bpo-45548: Remove _math.c workarounds for pre-C99 libm (GH-29179)

The :mod:`math` and :mod:`cmath` implementation now require a C99 compatible
``libm`` and no longer ship with workarounds for missing acosh, asinh,
expm1, and log1p functions.

The changeset also removes ``_math.c`` and moves the last remaining
workaround into ``_math.h``. This simplifies static builds with
``Modules/Setup`` and resolves symbol conflicts.

Co-authored-by: Mark Dickinson <mdickinson@enthought.com>
Co-authored-by: Brett Cannon <brett@python.org>
Signed-off-by: Christian Heimes <christian@python.org>

1 files changed, 0 insertions, 270 deletions

diff --git a/Modules/_math.c b/Modules/_math.c
deleted file mode 100644
index c1936a1088..0000000000
--- a/Modules/_math.c
+++ /dev/null
@@ -1,270 +0,0 @@
-/* Definitions of some C99 math library functions, for those platforms
-   that don't implement these functions already. */
-
-#ifndef Py_BUILD_CORE_BUILTIN
-#  define Py_BUILD_CORE_MODULE 1
-#endif
-
-#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 */
-}
-