Initial Spring 2016 commit.
This commit is contained in:
Geoffrey Challen
279
common/gcc-millicode/qdivrem.c
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279
common/gcc-millicode/qdivrem.c
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/*-
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* Copyright (c) 1992, 1993
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* The Regents of the University of California. All rights reserved.
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*
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* This software was developed by the Computer Systems Engineering group
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* at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
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* contributed to Berkeley.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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* 3. Neither the name of the University nor the names of its contributors
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* may be used to endorse or promote products derived from this software
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* without specific prior written permission.
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*
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* THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
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* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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* ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
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* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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* SUCH DAMAGE.
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*
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* From:
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* @(#)qdivrem.c 8.1 (Berkeley) 6/4/93
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* NetBSD: qdivrem.c,v 1.1 2005/12/20 19:28:51 christos Exp
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*/
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/*
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* Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
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* section 4.3.1, pp. 257--259.
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*/
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#include "longlong.h"
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#define B ((int)1 << HALF_BITS) /* digit base */
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/* Combine two `digits' to make a single two-digit number. */
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#define COMBINE(a, b) (((unsigned int)(a) << HALF_BITS) | (b))
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/* select a type for digits in base B: use unsigned short if they fit */
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#if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
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typedef unsigned short digit;
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#else
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typedef unsigned int digit;
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#endif
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static void shl(digit *p, int len, int sh);
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/*
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* __qdivrem(u, v, rem) returns u/v and, optionally, sets *rem to u%v.
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*
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* We do this in base 2-sup-HALF_BITS, so that all intermediate
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* products fit within unsigned int. As a consequence, the maximum
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* length dividend and divisor are 4 `digits' in this base (they are
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* shorter if they have leading zeros).
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*/
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unsigned long long
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__qdivrem(unsigned long long ull, unsigned long long vll,
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unsigned long long *arq)
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{
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union uu tmp;
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digit *u, *v, *q;
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digit v1, v2;
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unsigned int qhat, rhat, t;
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int m, n, d, j, i;
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digit uspace[5], vspace[5], qspace[5];
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/*
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* Take care of special cases: divide by zero, and u < v.
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*/
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if (vll == 0) {
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/* divide by zero. */
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static volatile const unsigned int zero = 0;
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tmp.ui[H] = tmp.ui[L] = 1 / zero;
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if (arq)
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*arq = ull;
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return (tmp.ll);
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}
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if (ull < vll) {
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if (arq)
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*arq = ull;
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return (0);
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}
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u = &uspace[0];
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v = &vspace[0];
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q = &qspace[0];
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/*
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* Break dividend and divisor into digits in base B, then
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* count leading zeros to determine m and n. When done, we
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* will have:
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* u = (u[1]u[2]...u[m+n]) sub B
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* v = (v[1]v[2]...v[n]) sub B
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* v[1] != 0
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* 1 < n <= 4 (if n = 1, we use a different division algorithm)
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* m >= 0 (otherwise u < v, which we already checked)
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* m + n = 4
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* and thus
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* m = 4 - n <= 2
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*/
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tmp.ull = ull;
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u[0] = 0;
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u[1] = (digit)HHALF(tmp.ui[H]);
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u[2] = (digit)LHALF(tmp.ui[H]);
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u[3] = (digit)HHALF(tmp.ui[L]);
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u[4] = (digit)LHALF(tmp.ui[L]);
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tmp.ull = vll;
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v[1] = (digit)HHALF(tmp.ui[H]);
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v[2] = (digit)LHALF(tmp.ui[H]);
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v[3] = (digit)HHALF(tmp.ui[L]);
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v[4] = (digit)LHALF(tmp.ui[L]);
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for (n = 4; v[1] == 0; v++) {
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if (--n == 1) {
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unsigned int rbj; /* r*B+u[j] (not root boy jim) */
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digit q1, q2, q3, q4;
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/*
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* Change of plan, per exercise 16.
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* r = 0;
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* for j = 1..4:
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* q[j] = floor((r*B + u[j]) / v),
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* r = (r*B + u[j]) % v;
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* We unroll this completely here.
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*/
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t = v[2]; /* nonzero, by definition */
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q1 = (digit)(u[1] / t);
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rbj = COMBINE(u[1] % t, u[2]);
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q2 = (digit)(rbj / t);
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rbj = COMBINE(rbj % t, u[3]);
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q3 = (digit)(rbj / t);
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rbj = COMBINE(rbj % t, u[4]);
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q4 = (digit)(rbj / t);
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if (arq)
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*arq = rbj % t;
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tmp.ui[H] = COMBINE(q1, q2);
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tmp.ui[L] = COMBINE(q3, q4);
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return (tmp.ll);
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}
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}
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/*
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* By adjusting q once we determine m, we can guarantee that
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* there is a complete four-digit quotient at &qspace[1] when
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* we finally stop.
