EmbeddedPkg: Extend NvVarStoreFormattedLib LIBRARY_CLASS

[mirror_edk2.git] / AppPkg / Applications / Python / Python-2.7.10 / Modules / _math.c

/* Definitions of some C99 math library functions, for those platforms\r
   that don't implement these functions already. */\r
\r
#include "Python.h"\r
#include <float.h>\r
#include "_math.h"\r
\r
/* The following copyright notice applies to the original\r
   implementations of acosh, asinh and atanh. */\r
\r
/*\r
 * ====================================================\r
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.\r
 *\r
 * Developed at SunPro, a Sun Microsystems, Inc. business.\r
 * Permission to use, copy, modify, and distribute this\r
 * software is freely granted, provided that this notice\r
 * is preserved.\r
 * ====================================================\r
 */\r
\r
static const double ln2 = 6.93147180559945286227E-01;\r
static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */\r
static const double two_pow_p28 = 268435456.0; /* 2**28 */\r
static const double zero = 0.0;\r
\r
/* acosh(x)\r
 * Method :\r
 *      Based on\r
 *            acosh(x) = log [ x + sqrt(x*x-1) ]\r
 *      we have\r
 *            acosh(x) := log(x)+ln2, if x is large; else\r
 *            acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else\r
 *            acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.\r
 *\r
 * Special cases:\r
 *      acosh(x) is NaN with signal if x<1.\r
 *      acosh(NaN) is NaN without signal.\r
 */\r
\r
double\r
_Py_acosh(double x)\r
{\r
    if (Py_IS_NAN(x)) {\r
        return x+x;\r
    }\r
    if (x < 1.) {                       /* x < 1;  return a signaling NaN */\r
        errno = EDOM;\r
#ifdef Py_NAN\r
        return Py_NAN;\r
#else\r
        return (x-x)/(x-x);\r
#endif\r
    }\r
    else if (x >= two_pow_p28) {        /* x > 2**28 */\r
        if (Py_IS_INFINITY(x)) {\r
            return x+x;\r
        }\r
        else {\r
            return log(x)+ln2;          /* acosh(huge)=log(2x) */\r
        }\r
    }\r
    else if (x == 1.) {\r
        return 0.0;                     /* acosh(1) = 0 */\r
    }\r
    else if (x > 2.) {                  /* 2 < x < 2**28 */\r
        double t = x*x;\r
        return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));\r
    }\r
    else {                              /* 1 < x <= 2 */\r
        double t = x - 1.0;\r
        return m_log1p(t + sqrt(2.0*t + t*t));\r
    }\r
}\r
\r
\r
/* asinh(x)\r
 * Method :\r
 *      Based on\r
 *              asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]\r
 *      we have\r
 *      asinh(x) := x  if  1+x*x=1,\r
 *               := sign(x)*(log(x)+ln2)) for large |x|, else\r
 *               := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else\r
 *               := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))\r
 */\r
\r
double\r
_Py_asinh(double x)\r
{\r
    double w;\r
    double absx = fabs(x);\r
\r
    if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {\r
        return x+x;\r
    }\r
    if (absx < two_pow_m28) {           /* |x| < 2**-28 */\r
        return x;                       /* return x inexact except 0 */\r
    }\r
    if (absx > two_pow_p28) {           /* |x| > 2**28 */\r
        w = log(absx)+ln2;\r
    }\r
    else if (absx > 2.0) {              /* 2 < |x| < 2**28 */\r
        w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));\r
    }\r
    else {                              /* 2**-28 <= |x| < 2= */\r
        double t = x*x;\r
        w = m_log1p(absx + t / (1.0 + sqrt(1.0 + t)));\r
    }\r
    return copysign(w, x);\r
\r
}\r
\r
/* atanh(x)\r
 * Method :\r
 *    1.Reduced x to positive by atanh(-x) = -atanh(x)\r
 *    2.For x>=0.5\r
 *                  1              2x                          x\r
 *      atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * -------)\r
 *                  2             1 - x                      1 - x\r
