qdivrem.c

/*-
 * 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.
 *
 * From:
 *      @(#)qdivrem.c   8.1 (Berkeley) 6/4/93
 *      NetBSD: qdivrem.c,v 1.1 2005/12/20 19:28:51 christos Exp
 */

/*
 * Multiprecision divide.  This algorithm is from Knuth vol. 2 (2nd ed),
 * section 4.3.1, pp. 257--259.
 */

#include "longlong.h"

#define B       ((int)1 << HALF_BITS)   /* digit base */

/* Combine two `digits' to make a single two-digit number. */
#define COMBINE(a, b) (((unsigned int)(a) << HALF_BITS) | (b))

/* select a type for digits in base B: use unsigned short if they fit */
#if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
typedef unsigned short digit;
#else
typedef unsigned int 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 unsigned int.  As a consequence, the maximum
 * length dividend and divisor are 4 `digits' in this base (they are
 * shorter if they have leading zeros).
 */
unsigned long long
__qdivrem(unsigned long long ull, unsigned long long vll,
          unsigned long long *arq)
{
        union uu tmp;
        digit *u, *v, *q;
        digit v1, v2;
        unsigned int 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 (vll == 0) {
                /* divide by zero. */
                static volatile const unsigned int zero = 0;

                tmp.ui[H] = tmp.ui[L] = 1 / zero;
                if (arq)
                        *arq = ull;
                return (tmp.ll);
        }
        if (ull < vll) {
                if (arq)
                        *arq = ull;
                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.ull = ull;
        u[0] = 0;
        u[1] = (digit)HHALF(tmp.ui[H]);
        u[2] = (digit)LHALF(tmp.ui[H]);
        u[3] = (digit)HHALF(tmp.ui[L]);
        u[4] = (digit)LHALF(tmp.ui[L]);
        tmp.ull = vll;
        v[1] = (digit)HHALF(tmp.ui[H]);
        v[2] = (digit)LHALF(tmp.ui[H]);
        v[3] = (digit)HHALF(tmp.ui[L]);
        v[4] = (digit)LHALF(tmp.ui[L]);
        for (n = 4; v[1] == 0; v++) {
                if (--n == 1) {
                        unsigned int 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 = (digit)(u[1] / t);
                        rbj = COMBINE(u[1] % t, u[2]);
                        q2 = (digit)(rbj / t);
                        rbj = COMBINE(rbj % t, u[3]);
                        q3 = (digit)(rbj / t);
                        rbj = COMBINE(rbj % t, u[4]);
                        q4 = (digit)(rbj / t);
                        if (arq)
                                *arq = rbj % t;
                        tmp.ui[H] = COMBINE(q1, q2);
                        tmp.ui[L] = COMBINE(q3, q4);
                        return (tmp.ll);
                }
        }

        /*
         * 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 {
                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 {
                        unsigned int nn = COMBINE(uj0, uj1);
                        qhat = nn / v1;
                        rhat = nn % 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] = (digit)LHALF(t);
                        t = (B - HHALF(t)) & (B - 1);
                }
                t = u[j] - t;
                u[j] = (digit)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] = (digit)LHALF(t);
                                t = HHALF(t);
                        }
                        u[j] = (digit)LHALF(u[j] + t);
                }
                q[j] = (digit)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] = (digit)(((unsigned int)u[i] >> d) |
                                    LHALF((unsigned int)u[i - 1] <<
                                          (HALF_BITS - d)));
                        u[i] = 0;
                }
                tmp.ui[H] = COMBINE(uspace[1], uspace[2]);
                tmp.ui[L] = COMBINE(uspace[3], uspace[4]);
                *arq = tmp.ll;
        }

        tmp.ui[H] = COMBINE(qspace[1], qspace[2]);
        tmp.ui[L] = COMBINE(qspace[3], qspace[4]);
        return (tmp.ll);
}

/*
 * 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(digit *p, int len, int sh)
{
        int i;

        for (i = 0; i < len; i++)
                p[i] = (digit)(LHALF((unsigned int)p[i] << sh) |
                    ((unsigned int)p[i + 1] >> (HALF_BITS - sh)));
        p[i] = (digit)(LHALF((unsigned int)p[i] << sh));
}