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*/
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for (m = 4 - n; u[1] == 0; u++)
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m--;
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for (i = 4 - m; --i >= 0;)
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q[i] = 0;
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q += 4 - m;
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/*
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* Here we run Program D, translated from MIX to C and acquiring
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* a few minor changes.
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*
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* D1: choose multiplier 1 << d to ensure v[1] >= B/2.
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*/
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d = 0;
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for (t = v[1]; t < B / 2; t <<= 1)
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d++;
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if (d > 0) {
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shl(&u[0], m + n, d); /* u <<= d */
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shl(&v[1], n - 1, d); /* v <<= d */
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}
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/*
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* D2: j = 0.
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*/
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j = 0;
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v1 = v[1]; /* for D3 -- note that v[1..n] are constant */
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v2 = v[2]; /* for D3 */
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do {
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digit uj0, uj1, uj2;
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/*
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* D3: Calculate qhat (\^q, in TeX notation).
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* Let qhat = min((u[j]*B + u[j+1])/v[1], B-1), and
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* let rhat = (u[j]*B + u[j+1]) mod v[1].
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* While rhat < B and v[2]*qhat > rhat*B+u[j+2],
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* decrement qhat and increase rhat correspondingly.
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* Note that if rhat >= B, v[2]*qhat < rhat*B.
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*/
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uj0 = u[j + 0]; /* for D3 only -- note that u[j+...] change */
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uj1 = u[j + 1]; /* for D3 only */
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uj2 = u[j + 2]; /* for D3 only */
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if (uj0 == v1) {
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qhat = B;
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rhat = uj1;
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goto qhat_too_big;
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} else {
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unsigned int nn = COMBINE(uj0, uj1);
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qhat = nn / v1;
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rhat = nn % v1;
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}
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while (v2 * qhat > COMBINE(rhat, uj2)) {
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qhat_too_big:
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qhat--;
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if ((rhat += v1) >= B)
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break;
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}
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/*
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* D4: Multiply and subtract.
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* The variable `t' holds any borrows across the loop.
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* We split this up so that we do not require v[0] = 0,
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* and to eliminate a final special case.
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*/
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for (t = 0, i = n; i > 0; i--) {
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t = u[i + j] - v[i] * qhat - t;
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u[i + j] = (digit)LHALF(t);
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t = (B - HHALF(t)) & (B - 1);
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}
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t = u[j] - t;
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u[j] = (digit)LHALF(t);
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/*
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* D5: test remainder.
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* There is a borrow if and only if HHALF(t) is nonzero;
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* in that (rare) case, qhat was too large (by exactly 1).
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* Fix it by adding v[1..n] to u[j..j+n].
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*/
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if (HHALF(t)) {
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qhat--;
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for (t = 0, i = n; i > 0; i--) { /* D6: add back. */
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t += u[i + j] + v[i];
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u[i + j] = (digit)LHALF(t);
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t = HHALF(t);
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}
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u[j] = (digit)LHALF(u[j] + t);
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}
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q[j] = (digit)qhat;
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} while (++j <= m); /* D7: loop on j. */
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/*
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* If caller wants the remainder, we have to calculate it as
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* u[m..m+n] >> d (this is at most n digits and thus fits in
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* u[m+1..m+n], but we may need more source digits).
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*/
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if (arq) {
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if (d) {
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for (i = m + n; i > m; --i)
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u[i] = (digit)(((unsigned int)u[i] >> d) |
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LHALF((unsigned int)u[i - 1] <<
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(HALF_BITS - d)));
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u[i] = 0;
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}
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tmp.ui[H] = COMBINE(uspace[1], uspace[2]);
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tmp.ui[L] = COMBINE(uspace[3], uspace[4]);
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*arq = tmp.ll;
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}
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tmp.ui[H] = COMBINE(qspace[1], qspace[2]);
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tmp.ui[L] = COMBINE(qspace[3], qspace[4]);
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return (tmp.ll);
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}
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/*
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* Shift p[0]..p[len] left `sh' bits, ignoring any bits that
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* `fall out' the left (there never will be any such anyway).
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* We may assume len >= 0. NOTE THAT THIS WRITES len+1 DIGITS.
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*/
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static void
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shl(digit *p, int len, int sh)
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{
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int i;
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for (i = 0; i < len; i++)
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p[i] = (digit)(LHALF((unsigned int)p[i] << sh) |
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((unsigned int)p[i + 1] >> (HALF_BITS - sh)));
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p[i] = (digit)(LHALF((unsigned int)p[i] << sh));
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}
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