 *\r
 *      For x<0.5\r
 *      atanh(x) = 0.5*log1p(2x+2x*x/(1-x))\r
 *\r
 * Special cases:\r
 *      atanh(x) is NaN if |x| >= 1 with signal;\r
 *      atanh(NaN) is that NaN with no signal;\r
 *\r
 */\r
\r
double\r
_Py_atanh(double x)\r
{\r
    double absx;\r
    double t;\r
\r
    if (Py_IS_NAN(x)) {\r
        return x+x;\r
    }\r
    absx = fabs(x);\r
    if (absx >= 1.) {                   /* |x| >= 1 */\r
        errno = EDOM;\r
#ifdef Py_NAN\r
        return Py_NAN;\r
#else\r
        return x/zero;\r
#endif\r
    }\r
    if (absx < two_pow_m28) {           /* |x| < 2**-28 */\r
        return x;\r
    }\r
    if (absx < 0.5) {                   /* |x| < 0.5 */\r
        t = absx+absx;\r
        t = 0.5 * m_log1p(t + t*absx / (1.0 - absx));\r
    }\r
    else {                              /* 0.5 <= |x| <= 1.0 */\r
        t = 0.5 * m_log1p((absx + absx) / (1.0 - absx));\r
    }\r
    return copysign(t, x);\r
}\r
\r
/* Mathematically, expm1(x) = exp(x) - 1.  The expm1 function is designed\r
   to avoid the significant loss of precision that arises from direct\r
   evaluation of the expression exp(x) - 1, for x near 0. */\r
\r
double\r
_Py_expm1(double x)\r
{\r
    /* For abs(x) >= log(2), it's safe to evaluate exp(x) - 1 directly; this\r
       also works fine for infinities and nans.\r
\r
       For smaller x, we can use a method due to Kahan that achieves close to\r
       full accuracy.\r
    */\r
\r
    if (fabs(x) < 0.7) {\r
        double u;\r
        u = exp(x);\r
        if (u == 1.0)\r
            return x;\r
        else\r
            return (u - 1.0) * x / log(u);\r
    }\r
    else\r
        return exp(x) - 1.0;\r
}\r
\r
/* log1p(x) = log(1+x).  The log1p function is designed to avoid the\r
   significant loss of precision that arises from direct evaluation when x is\r
   small. */\r
\r
#ifdef HAVE_LOG1P\r
\r
double\r
_Py_log1p(double x)\r
{\r
    /* Some platforms supply a log1p function but don't respect the sign of\r
       zero:  log1p(-0.0) gives 0.0 instead of the correct result of -0.0.\r
\r
       To save fiddling with configure tests and platform checks, we handle the\r
       special case of zero input directly on all platforms.\r
    */\r
    if (x == 0.0) {\r
        return x;\r
    }\r
    else {\r
        return log1p(x);\r
    }\r
}\r
\r
#else\r
\r
double\r
_Py_log1p(double x)\r
{\r
    /* For x small, we use the following approach.  Let y be the nearest float\r
       to 1+x, then\r
\r
         1+x = y * (1 - (y-1-x)/y)\r
\r
       so log(1+x) = log(y) + log(1-(y-1-x)/y).  Since (y-1-x)/y is tiny, the\r
       second term is well approximated by (y-1-x)/y.  If abs(x) >=\r
       DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest\r
       then y-1-x will be exactly representable, and is computed exactly by\r
       (y-1)-x.\r
\r
       If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be\r
       round-to-nearest then this method is slightly dangerous: 1+x could be\r
       rounded up to 1+DBL_EPSILON instead of down to 1, and in that case\r
       y-1-x will not be exactly representable any more and the result can be\r
       off by many ulps.  But this is easily fixed: for a floating-point\r
       number |x| < DBL_EPSILON/2., the closest floating-point number to\r
       log(1+x) is exactly x.\r
    */\r
\r
    double y;\r
    if (fabs(x) < DBL_EPSILON/2.) {\r
        return x;\r
    }\r
    else if (-0.5 <= x && x <= 1.) {\r
        /* WARNING: it's possible than an overeager compiler\r
           will incorrectly optimize the following two lines\r
           to the equivalent of "return log(1.+x)". If this\r
           happens, then results from log1p will be inaccurate\r
           for small x. */\r
        y = 1.+x;\r
        return log(y)-((y-1.)-x)/y;\r
    }\r
    else {\r
        /* NaNs and infinities should end up here */\r
        return log(1.+x);\r
    }\r
}\r
\r
#endif /* ifdef HAVE_LOG1P */\r