mirror of
https://github.com/python/cpython.git
synced 2024-11-28 08:20:55 +01:00
7abf8d4066
problems for binary distributions of Python in situations where the build machine has SSE2 but the target machine does not. Therefore, don't enable SSE2 instructions automatically on x86.
279 lines
6.2 KiB
C
279 lines
6.2 KiB
C
#include "Python.h"
|
|
|
|
#ifdef X87_DOUBLE_ROUNDING
|
|
/* On x86 platforms using an x87 FPU, this function is called from the
|
|
Py_FORCE_DOUBLE macro (defined in pymath.h) to force a floating-point
|
|
number out of an 80-bit x87 FPU register and into a 64-bit memory location,
|
|
thus rounding from extended precision to double precision. */
|
|
double _Py_force_double(double x)
|
|
{
|
|
volatile double y;
|
|
y = x;
|
|
return y;
|
|
}
|
|
#endif
|
|
|
|
#ifdef HAVE_GCC_ASM_FOR_X87
|
|
|
|
/* inline assembly for getting and setting the 387 FPU control word on
|
|
gcc/x86 */
|
|
|
|
unsigned short _Py_get_387controlword(void) {
|
|
unsigned short cw;
|
|
__asm__ __volatile__ ("fnstcw %0" : "=m" (cw));
|
|
return cw;
|
|
}
|
|
|
|
void _Py_set_387controlword(unsigned short cw) {
|
|
__asm__ __volatile__ ("fldcw %0" : : "m" (cw));
|
|
}
|
|
|
|
#endif
|
|
|
|
|
|
#ifndef HAVE_HYPOT
|
|
double hypot(double x, double y)
|
|
{
|
|
double yx;
|
|
|
|
x = fabs(x);
|
|
y = fabs(y);
|
|
if (x < y) {
|
|
double temp = x;
|
|
x = y;
|
|
y = temp;
|
|
}
|
|
if (x == 0.)
|
|
return 0.;
|
|
else {
|
|
yx = y/x;
|
|
return x*sqrt(1.+yx*yx);
|
|
}
|
|
}
|
|
#endif /* HAVE_HYPOT */
|
|
|
|
#ifndef HAVE_COPYSIGN
|
|
double
|
|
copysign(double x, double y)
|
|
{
|
|
/* use atan2 to distinguish -0. from 0. */
|
|
if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {
|
|
return fabs(x);
|
|
} else {
|
|
return -fabs(x);
|
|
}
|
|
}
|
|
#endif /* HAVE_COPYSIGN */
|
|
|
|
#ifndef HAVE_ROUND
|
|
double
|
|
round(double x)
|
|
{
|
|
double absx, y;
|
|
absx = fabs(x);
|
|
y = floor(absx);
|
|
if (absx - y >= 0.5)
|
|
y += 1.0;
|
|
return copysign(y, x);
|
|
}
|
|
#endif /* HAVE_ROUND */
|
|
|
|
#ifndef HAVE_LOG1P
|
|
#include <float.h>
|
|
|
|
double
|
|
log1p(double x)
|
|
{
|
|
/* For x small, we use the following approach. Let y be the nearest
|
|
float to 1+x, then
|
|
|
|
1+x = y * (1 - (y-1-x)/y)
|
|
|
|
so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny,
|
|
the second term is well approximated by (y-1-x)/y. If abs(x) >=
|
|
DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
|
|
then y-1-x will be exactly representable, and is computed exactly
|
|
by (y-1)-x.
|
|
|
|
If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
|
|
round-to-nearest then this method is slightly dangerous: 1+x could
|
|
be rounded up to 1+DBL_EPSILON instead of down to 1, and in that
|
|
case y-1-x will not be exactly representable any more and the
|
|
result can be off by many ulps. But this is easily fixed: for a
|
|
floating-point number |x| < DBL_EPSILON/2., the closest
|
|
floating-point number to log(1+x) is exactly x.
|
|
*/
|
|
|
|
double y;
|
|
if (fabs(x) < DBL_EPSILON/2.) {
|
|
return x;
|
|
} else if (-0.5 <= x && x <= 1.) {
|
|
/* WARNING: it's possible than an overeager compiler
|
|
will incorrectly optimize the following two lines
|
|
to the equivalent of "return log(1.+x)". If this
|
|
happens, then results from log1p will be inaccurate
|
|
for small x. */
|
|
y = 1.+x;
|
|
return log(y)-((y-1.)-x)/y;
|
|
} else {
|
|
/* NaNs and infinities should end up here */
|
|
return log(1.+x);
|
|
}
|
|
}
|
|
#endif /* HAVE_LOG1P */
|
|
|
|
/*
|
|
* ====================================================
|
|
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
|
|
*
|
|
* Developed at SunPro, a Sun Microsystems, Inc. business.
|
|
* Permission to use, copy, modify, and distribute this
|
|
* software is freely granted, provided that this notice
|
|
* is preserved.
