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+/*-
+ * SPDX-License-Identifier: BSD-3-Clause
+ *
+ * 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)
+static char sccsid[] = "@(#)qdivrem.c 8.1 (Berkeley) 6/4/93";
+#endif /* LIBC_SCCS and not lint */
+#include <sys/cdefs.h>
+__FBSDID("$FreeBSD$");
+
+/*
+ * Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
+ * section 4.3.1, pp. 257--259.
+ */
+
+#include "quad.h"
+
+#define B (1L << 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
+
+/*
+ * 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] = LHALF(p[i] << sh) | (p[i + 1] >> (HALF_BITS - sh));
+ p[i] = LHALF(p[i] << 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(u_quad_t uq, u_quad_t vq, u_quad_t *arq)
+{
+ union uu tmp;
+ digit *u, *v, *q;
+ 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 {
+ 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);
+}