/*-
* 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.
* 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.
*/
#ifndef _LIBKERN_QUAD_H_
#define _LIBKERN_QUAD_H_
/*
* Quad arithmetic.
*
* This library makes the following assumptions:
*
* - The type long long (aka quad_t) exists.
*
* - A quad variable is exactly twice as long as `long'.
*
* - The machine's arithmetic is two's complement.
*
* This library can provide 128-bit arithmetic on a machine with 128-bit
* quads and 64-bit longs, for instance, or 96-bit arithmetic on machines
* with 48-bit longs.
*/
/*
#include <sys/cdefs.h>
#include <sys/types.h>
#include <sys/limits.h>
#include <sys/syslimits.h>
*/
#include <limits.h>
typedef long long quad_t;
typedef unsigned long long u_quad_t;
typedef unsigned long u_long;
#define CHAR_BIT __CHAR_BIT__
/*
* Define the order of 32-bit words in 64-bit words.
* For little endian only.
*/
#define _QUAD_HIGHWORD 1
#define _QUAD_LOWWORD 0
/*
* Depending on the desired operation, we view a `long long' (aka quad_t) in
* one or more of the following formats.
*/
union uu {
quad_t q; /* as a (signed) quad */
quad_t uq; /* as an unsigned quad */
long sl[2]; /* as two signed longs */
u_long ul[2]; /* as two unsigned longs */
};
/*
* Define high and low longwords.
*/
#define H _QUAD_HIGHWORD
#define L _QUAD_LOWWORD
/*
* Total number of bits in a quad_t and in the pieces that make it up.
* These are used for shifting, and also below for halfword extraction
* and assembly.
*/
#define QUAD_BITS (sizeof(quad_t) * CHAR_BIT)
#define LONG_BITS (sizeof(long) * CHAR_BIT)
#define HALF_BITS (sizeof(long) * CHAR_BIT / 2)
/*
* Extract high and low shortwords from longword, and move low shortword of
* longword to upper half of long, i.e., produce the upper longword of
* ((quad_t)(x) << (number_of_bits_in_long/2)). (`x' must actually be u_long.)
*
* These are used in the multiply code, to split a longword into upper
* and lower halves, and to reassemble a product as a quad_t, shifted left
* (sizeof(long)*CHAR_BIT/2).
*/
#define HHALF(x) ((x) >> HALF_BITS)
#define LHALF(x) ((x) & ((1 << HALF_BITS) - 1))
#define LHUP(x) ((x) << HALF_BITS)
typedef unsigned int qshift_t;
quad_t __ashldi3(quad_t, qshift_t);
quad_t __ashrdi3(quad_t, qshift_t);
int __cmpdi2(quad_t a, quad_t b);
quad_t __divdi3(quad_t a, quad_t b);
quad_t __lshrdi3(quad_t, qshift_t);
quad_t __moddi3(quad_t a, quad_t b);
u_quad_t __qdivrem(u_quad_t u, u_quad_t v, u_quad_t *rem);
u_quad_t __udivdi3(u_quad_t a, u_quad_t b);
u_quad_t __umoddi3(u_quad_t a, u_quad_t b);
int __ucmpdi2(u_quad_t a, u_quad_t b);
quad_t __divmoddi4(quad_t a, quad_t b, quad_t *rem);
#endif /* !_LIBKERN_QUAD_H_ */
#if defined (_X86_) && !defined (__x86_64__)
/*
* Shift a (signed) quad value left (arithmetic shift left).
* This is the same as logical shift left!
*/
quad_t
__ashldi3(a, shift)
quad_t a;
qshift_t shift;
{
union uu aa;
aa.q = a;
if (shift >= LONG_BITS) {
aa.ul[H] = shift >= QUAD_BITS ? 0 :
aa.ul[L] << (shift - LONG_BITS);
aa.ul[L] = 0;
} else if (shift > 0) {
aa.ul[H] = (aa.ul[H] << shift) |
(aa.ul[L] >> (LONG_BITS - shift));
aa.ul[L] <<= shift;
}
return (aa.q);
}
/*
* Shift a (signed) quad value right (arithmetic shift right).
*/
quad_t
__ashrdi3(a, shift)
quad_t a;
qshift_t shift;
{
union uu aa;
aa.q = a;
if (shift >= LONG_BITS) {
long s;
/*
* Smear bits rightward using the machine's right-shift
* method, whether that is sign extension or zero fill,
* to get the `sign word' s. Note that shifting by
* LONG_BITS is undefined, so we shift (LONG_BITS-1),
* then 1 more, to get our answer.
