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	 b08a53a99d
			
		
	
	
		b08a53a99d
		
	
	
	
	
		
			
			- incorporate and adapt David Gay's dtoa and strtod into the Python core - on platforms where we can use Gay's code (almost all!), repr(float) is based on the shortest sequence of decimal digits that rounds correctly. - add sys.float_repr_style attribute to indicate whether we're using Gay's code or not - add autoconf magic to detect and enable SSE2 instructions on x86/gcc - slight change to repr and str: repr switches to exponential notation at 1e16 instead of 1e17, str switches at 1e11 instead of 1e12
		
			
				
	
	
		
			269 lines
		
	
	
	
		
			6.1 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
			
		
		
	
	
			269 lines
		
	
	
	
		
			6.1 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
| #include "Python.h"
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| 
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| #ifdef X87_DOUBLE_ROUNDING
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| /* On x86 platforms using an x87 FPU, this function is called from the
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|    Py_FORCE_DOUBLE macro (defined in pymath.h) to force a floating-point
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|    number out of an 80-bit x87 FPU register and into a 64-bit memory location,
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|    thus rounding from extended precision to double precision. */
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| double _Py_force_double(double x)
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| {
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| 	volatile double y;
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| 	y = x;
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| 	return y;
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| }
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| #endif
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| 
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| #ifdef USING_X87_FPU
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| #  ifdef HAVE_GCC_ASM_FOR_X87
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| 
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| /* inline assembly for getting and setting the 387 FPU control word on
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|    gcc/x86 */
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| 
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| unsigned short _Py_get_387controlword(void) {
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|     unsigned short cw;
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|     __asm__ __volatile__ ("fnstcw %0" : "=m" (cw));
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|     return cw;
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| }
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| 
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| void _Py_set_387controlword(unsigned short cw) {
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|     __asm__ __volatile__ ("fldcw %0" : : "m" (cw));
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| }
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| 
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| #  else
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| #    error "Unable to get and set x87 control word"
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| #  endif
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| #endif
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| 
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| 
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| #ifndef HAVE_HYPOT
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| double hypot(double x, double y)
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| {
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| 	double yx;
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| 
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| 	x = fabs(x);
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| 	y = fabs(y);
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| 	if (x < y) {
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| 		double temp = x;
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| 		x = y;
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| 		y = temp;
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| 	}
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| 	if (x == 0.)
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| 		return 0.;
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| 	else {
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| 		yx = y/x;
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| 		return x*sqrt(1.+yx*yx);
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| 	}
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| }
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| #endif /* HAVE_HYPOT */
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| 
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| #ifndef HAVE_COPYSIGN
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| static double
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| copysign(double x, double y)
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| {
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| 	/* use atan2 to distinguish -0. from 0. */
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| 	if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {
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| 		return fabs(x);
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| 	} else {
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| 		return -fabs(x);
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| 	}
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| }
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| #endif /* HAVE_COPYSIGN */
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| 
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| #ifndef HAVE_LOG1P
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| #include <float.h>
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| 
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| double
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| log1p(double x)
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| {
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| 	/* For x small, we use the following approach.  Let y be the nearest
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| 	   float to 1+x, then
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| 
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| 	     1+x = y * (1 - (y-1-x)/y)
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| 
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| 	   so log(1+x) = log(y) + log(1-(y-1-x)/y).  Since (y-1-x)/y is tiny,
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| 	   the second term is well approximated by (y-1-x)/y.  If abs(x) >=
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| 	   DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
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| 	   then y-1-x will be exactly representable, and is computed exactly
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| 	   by (y-1)-x.
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| 
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| 	   If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
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| 	   round-to-nearest then this method is slightly dangerous: 1+x could
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| 	   be rounded up to 1+DBL_EPSILON instead of down to 1, and in that
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| 	   case y-1-x will not be exactly representable any more and the
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| 	   result can be off by many ulps.  But this is easily fixed: for a
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| 	   floating-point number |x| < DBL_EPSILON/2., the closest
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| 	   floating-point number to log(1+x) is exactly x.
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| 	*/
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| 
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| 	double y;
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| 	if (fabs(x) < DBL_EPSILON/2.) {
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| 		return x;
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| 	} else if (-0.5 <= x && x <= 1.) {
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| 		/* WARNING: it's possible than an overeager compiler
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| 		   will incorrectly optimize the following two lines
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| 		   to the equivalent of "return log(1.+x)". If this
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| 		   happens, then results from log1p will be inaccurate
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| 		   for small x. */
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| 		y = 1.+x;
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| 		return log(y)-((y-1.)-x)/y;
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| 	} else {
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| 		/* NaNs and infinities should end up here */
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| 		return log(1.+x);
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| 	}
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| }
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| #endif /* HAVE_LOG1P */
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| 
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| /*
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|  * ====================================================
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|  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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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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|  * is preserved.
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|  * ====================================================
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|  */
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| 
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| static const double ln2 = 6.93147180559945286227E-01;
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| static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
