280 lines
7.8 KiB
C
280 lines
7.8 KiB
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. All advertising materials mentioning features or use of this software
|
|
* must display the following acknowledgement:
|
|
* This product includes software developed by the University of
|
|
* California, Berkeley and its contributors.
|
|
* 4. 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 */
|
|
|
|
/*
|
|
* Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
|
|
* section 4.3.1, pp. 257--259.
|
|
*/
|
|
|
|
#include "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(register digit *p, register int len, register int sh)
|
|
{
|
|
register 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(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);
|
|
}
|