Staging
v0.5.1
v0.5.1
Revision b8e682427a80798fec90dce31392beaf616c3e37 authored by Miss Islington (bot) on 14 August 2019, 22:02:12 UTC, committed by GitHub on 14 August 2019, 22:02:12 UTC
faulthandler now allocates a dedicated stack of SIGSTKSZ*2 bytes, instead of just SIGSTKSZ bytes. Calling the previous signal handler in faulthandler signal handler uses more than SIGSTKSZ bytes of stack memory on some platforms. (cherry picked from commit ac827edc493d3ac3f5b9b0cc353df1d4b418a9aa) Co-authored-by: Victor Stinner <vstinner@redhat.com>
1 parent 123f6c4
_math.c
/* Definitions of some C99 math library functions, for those platforms
that don't implement these functions already. */
#include "Python.h"
#include <float.h>
#include "_math.h"
/* The following copyright notice applies to the original
implementations of acosh, asinh and atanh. */
/*
* ====================================================
* 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.
* ====================================================
*/
#if !defined(HAVE_ACOSH) || !defined(HAVE_ASINH)
static const double ln2 = 6.93147180559945286227E-01;
static const double two_pow_p28 = 268435456.0; /* 2**28 */
#endif
#if !defined(HAVE_ASINH) || !defined(HAVE_ATANH)
static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
#endif
#if !defined(HAVE_ATANH) && !defined(Py_NAN)
static const double zero = 0.0;
#endif
#ifndef HAVE_ACOSH
/* 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.
*/
double
_Py_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 m_log1p(t + sqrt(2.0 * t + t * t));
}
}
#endif /* HAVE_ACOSH */
#ifndef HAVE_ASINH
/* 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)))
*/
double
_Py_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 = m_log1p(absx + t / (1.0 + sqrt(1.0 + t)));
}
return copysign(w, x);
}
#endif /* HAVE_ASINH */
#ifndef HAVE_ATANH
/* 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;
*
*/
double
_Py_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 * m_log1p(t + t*absx / (1.0 - absx));
}
else { /* 0.5 <= |x| <= 1.0 */
t = 0.5 * m_log1p((absx + absx) / (1.0 - absx));
}
return copysign(t, x);
}
#endif /* HAVE_ATANH */
#ifndef HAVE_EXPM1
/* Mathematically, expm1(x) = exp(x) - 1. The expm1 function is designed
to avoid the significant loss of precision that arises from direct
evaluation of the expression exp(x) - 1, for x near 0. */
double
_Py_expm1(double x)
{
/* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this
also works fine for infinities and nans.
For smaller x, we can use a method due to Kahan that achieves close to
full accuracy.
*/
if (fabs(x) < 0.7) {
double u;
u = exp(x);
if (u == 1.0)
return x;
else
return (u - 1.0) * x / log(u);
}
else
return exp(x) - 1.0;
}
#endif /* HAVE_EXPM1 */
/* log1p(x) = log(1+x). The log1p function is designed to avoid the
significant loss of precision that arises from direct evaluation when x is
small. */
double
_Py_log1p(double x)
{
#ifdef HAVE_LOG1P
/* Some platforms supply a log1p function but don't respect the sign of
zero: log1p(-0.0) gives 0.0 instead of the correct result of -0.0.
To save fiddling with configure tests and platform checks, we handle the
special case of zero input directly on all platforms.
*/
if (x == 0.0) {
return x;
}
else {
return log1p(x);
}
#else
/* 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 that 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 /* ifdef HAVE_LOG1P */
}
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