/Python/pymath.c
http://unladen-swallow.googlecode.com/ · C · 247 lines · 150 code · 18 blank · 79 comment · 38 complexity · d92ee247d5047f14cd57bf2169354d21 MD5 · raw file
- #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 */