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Christian Heimes53876d92008-04-19 00:31:39 +00001#include "Python.h"
2
Mark Dickinson87ec0852009-02-09 17:15:59 +00003#ifdef X87_DOUBLE_ROUNDING
4/* On x86 platforms using an x87 FPU, this function is called from the
5 Py_FORCE_DOUBLE macro (defined in pymath.h) to force a floating-point
6 number out of an 80-bit x87 FPU register and into a 64-bit memory location,
7 thus rounding from extended precision to double precision. */
8double _Py_force_double(double x)
9{
10 volatile double y;
11 y = x;
12 return y;
13}
14#endif
15
Mark Dickinson7abf8d42009-04-18 20:17:52 +000016#ifdef HAVE_GCC_ASM_FOR_X87
Mark Dickinsonb08a53a2009-04-16 19:52:09 +000017
18/* inline assembly for getting and setting the 387 FPU control word on
19 gcc/x86 */
20
21unsigned short _Py_get_387controlword(void) {
22 unsigned short cw;
23 __asm__ __volatile__ ("fnstcw %0" : "=m" (cw));
24 return cw;
25}
26
27void _Py_set_387controlword(unsigned short cw) {
28 __asm__ __volatile__ ("fldcw %0" : : "m" (cw));
29}
30
Mark Dickinsonb08a53a2009-04-16 19:52:09 +000031#endif
32
33
Christian Heimes53876d92008-04-19 00:31:39 +000034#ifndef HAVE_HYPOT
35double hypot(double x, double y)
36{
37 double yx;
38
39 x = fabs(x);
40 y = fabs(y);
41 if (x < y) {
42 double temp = x;
43 x = y;
44 y = temp;
45 }
46 if (x == 0.)
47 return 0.;
48 else {
49 yx = y/x;
50 return x*sqrt(1.+yx*yx);
51 }
52}
53#endif /* HAVE_HYPOT */
54
55#ifndef HAVE_COPYSIGN
Mark Dickinson23b62862009-04-18 14:14:48 +000056double
Christian Heimes53876d92008-04-19 00:31:39 +000057copysign(double x, double y)
58{
59 /* use atan2 to distinguish -0. from 0. */
60 if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {
61 return fabs(x);
62 } else {
63 return -fabs(x);
64 }
65}
66#endif /* HAVE_COPYSIGN */
67
Mark Dickinsonf2537862009-04-18 13:58:18 +000068#ifndef HAVE_ROUND
69double
70round(double x)
71{
72 double absx, y;
73 absx = fabs(x);
74 y = floor(absx);
75 if (absx - y >= 0.5)
76 y += 1.0;
77 return copysign(y, x);
78}
79#endif /* HAVE_ROUND */
80
Christian Heimes53876d92008-04-19 00:31:39 +000081#ifndef HAVE_LOG1P
Andrew MacIntyre45612572008-09-22 14:49:01 +000082#include <float.h>
83
Christian Heimes53876d92008-04-19 00:31:39 +000084double
85log1p(double x)
86{
87 /* For x small, we use the following approach. Let y be the nearest
88 float to 1+x, then
89
90 1+x = y * (1 - (y-1-x)/y)
91
92 so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny,
93 the second term is well approximated by (y-1-x)/y. If abs(x) >=
94 DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
95 then y-1-x will be exactly representable, and is computed exactly
96 by (y-1)-x.
97
98 If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
99 round-to-nearest then this method is slightly dangerous: 1+x could
100 be rounded up to 1+DBL_EPSILON instead of down to 1, and in that
101 case y-1-x will not be exactly representable any more and the
102 result can be off by many ulps. But this is easily fixed: for a
103 floating-point number |x| < DBL_EPSILON/2., the closest
104 floating-point number to log(1+x) is exactly x.
105 */
106
107 double y;
108 if (fabs(x) < DBL_EPSILON/2.) {
109 return x;
110 } else if (-0.5 <= x && x <= 1.) {
111 /* WARNING: it's possible than an overeager compiler
112 will incorrectly optimize the following two lines
113 to the equivalent of "return log(1.+x)". If this
114 happens, then results from log1p will be inaccurate
115 for small x. */
116 y = 1.+x;
117 return log(y)-((y-1.)-x)/y;
118 } else {
119 /* NaNs and infinities should end up here */
120 return log(1.+x);
121 }
122}
123#endif /* HAVE_LOG1P */
124
125/*
126 * ====================================================
127 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
128 *
129 * Developed at SunPro, a Sun Microsystems, Inc. business.
