Restoring authorship annotation for <[email protected]>. Commit 1 of 2.

1 files changed, 265 insertions, 265 deletions

diff --git a/contrib/tools/python3/src/Modules/_math.c b/contrib/tools/python3/src/Modules/_math.c
index 68e3a234692..2ab022c0933 100644
--- a/contrib/tools/python3/src/Modules/_math.c
+++ b/contrib/tools/python3/src/Modules/_math.c
@@ -1,266 +1,266 @@
-/* Definitions of some C99 math library functions, for those platforms
-   that don't implement these functions already. */
-
-#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,
+/* Definitions of some C99 math library functions, for those platforms 
+   that don't implement these functions already. */ 
+ 
+#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 */
-}
-
+ *               := 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 */ 
+} 
+