[multiple changes]
Fri May 28 22:20:03 1999 Anthony Green <green@cygnus.com> * java/lang/fdlibm.h: Don't use __uint32_t. Include mprec.h. * java/lang/e_log.c: Don't use __uint32_t. 1999-05-27 Eric Christopher <echristo@cygnus.com> * configure: Rebuilt * configure.in: Fixed ISO C9X and namespace collision with __uint32_t * acconfig.h: Rebuilt * include/config.h.in: Rebuilt * java/lang/mprec.h, java/lang/e_acos.c, java/lang/e_asin.c, java/lang/e_atan2.c, java/lang/e_exp.c, java/lang/e_fmod.c, e_log.c, java/lang/e_pow.c, java/lang/e_rem_pio2.c, java/lang/e_remainder.c, java/lang/e_sqrt.c, java/lang/fdlibm.h, k_tan.c, java/lang/mprec.h, java/lang/s_atan.c, java/lang/s_ceil.c, java/lang/s_copysign.c, java/lang/s_fabs.c, s_floor.c, java/lang/s_rint.c, java/lang/sf_rint.c: Fixed ISO C9X and namespace collision with __uint32_t From-SVN: r27729
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31 changed files with 614 additions and 526 deletions
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@ -6,7 +6,7 @@
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*
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* Developed at SunPro, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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@ -15,25 +15,25 @@
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* kernel tan function on [-pi/4, pi/4], pi/4 ~ 0.7854
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* Input x is assumed to be bounded by ~pi/4 in magnitude.
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* Input y is the tail of x.
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* Input k indicates whether tan (if k=1) or
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* Input k indicates whether tan (if k=1) or
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* -1/tan (if k= -1) is returned.
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*
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* Algorithm
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* 1. Since tan(-x) = -tan(x), we need only to consider positive x.
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* 1. Since tan(-x) = -tan(x), we need only to consider positive x.
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* 2. if x < 2^-28 (hx<0x3e300000 0), return x with inexact if x!=0.
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* 3. tan(x) is approximated by a odd polynomial of degree 27 on
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* [0,0.67434]
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* 3 27
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* tan(x) ~ x + T1*x + ... + T13*x
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* where
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*
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*
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* |tan(x) 2 4 26 | -59.2
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* |----- - (1+T1*x +T2*x +.... +T13*x )| <= 2
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* | x |
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*
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* | x |
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*
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* Note: tan(x+y) = tan(x) + tan'(x)*y
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* ~ tan(x) + (1+x*x)*y
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* Therefore, for better accuracy in computing tan(x+y), let
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* Therefore, for better accuracy in computing tan(x+y), let
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* 3 2 2 2 2
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* r = x *(T2+x *(T3+x *(...+x *(T12+x *T13))))
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* then
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@ -50,9 +50,9 @@
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#ifndef _DOUBLE_IS_32BITS
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#ifdef __STDC__
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static const double
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static const double
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#else
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static double
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static double
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#endif
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one = 1.00000000000000000000e+00, /* 0x3FF00000, 0x00000000 */
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pio4 = 7.85398163397448278999e-01, /* 0x3FE921FB, 0x54442D18 */
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@ -81,12 +81,12 @@ T[] = {
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#endif
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{
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double z,r,v,w,s;
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__int32_t ix,hx;
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int32_t ix,hx;
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GET_HIGH_WORD(hx,x);
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ix = hx&0x7fffffff; /* high word of |x| */
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if(ix<0x3e300000) /* x < 2**-28 */
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{if((int)x==0) { /* generate inexact */
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__uint32_t low;
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uint32_t low;
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GET_LOW_WORD(low,x);
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if(((ix|low)|(iy+1))==0) return one/fabs(x);
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else return (iy==1)? x: -one/x;
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@ -115,7 +115,7 @@ T[] = {
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return (double)(1-((hx>>30)&2))*(v-2.0*(x-(w*w/(w+v)-r)));
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}
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if(iy==1) return w;
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else { /* if allow error up to 2 ulp,
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else { /* if allow error up to 2 ulp,
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simply return -1.0/(x+r) here */
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/* compute -1.0/(x+r) accurately */
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double a,t;
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