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| static const double two_pow_p28 = 268435456.0; /* 2**28 */
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| static const double zero = 0.0;
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| 
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| /* asinh(x)
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|  * Method :
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|  *	Based on 
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|  *		asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
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|  *	we have
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|  *	asinh(x) := x  if  1+x*x=1,
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|  *		 := sign(x)*(log(x)+ln2)) for large |x|, else
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|  *		 := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
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|  *		 := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))  
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|  */
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| 
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| #ifndef HAVE_ASINH
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| double
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| asinh(double x)
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| {	
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| 	double w;
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| 	double absx = fabs(x);
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| 
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| 	if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
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| 		return x+x;
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| 	}
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| 	if (absx < two_pow_m28) {	/* |x| < 2**-28 */
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| 		return x;	/* return x inexact except 0 */
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| 	} 
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| 	if (absx > two_pow_p28) {	/* |x| > 2**28 */
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| 		w = log(absx)+ln2;
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| 	}
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| 	else if (absx > 2.0) {		/* 2 < |x| < 2**28 */
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| 		w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
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| 	}
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| 	else {				/* 2**-28 <= |x| < 2= */
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| 		double t = x*x;
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| 		w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
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| 	}
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| 	return copysign(w, x);
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| 	
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| }
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| #endif /* HAVE_ASINH */
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| 
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| /* acosh(x)
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|  * Method :
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|  *      Based on
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|  *	      acosh(x) = log [ x + sqrt(x*x-1) ]
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|  *      we have
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|  *	      acosh(x) := log(x)+ln2, if x is large; else
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|  *	      acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
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|  *	      acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
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|  *
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|  * Special cases:
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|  *      acosh(x) is NaN with signal if x<1.
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|  *      acosh(NaN) is NaN without signal.
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|  */
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| 
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| #ifndef HAVE_ACOSH
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| double
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| acosh(double x)
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| {
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| 	if (Py_IS_NAN(x)) {
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| 		return x+x;
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| 	}
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| 	if (x < 1.) {			/* x < 1;  return a signaling NaN */
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| 		errno = EDOM;
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| #ifdef Py_NAN
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| 		return Py_NAN;
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| #else
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| 		return (x-x)/(x-x);
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| #endif
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| 	}
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| 	else if (x >= two_pow_p28) {	/* x > 2**28 */
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| 		if (Py_IS_INFINITY(x)) {
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| 			return x+x;
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| 		} else {
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| 			return log(x)+ln2;	/* acosh(huge)=log(2x) */
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| 		}
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| 	}
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| 	else if (x == 1.) {
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| 		return 0.0;			/* acosh(1) = 0 */
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| 	}
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| 	else if (x > 2.) {			/* 2 < x < 2**28 */
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| 		double t = x*x;
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| 		return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
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| 	}
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| 	else {				/* 1 < x <= 2 */
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| 		double t = x - 1.0;
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| 		return log1p(t + sqrt(2.0*t + t*t));
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| 	}
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| }
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| #endif /* HAVE_ACOSH */
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| 
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| /* atanh(x)
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|  * Method :
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|  *    1.Reduced x to positive by atanh(-x) = -atanh(x)
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|  *    2.For x>=0.5
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|  *		  1	      2x			  x
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|  *      atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)
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|  *		  2	     1 - x		      1 - x
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|  *
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|  *      For x<0.5
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|  *      atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
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|  *
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|  * Special cases:
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|  *      atanh(x) is NaN if |x| >= 1 with signal;
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|  *      atanh(NaN) is that NaN with no signal;
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|  *
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|  */
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| 
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| #ifndef HAVE_ATANH
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| double
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| atanh(double x)
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| {
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| 	double absx;
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| 	double t;
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| 
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| 	if (Py_IS_NAN(x)) {
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| 		return x+x;
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| 	}
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| 	absx = fabs(x);
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| 	if (absx >= 1.) {		/* |x| >= 1 */
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| 		errno = EDOM;
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| #ifdef Py_NAN
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| 		return Py_NAN;
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| #else
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| 		return x/zero;
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| #endif
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| 	}
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| 	if (absx < two_pow_m28) {	/* |x| < 2**-28 */
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| 		return x;
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| 	}
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| 	if (absx < 0.5) {		/* |x| < 0.5 */
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| 		t = absx+absx;
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| 		t = 0.5 * log1p(t + t*absx / (1.0 - absx));
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| 	} 
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| 	else {				/* 0.5 <= |x| <= 1.0 */
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| 		t = 0.5 * log1p((absx + absx) / (1.0 - absx));
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| 	}
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| 	return copysign(t, x);
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| }
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| #endif /* HAVE_ATANH */
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