<= d */
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| 181 | shl(&v[1], n - 1, d); /* v <<= d */
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| 182 | }
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| 183 | /*
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| 184 | * D2: j = 0.
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| 185 | */
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| 186 | j = 0;
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| 187 | v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
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| 188 | v2 = v[2]; /* for D3 */
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| 189 | do {
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| 190 | digit uj0, uj1, uj2;
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| 191 |
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| 192 | /*
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| 193 | * D3: Calculate qhat (\^q, in TeX notation).
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| 194 | * Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
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| 195 | * let rhat = (u[j]*B + u[j+1]) mod v[1].
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| 196 | * While rhat < B and v[2]*qhat > rhat*B+u[j+2],
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| 197 | * decrement qhat and increase rhat correspondingly.
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| 198 | * Note that if rhat >= B, v[2]*qhat < rhat*B.
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| 199 | */
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| 200 | uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
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| 201 | uj1 = u[j + 1]; /* for D3 only */
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| 202 | uj2 = u[j + 2]; /* for D3 only */
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| 203 | if (uj0 == v1) {
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| 204 | qhat = B;
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| 205 | rhat = uj1;
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| 206 | goto qhat_too_big;
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| 207 | } else {
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| 208 | u_int nn = COMBINE(uj0, uj1);
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| 209 | qhat = nn / v1;
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| 210 | rhat = nn % v1;
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| 211 | }
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| 212 | while (v2 * qhat > COMBINE(rhat, uj2)) {
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| 213 | qhat_too_big:
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| 214 | qhat--;
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| 215 | if ((rhat += v1) >= B)
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| 216 | break;
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| 217 | }
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| 218 | /*
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| 219 | * D4: Multiply and subtract.
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| 220 | * The variable `t' holds any borrows across the loop.
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| 221 | * We split this up so that we do not require v[0] = 0,
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| 222 | * and to eliminate a final special case.
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| 223 | */
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| 224 | for (t = 0, i = n; i > 0; i--) {
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| 225 | t = u[i + j] - v[i] * qhat - t;
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| 226 | u[i + j] = (digit)LHALF(t);
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| 227 | t = (B - HHALF(t)) & (B - 1);
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| 228 | }
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| 229 | t = u[j] - t;
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| 230 | u[j] = (digit)LHALF(t);
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| 231 | /*
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| 232 | * D5: test remainder.
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| 233 | * There is a borrow if and only if HHALF(t) is nonzero;
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| 234 | * in that (rare) case, qhat was too large (by exactly 1).
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| 235 | * Fix it by adding v[1..n] to u[j..j+n].
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| 236 | */
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| 237 | if (HHALF(t)) {
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| 238 | qhat--;
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| 239 | for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
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| 240 | t += u[i + j] + v[i];
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| 241 | u[i + j] = (digit)LHALF(t);
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| 242 | t = HHALF(t);
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| 243 | }
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| 244 | u[j] = (digit)LHALF(u[j] + t);
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| 245 | }
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| 246 | q[j] = (digit)qhat;
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| 247 | } while (++j <= m); /* D7: loop on j. */
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| 248 |
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| 249 | /*
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| 250 | * If caller wants the remainder, we have to calculate it as
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| 251 | * u[m..m+n] >> d (this is at most n digits and thus fits in
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| 252 | * u[m+1..m+n], but we may need more source digits).
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| 253 | */
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| 254 | if (arq) {
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| 255 | if (d) {
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| 256 | for (i = m + n; i > m; --i)
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| 257 | u[i] = (digit)(((u_int)u[i] >> d) |
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| 258 | LHALF((u_int)u[i - 1] << (HALF_BITS - d)));
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| 259 | u[i] = 0;
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| 260 | }
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| 261 | tmp.ul[H] = COMBINE(uspace[1], uspace[2]);
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| 262 | tmp.ul[L] = COMBINE(uspace[3], uspace[4]);
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| 263 | *arq = tmp.q;
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| 264 | }
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| 265 |
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| 266 | tmp.ul[H] = COMBINE(qspace[1], qspace[2]);
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| 267 | tmp.ul[L] = COMBINE(qspace[3], qspace[4]);
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| 268 | return (tmp.q);
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| 269 | }
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| 270 |
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| 271 | /*
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| 272 | * Shift p[0]..p[len] left `sh' bits, ignoring any bits that
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| 273 | * `fall out' the left (there never will be any such anyway).
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| 274 | * We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
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| 275 | */
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| 276 | static void
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| 277 | shl(digit *p, int len, int sh)
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| 278 | {
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| 279 | int i;
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| 280 |
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| 281 | for (i = 0; i < len; i++)
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| 282 | p[i] = (digit)(LHALF((u_int)p[i] << sh) |
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| 283 | ((u_int)p[i + 1] >> (HALF_BITS - sh)));
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| 284 | p[i] = (digit)(LHALF((u_int)p[i] << sh));
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| 285 | }
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