130 * Permission to use, copy, modify, and distribute this
131 * software is freely granted, provided that this notice
132 * is preserved.
133 * ====================================================
134 */
135
136static const double ln2 = 6.93147180559945286227E-01;
137static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
138static const double two_pow_p28 = 268435456.0; /* 2**28 */
139static const double zero = 0.0;
140
141/* asinh(x)
142 * Method :
143 * Based on
144 * asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
145 * we have
146 * asinh(x) := x if 1+x*x=1,
147 * := sign(x)*(log(x)+ln2)) for large |x|, else
148 * := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
149 * := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
150 */
151
152#ifndef HAVE_ASINH
153double
154asinh(double x)
155{
156 double w;
157 double absx = fabs(x);
158
159 if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
160 return x+x;
161 }
162 if (absx < two_pow_m28) { /* |x| < 2**-28 */
163 return x; /* return x inexact except 0 */
164 }
165 if (absx > two_pow_p28) { /* |x| > 2**28 */
166 w = log(absx)+ln2;
167 }
168 else if (absx > 2.0) { /* 2 < |x| < 2**28 */
169 w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
170 }
171 else { /* 2**-28 <= |x| < 2= */
172 double t = x*x;
173 w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
174 }
175 return copysign(w, x);
176
177}
178#endif /* HAVE_ASINH */
179
180/* acosh(x)
181 * Method :
182 * Based on
183 * acosh(x) = log [ x + sqrt(x*x-1) ]
184 * we have
185 * acosh(x) := log(x)+ln2, if x is large; else
186 * acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
187 * acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
188 *
189 * Special cases:
190 * acosh(x) is NaN with signal if x<1.
191 * acosh(NaN) is NaN without signal.
192 */
193
194#ifndef HAVE_ACOSH
195double
196acosh(double x)
197{
198 if (Py_IS_NAN(x)) {
199 return x+x;
200 }
201 if (x < 1.) { /* x < 1; return a signaling NaN */
202 errno = EDOM;
203#ifdef Py_NAN
204 return Py_NAN;
205#else
206 return (x-x)/(x-x);
207#endif
208 }
209 else if (x >= two_pow_p28) { /* x > 2**28 */
210 if (Py_IS_INFINITY(x)) {
211 return x+x;
212 } else {
213 return log(x)+ln2; /* acosh(huge)=log(2x) */
214 }
215 }
216 else if (x == 1.) {
217 return 0.0; /* acosh(1) = 0 */
218 }
219 else if (x > 2.) { /* 2 < x < 2**28 */
220 double t = x*x;
221 return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
222 }
223 else { /* 1 < x <= 2 */
224 double t = x - 1.0;
225 return log1p(t + sqrt(2.0*t + t*t));
226 }
227}
228#endif /* HAVE_ACOSH */
229
230/* atanh(x)
231 * Method :
232 * 1.Reduced x to positive by atanh(-x) = -atanh(x)
233 * 2.For x>=0.5
234 * 1 2x x
235 * atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)
236 * 2 1 - x 1 - x
237 *
238 * For x<0.5
239 * atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
240 *
241 * Special cases:
242 * atanh(x) is NaN if |x| >= 1 with signal;
243 * atanh(NaN) is that NaN with no signal;
244 *
245 */
246
247#ifndef HAVE_ATANH
248double
249atanh(double x)
250{
251 double absx;
252 double t;
253
254 if (Py_IS_NAN(x)) {
255 return x+x;
256 }
257 absx = fabs(x);
258 if (absx >= 1.) { /* |x| >= 1 */
259 errno = EDOM;
260#ifdef Py_NAN
261 return Py_NAN;
262#else
263 return x/zero;
264#endif
265 }
266 if (absx < two_pow_m28) { /* |x| < 2**-28 */
267 return x;
268 }
269 if (absx < 0.5) { /* |x| < 0.5 */
270 t = absx+absx;
271 t = 0.5 * log1p(t + t*absx / (1.0 - absx));
272 }
273 else { /* 0.5 <= |x| <= 1.0 */
274 t = 0.5 * log1p((absx + absx) / (1.0 - absx));
275 }
276 return copysign(t, x);
277}
278#endif /* HAVE_ATANH */