#include <quad.h>

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);
}

