Adding upstream version 6.0.4-dfsg.upstream/6.0.4-dfsgupstream

Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>

1 files changed, 285 insertions, 0 deletions

diff --git a/src/VBox/Runtime/common/math/gcc/qdivrem.c b/src/VBox/Runtime/common/math/gcc/qdivrem.c
new file mode 100644
index 00000000..7ca2d38c
--- /dev/null
+++ b/src/VBox/Runtime/common/math/gcc/qdivrem.c
@@ -0,0 +1,285 @@
+/*	$NetBSD: qdivrem.c,v 1.12 2005/12/11 12:24:37 christos Exp $	*/
+
+/*-
+ * 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>
+#if defined(LIBC_SCCS) && !defined(lint)
+#if 0
+static char sccsid[] = "@(#)qdivrem.c	8.1 (Berkeley) 6/4/93";
+#else
+__RCSID("$NetBSD: qdivrem.c,v 1.12 2005/12/11 12:24:37 christos Exp $");
+#endif
+#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	((int)1 << HALF_BITS)	/* digit base */
+
+/* Combine two `digits' to make a single two-digit number. */
+#define	COMBINE(a, b) (((u_int)(a) << HALF_BITS) | (b))
+
+/* select a type for digits in base B: use unsigned short if they fit */
+#if UINT_MAX == 0xffffffffU && USHRT_MAX >= 0xffff
+typedef unsigned short digit;
+#else
+typedef u_int digit;
+#endif
+
+static void shl __P((digit *p, int len, int 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_int.  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;
+	digit v1, v2;
+	u_int 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] = (digit)HHALF(tmp.ul[H]);
+	u[2] = (digit)LHALF(tmp.ul[H]);
+	u[3] = (digit)HHALF(tmp.ul[L]);
+	u[4] = (digit)LHALF(tmp.ul[L]);
+	tmp.uq = vq;
+	v[1] = (digit)HHALF(tmp.ul[H]);
+	v[2] = (digit)LHALF(tmp.ul[H]);
+	v[3] = (digit)HHALF(tmp.ul[L]);
+	v[4] = (digit)LHALF(tmp.ul[L]);
+	for (n = 4; v[1] == 0; v++) {
+		if (--n == 1) {
+			u_int 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 = (digit)(u[1] / t);
+			rbj = COMBINE(u[1] % t, u[2]);
+			q2 = (digit)(rbj / t);
+			rbj = COMBINE(rbj % t, u[3]);
+			q3 = (digit)(rbj / t);
+			rbj = COMBINE(rbj % t, u[4]);
+			q4 = (digit)(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_int 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] = (digit)LHALF(t);
+			t = (B - HHALF(t)) & (B - 1);
+		}
+		t = u[j] - t;
+		u[j] = (digit)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] = (digit)LHALF(t);
+				t = HHALF(t);
+			}
+			u[j] = (digit)LHALF(u[j] + t);
+		}
+		q[j] = (digit)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] = (digit)(((u_int)u[i] >> d) |
+				    LHALF((u_int)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);
+}
+
+/*
+ * 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] = (digit)(LHALF((u_int)p[i] << sh) |
+		    ((u_int)p[i + 1] >> (HALF_BITS - sh)));
+	p[i] = (digit)(LHALF((u_int)p[i] << sh));
+}