/* $OpenBSD: qdivrem.c,v 1.7 2005/08/08 08:05:35 espie 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.
 */

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

#include "quad.h"

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

/* Combine two `digits' to make a single two-digit number. */
#define COMBINE(a, b) (((u_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 u_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 u_int.  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(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
{
    union uu tmp;
    digit *u, *v, *q;
    digit v1, v2;
    u_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 (vq == 0) {
        /* divide by zero. */
        static volatile const unsigned int zero = 0;

        /* cppcheck-suppress zerodiv
         * Divission by zero is on purpose here */
        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] = (digit) HHALF(tmp.ul[H]);
    u[2] = (digit) LHALF(tmp.ul[H]);
    u[3] = (digit) HHALF(tmp.ul[L]);
    u[4] = (digit) LHALF(tmp.ul[L]);
    tmp.uq = vq;
    v[1] = (digit) HHALF(tmp.ul[H]);
    v[2] = (digit) LHALF(tmp.ul[H]);
    v[3] = (digit) HHALF(tmp.ul[L]);
    v[4] = (digit) LHALF(tmp.ul[L]);

    for (n = 4; v[1] == 0; v++) {
        if (--n == 1) {
            u_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.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 {
        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_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)(((u_int)u[i] >> d) | LHALF((u_int)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(digit *p, int len, int sh)
{
    int i;

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

    p[i] = (digit)(LHALF((u_int)p[i] << sh));
}