Source

cpython / Python / pymath.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 */
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