248 lines
6.6 KiB
C
248 lines
6.6 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
|
|
|
|
#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_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 */
|