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📄 s_expm1l.c

📁 Glibc 2.3.2源代码(解压后有100多M)
💻 C
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/*							expm1l.c * *	Exponential function, minus 1 *      128-bit long double precision * * * * SYNOPSIS: * * long double x, y, expm1l(); * * y = expm1l( x ); * * * * DESCRIPTION: * * Returns e (2.71828...) raised to the x power, minus one. * * Range reduction is accomplished by separating the argument * into an integer k and fraction f such that * *     x    k  f *    e  = 2  e. * * An expansion x + .5 x^2 + x^3 R(x) approximates exp(f) - 1 * in the basic range [-0.5 ln 2, 0.5 ln 2]. * * * ACCURACY: * *                      Relative error: * arithmetic   domain     # trials      peak         rms *    IEEE    -79,+MAXLOG    100,000     1.7e-34     4.5e-35 * *//* Copyright 2001 by Stephen L. Moshier     This library is free software; you can redistribute it and/or    modify it under the terms of the GNU Lesser General Public    License as published by the Free Software Foundation; either    version 2.1 of the License, or (at your option) any later version.    This library is distributed in the hope that it will be useful,    but WITHOUT ANY WARRANTY; without even the implied warranty of    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU    Lesser General Public License for more details.    You should have received a copy of the GNU Lesser General Public    License along with this library; if not, write to the Free Software    Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307  USA */#include "math.h"#include "math_private.h"/* exp(x) - 1 = x + 0.5 x^2 + x^3 P(x)/Q(x)   -.5 ln 2  <  x  <  .5 ln 2   Theoretical peak relative error = 8.1e-36  */static const long double  P0 = 2.943520915569954073888921213330863757240E8L,  P1 = -5.722847283900608941516165725053359168840E7L,  P2 = 8.944630806357575461578107295909719817253E6L,  P3 = -7.212432713558031519943281748462837065308E5L,  P4 = 4.578962475841642634225390068461943438441E4L,  P5 = -1.716772506388927649032068540558788106762E3L,  P6 = 4.401308817383362136048032038528753151144E1L,  P7 = -4.888737542888633647784737721812546636240E-1L,  Q0 = 1.766112549341972444333352727998584753865E9L,  Q1 = -7.848989743695296475743081255027098295771E8L,  Q2 = 1.615869009634292424463780387327037251069E8L,  Q3 = -2.019684072836541751428967854947019415698E7L,  Q4 = 1.682912729190313538934190635536631941751E6L,  Q5 = -9.615511549171441430850103489315371768998E4L,  Q6 = 3.697714952261803935521187272204485251835E3L,  Q7 = -8.802340681794263968892934703309274564037E1L,  /* Q8 = 1.000000000000000000000000000000000000000E0 *//* C1 + C2 = ln 2 */  C1 = 6.93145751953125E-1L,  C2 = 1.428606820309417232121458176568075500134E-6L,/* ln (2^16384 * (1 - 2^-113)) */  maxlog = 1.1356523406294143949491931077970764891253E4L,/* ln 2^-114 */  minarg = -7.9018778583833765273564461846232128760607E1L, big = 2e4932L;long double__expm1l (long double x){  long double px, qx, xx;  int32_t ix, sign;  ieee854_long_double_shape_type u;  int k;  /* Detect infinity and NaN.  */  u.value = x;  ix = u.parts32.w0;  sign = ix & 0x80000000;  ix &= 0x7fffffff;  if (ix >= 0x7fff0000)    {      /* Infinity. */      if (((ix & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3) == 0)	{	  if (sign)	    return -1.0L;	  else	    return x;	}      /* NaN. No invalid exception. */      return x;    }  /* expm1(+- 0) = +- 0.  */  if ((ix == 0) && (u.parts32.w1 | u.parts32.w2 | u.parts32.w3) == 0)    return x;  /* Overflow.  */  if (x > maxlog)    return (big * big);  /* Minimum value.  */  if (x < minarg)    return (4.0/big - 1.0L);  /* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */  xx = C1 + C2;			/* ln 2. */  px = __floorl (0.5 + x / xx);  k = px;  /* remainder times ln 2 */  x -= px * C1;  x -= px * C2;  /* Approximate exp(remainder ln 2).  */  px = (((((((P7 * x	      + P6) * x	     + P5) * x + P4) * x + P3) * x + P2) * x + P1) * x + P0) * x;  qx = (((((((x	      + Q7) * x	     + Q6) * x + Q5) * x + Q4) * x + Q3) * x + Q2) * x + Q1) * x + Q0;  xx = x * x;  qx = x + (0.5 * xx + xx * px / qx);  /* exp(x) = exp(k ln 2) exp(remainder ln 2) = 2^k exp(remainder ln 2).  We have qx = exp(remainder ln 2) - 1, so  exp(x) - 1 = 2^k (qx + 1) - 1             = 2^k qx + 2^k - 1.  */  px = ldexpl (1.0L, k);  x = px * qx + (px - 1.0);  return x;}libm_hidden_def (__expm1l)weak_alias (__expm1l, expm1l)

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