freebsd-nq/sys/libkern/arm/muldi3.c
Pedro F. Giffuni 51369649b0 sys: further adoption of SPDX licensing ID tags.
Mainly focus on files that use BSD 3-Clause license.

The Software Package Data Exchange (SPDX) group provides a specification
to make it easier for automated tools to detect and summarize well known
opensource licenses. We are gradually adopting the specification, noting
that the tags are considered only advisory and do not, in any way,
superceed or replace the license texts.

Special thanks to Wind River for providing access to "The Duke of
Highlander" tool: an older (2014) run over FreeBSD tree was useful as a
starting point.
2017-11-20 19:43:44 +00:00

252 lines
7.0 KiB
C

/* $NetBSD: muldi3.c,v 1.8 2003/08/07 16:32:09 agc Exp $ */
/*-
* 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>
#if defined(LIBC_SCCS) && !defined(lint)
#if 0
static char sccsid[] = "@(#)muldi3.c 8.1 (Berkeley) 6/4/93";
#else
__FBSDID("$FreeBSD$");
#endif
#endif /* LIBC_SCCS and not lint */
#include <libkern/quad.h>
/*
* Multiply two quads.
*
* Our algorithm is based on the following. Split incoming quad values
* u and v (where u,v >= 0) into
*
* u = 2^n u1 * u0 (n = number of bits in `u_int', usu. 32)
*
* and
*
* v = 2^n v1 * v0
*
* Then
*
* uv = 2^2n u1 v1 + 2^n u1 v0 + 2^n v1 u0 + u0 v0
* = 2^2n u1 v1 + 2^n (u1 v0 + v1 u0) + u0 v0
*
* Now add 2^n u1 v1 to the first term and subtract it from the middle,
* and add 2^n u0 v0 to the last term and subtract it from the middle.
* This gives:
*
* uv = (2^2n + 2^n) (u1 v1) +
* (2^n) (u1 v0 - u1 v1 + u0 v1 - u0 v0) +
* (2^n + 1) (u0 v0)
*
* Factoring the middle a bit gives us:
*
* uv = (2^2n + 2^n) (u1 v1) + [u1v1 = high]
* (2^n) (u1 - u0) (v0 - v1) + [(u1-u0)... = mid]
* (2^n + 1) (u0 v0) [u0v0 = low]
*
* The terms (u1 v1), (u1 - u0) (v0 - v1), and (u0 v0) can all be done
* in just half the precision of the original. (Note that either or both
* of (u1 - u0) or (v0 - v1) may be negative.)
*
* This algorithm is from Knuth vol. 2 (2nd ed), section 4.3.3, p. 278.
*
* Since C does not give us a `int * int = quad' operator, we split
* our input quads into two ints, then split the two ints into two
* shorts. We can then calculate `short * short = int' in native
* arithmetic.
*
* Our product should, strictly speaking, be a `long quad', with 128
* bits, but we are going to discard the upper 64. In other words,
* we are not interested in uv, but rather in (uv mod 2^2n). This
* makes some of the terms above vanish, and we get:
*
* (2^n)(high) + (2^n)(mid) + (2^n + 1)(low)
*
* or
*
* (2^n)(high + mid + low) + low
*
* Furthermore, `high' and `mid' can be computed mod 2^n, as any factor
* of 2^n in either one will also vanish. Only `low' need be computed
* mod 2^2n, and only because of the final term above.
*/
static quad_t __lmulq(u_int, u_int);
quad_t __muldi3(quad_t, quad_t);
quad_t
__muldi3(quad_t a, quad_t b)
{
union uu u, v, low, prod;
u_int high, mid, udiff, vdiff;
int negall, negmid;
#define u1 u.ul[H]
#define u0 u.ul[L]
#define v1 v.ul[H]
#define v0 v.ul[L]
/*
* Get u and v such that u, v >= 0. When this is finished,
* u1, u0, v1, and v0 will be directly accessible through the
* int fields.
*/
if (a >= 0)
u.q = a, negall = 0;
else
u.q = -a, negall = 1;
if (b >= 0)
v.q = b;
else
v.q = -b, negall ^= 1;
if (u1 == 0 && v1 == 0) {
/*
* An (I hope) important optimization occurs when u1 and v1
* are both 0. This should be common since most numbers
* are small. Here the product is just u0*v0.
*/
prod.q = __lmulq(u0, v0);
} else {
/*
* Compute the three intermediate products, remembering
* whether the middle term is negative. We can discard
* any upper bits in high and mid, so we can use native
* u_int * u_int => u_int arithmetic.
*/
low.q = __lmulq(u0, v0);
if (u1 >= u0)
negmid = 0, udiff = u1 - u0;
else
negmid = 1, udiff = u0 - u1;
if (v0 >= v1)
vdiff = v0 - v1;
else
vdiff = v1 - v0, negmid ^= 1;
mid = udiff * vdiff;
high = u1 * v1;
/*
* Assemble the final product.
*/
prod.ul[H] = high + (negmid ? -mid : mid) + low.ul[L] +
low.ul[H];
prod.ul[L] = low.ul[L];
}
return (negall ? -prod.q : prod.q);
#undef u1
#undef u0
#undef v1
#undef v0
}
/*
* Multiply two 2N-bit ints to produce a 4N-bit quad, where N is half
* the number of bits in an int (whatever that is---the code below
* does not care as long as quad.h does its part of the bargain---but
* typically N==16).
*
* We use the same algorithm from Knuth, but this time the modulo refinement
* does not apply. On the other hand, since N is half the size of an int,
* we can get away with native multiplication---none of our input terms
* exceeds (UINT_MAX >> 1).
*
* Note that, for u_int l, the quad-precision result
*
* l << N
*
* splits into high and low ints as HHALF(l) and LHUP(l) respectively.
*/
static quad_t
__lmulq(u_int u, u_int v)
{
u_int u1, u0, v1, v0, udiff, vdiff, high, mid, low;
u_int prodh, prodl, was;
union uu prod;
int neg;
u1 = HHALF(u);
u0 = LHALF(u);
v1 = HHALF(v);
v0 = LHALF(v);
low = u0 * v0;
/* This is the same small-number optimization as before. */
if (u1 == 0 && v1 == 0)
return (low);
if (u1 >= u0)
udiff = u1 - u0, neg = 0;
else
udiff = u0 - u1, neg = 1;
if (v0 >= v1)
vdiff = v0 - v1;
else
vdiff = v1 - v0, neg ^= 1;
mid = udiff * vdiff;
high = u1 * v1;
/* prod = (high << 2N) + (high << N); */
prodh = high + HHALF(high);
prodl = LHUP(high);
/* if (neg) prod -= mid << N; else prod += mid << N; */
if (neg) {
was = prodl;
prodl -= LHUP(mid);
prodh -= HHALF(mid) + (prodl > was);
} else {
was = prodl;
prodl += LHUP(mid);
prodh += HHALF(mid) + (prodl < was);
}
/* prod += low << N */
was = prodl;
prodl += LHUP(low);
prodh += HHALF(low) + (prodl < was);
/* ... + low; */
if ((prodl += low) < low)
prodh++;
/* return 4N-bit product */
prod.ul[H] = prodh;
prod.ul[L] = prodl;
return (prod.q);
}