|
|
* ====================================================
|
|
*/
|
|
|
|
static const double ln2 = 6.93147180559945286227E-01;
|
|
static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
|
|
static const double two_pow_p28 = 268435456.0; /* 2**28 */
|
|
static const double zero = 0.0;
|
|
|
|
/* asinh(x)
|
|
* Method :
|
|
* Based on
|
|
* asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
|
|
* we have
|
|
* asinh(x) := x if 1+x*x=1,
|
|
* := sign(x)*(log(x)+ln2)) for large |x|, else
|
|
* := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
|
|
* := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
|
|
*/
|
|
|
|
#ifndef HAVE_ASINH
|
|
double
|
|
asinh(double x)
|
|
{
|
|
double w;
|
|
double absx = fabs(x);
|
|
|
|
if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
|
|
return x+x;
|
|
}
|
|
if (absx < two_pow_m28) { /* |x| < 2**-28 */
|
|
return x; /* return x inexact except 0 */
|
|
}
|
|
if (absx > two_pow_p28) { /* |x| > 2**28 */
|
|
w = log(absx)+ln2;
|
|
}
|
|
else if (absx > 2.0) { /* 2 < |x| < 2**28 */
|
|
w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
|
|
}
|
|
else { /* 2**-28 <= |x| < 2= */
|
|
double t = x*x;
|
|
w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
|
|
}
|
|
return copysign(w, x);
|
|
|
|
}
|
|
#endif /* HAVE_ASINH */
|
|
|
|
/* acosh(x)
|
|
* Method :
|
|
* Based on
|
|
* acosh(x) = log [ x + sqrt(x*x-1) ]
|
|
* we have
|
|
* acosh(x) := log(x)+ln2, if x is large; else
|
|
* acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
|
|
* acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
|
|
*
|
|
* Special cases:
|
|
* acosh(x) is NaN with signal if x<1.
|
|
* acosh(NaN) is NaN without signal.
|
|
*/
|
|
|
|
#ifndef HAVE_ACOSH
|
|
double
|
|
acosh(double x)
|
|
{
|
|
if (Py_IS_NAN(x)) {
|
|
return x+x;
|
|
}
|
|
if (x < 1.) { /* x < 1; return a signaling NaN */
|
|
errno = EDOM;
|
|
#ifdef Py_NAN
|
|
return Py_NAN;
|
|
#else
|
|
return (x-x)/(x-x);
|
|
#endif
|
|
}
|
|
else if (x >= two_pow_p28) { /* x > 2**28 */
|
|
if (Py_IS_INFINITY(x)) {
|
|
return x+x;
|
|
} else {
|
|
return log(x)+ln2; /* acosh(huge)=log(2x) */
|
|
}
|
|
}
|
|
else if (x == 1.) {
|
|
return 0.0; /* acosh(1) = 0 */
|
|
}
|
|
else if (x > 2.) { /* 2 < x < 2**28 */
|
|
double t = x*x;
|
|
return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
|
|
}
|
|
else { /* 1 < x <= 2 */
|
|
double t = x - 1.0;
|
|
return log1p(t + sqrt(2.0*t + t*t));
|
|
}
|
|
}
|
|
#endif /* HAVE_ACOSH */
|
|
|
|
/* atanh(x)
|
|
* Method :
|
|
* 1.Reduced x to positive by atanh(-x) = -atanh(x)
|
|
* 2.For x>=0.5
|
|
* 1 2x x
|
|
* atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)
|
|
* 2 1 - x 1 - x
|
|
*
|
|
* For x<0.5
|
|
* atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
|
|
*
|
|
* Special cases:
|
|
* atanh(x) is NaN if |x| >= 1 with signal;
|
|
* atanh(NaN) is that NaN with no signal;
|
|
*
|
|
*/
|
|
|
|
#ifndef HAVE_ATANH
|
|
double
|
|
atanh(double x)
|
|
{
|
|
double absx;
|
|
double t;
|
|
|
|
if (Py_IS_NAN(x)) {
|
|
return x+x;
|
|
}
|
|
absx = fabs(x);
|
|
if (absx >= 1.) { /* |x| >= 1 */
|
|
errno = EDOM;
|
|
#ifdef Py_NAN
|
|
return Py_NAN;
|
|
#else
|
|
return x/zero;
|
|
#endif
|
|
}
|
|
if (absx < two_pow_m28) { /* |x| < 2**-28 */
|
|
return x;
|
|
}
|
|
if (absx < 0.5) { /* |x| < 0.5 */
|
|
t = absx+absx;
|
|
t = 0.5 * log1p(t + t*absx / (1.0 - absx));
|
|
}
|
|
else { /* 0.5 <= |x| <= 1.0 */
|
|
t = 0.5 * log1p((absx + absx) / (1.0 - absx));
|
|
}
|
|
return copysign(t, x);
|
|
}
|
|
#endif /* HAVE_ATANH */
|