--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/symbian-qemu-0.9.1-12/python-2.6.1/Python/pymath.c Fri Jul 31 15:01:17 2009 +0100
@@ -0,0 +1,234 @@
+#include "Python.h"
+
+#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
+static 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 */