/*-
* 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.
*/
#include <sys/cdefs.h>
/*
* Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
* section 4.3.1, pp. 257--259.
*/
#include <libkern/quad.h>
#define B (1 << 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 (__predict_false(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 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] = 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);
}
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