*/
s = (aa.sl[H] >> (LONG_BITS - 1)) >> 1;
aa.ul[L] = shift >= QUAD_BITS ? s :
aa.sl[H] >> (shift - LONG_BITS);
aa.ul[H] = s;
} else if (shift > 0) {
aa.ul[L] = (aa.ul[L] >> shift) |
(aa.ul[H] << (LONG_BITS - shift));
aa.sl[H] >>= shift;
}
return (aa.q);
}
/*
* Return 0, 1, or 2 as a <, =, > b respectively.
* Both a and b are considered signed---which means only the high word is
* signed.
*/
int
__cmpdi2(a, b)
quad_t a, b;
{
union uu aa, bb;
aa.q = a;
bb.q = b;
return (aa.sl[H] < bb.sl[H] ? 0 : aa.sl[H] > bb.sl[H] ? 2 :
aa.ul[L] < bb.ul[L] ? 0 : aa.ul[L] > bb.ul[L] ? 2 : 1);
}
/*
* Divide two signed quads.
* ??? if -1/2 should produce -1 on this machine, this code is wrong
*/
quad_t
__divdi3(a, b)
quad_t a, b;
{
u_quad_t ua, ub, uq;
int neg;
if (a < 0)
ua = -(u_quad_t)a, neg = 1;
else
ua = a, neg = 0;
if (b < 0)
ub = -(u_quad_t)b, neg ^= 1;
else
ub = b;
uq = __qdivrem(ua, ub, (u_quad_t *)0);
return (neg ? -uq : uq);
}
/*
* Shift an (unsigned) quad value right (logical shift right).
*/
quad_t
__lshrdi3(a, shift)
quad_t a;
qshift_t shift;
{
union uu aa;
aa.q = a;
if (shift >= LONG_BITS) {
aa.ul[L] = shift >= QUAD_BITS ? 0 :
aa.ul[H] >> (shift - LONG_BITS);
aa.ul[H] = 0;
} else if (shift > 0) {
aa.ul[L] = (aa.ul[L] >> shift) |
(aa.ul[H] << (LONG_BITS - shift));
aa.ul[H] >>= shift;
}
return (aa.q);
}
/*
* Return remainder after dividing two signed quads.
*
* XXX
* If -1/2 should produce -1 on this machine, this code is wrong.
*/
quad_t
__moddi3(a, b)
quad_t a, b;
{
u_quad_t ua, ub, ur;
int neg;
if (a < 0)
ua = -(u_quad_t)a, neg = 1;
else
ua = a, neg = 0;
if (b < 0)
ub = -(u_quad_t)b;
else
ub = b;
(void)__qdivrem(ua, ub, &ur);
return (neg ? -ur : ur);
}
/*
* Multiprecision divide. This algorithm is from Knuth vol. 2 (2nd ed),
* section 4.3.1, pp. 257--259.
*/
#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 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);
}
/*
* Return 0, 1, or 2 as a <, =, > b respectively.
* Neither a nor b are considered signed.
*/
int
__ucmpdi2(a, b)
u_quad_t a, b;
{
union uu aa, bb;
aa.uq = a;
bb.uq = b;
return (aa.ul[H] < bb.ul[H] ? 0 : aa.ul[H] > bb.ul[H] ? 2 :
aa.ul[L] < bb.ul[L] ? 0 : aa.ul[L] > bb.ul[L] ? 2 : 1);
}
/*
* Divide two unsigned quads.
*/
u_quad_t
__udivdi3(a, b)
u_quad_t a, b;
{
return (__qdivrem(a, b, (u_quad_t *)0));
}
/*
* Return remainder after dividing two unsigned quads.
*/
u_quad_t
__umoddi3(a, b)
u_quad_t a, b;
{
u_quad_t r;
(void)__qdivrem(a, b, &r);
return (r);
}
/*
* Divide two signed quads.
* This function is new in GCC 7.
*/
quad_t
__divmoddi4(a, b, rem)
quad_t a, b, *rem;
{
u_quad_t ua, ub, uq, ur;
int negq, negr;
if (a < 0)
ua = -(u_quad_t)a, negq = 1, negr = 1;
else
ua = a, negq = 0, negr = 0;
if (b < 0)
ub = -(u_quad_t)b, negq ^= 1;
else
ub = b;
uq = __qdivrem(ua, ub, &ur);
if (rem)
*rem = (negr ? -ur : ur);
return (negq ? -uq : uq);
}
#else
static int __attribute__((unused)) dummy;
#endif /*deined (_X86_) && !defined (__x86_64__)*/