 b6cbf7203b
			
		
	
	
		b6cbf7203b
		
	
	
	
	
		
			
			This patch imports the unmodified current version of NetBSD libc. The NetBSD includes are in /nbsd_include, while the libc code itself is split between lib/nbsd_libc and common/lib/libc.
		
			
				
	
	
		
			104 lines
		
	
	
		
			3.2 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
			
		
		
	
	
			104 lines
		
	
	
		
			3.2 KiB
		
	
	
	
		
			C
		
	
	
	
	
	
| /* $NetBSD: compat_modf_ieee754.c,v 1.4 2010/04/23 19:04:54 drochner Exp $ */
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| 
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| /*
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|  * Copyright (c) 1994, 1995 Carnegie-Mellon University.
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|  * All rights reserved.
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|  *
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|  * Author: Chris G. Demetriou
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|  * 
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|  * Permission to use, copy, modify and distribute this software and
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|  * its documentation is hereby granted, provided that both the copyright
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|  * notice and this permission notice appear in all copies of the
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|  * software, derivative works or modified versions, and any portions
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|  * thereof, and that both notices appear in supporting documentation.
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|  * 
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|  * CARNEGIE MELLON ALLOWS FREE USE OF THIS SOFTWARE IN ITS "AS IS" 
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|  * CONDITION.  CARNEGIE MELLON DISCLAIMS ANY LIABILITY OF ANY KIND 
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|  * FOR ANY DAMAGES WHATSOEVER RESULTING FROM THE USE OF THIS SOFTWARE.
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|  * 
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|  * Carnegie Mellon requests users of this software to return to
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|  *
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|  *  Software Distribution Coordinator  or  Software.Distribution@CS.CMU.EDU
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|  *  School of Computer Science
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|  *  Carnegie Mellon University
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|  *  Pittsburgh PA 15213-3890
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|  *
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|  * any improvements or extensions that they make and grant Carnegie the
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|  * rights to redistribute these changes.
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|  */
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| 
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| #include <sys/types.h>
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| #include <machine/ieee.h>
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| #include <errno.h>
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| 
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| double modf(double, double *);
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| 
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| /*
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|  * double modf(double val, double *iptr)
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|  * returns: f and i such that |f| < 1.0, (f + i) = val, and
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|  *	sign(f) == sign(i) == sign(val).
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|  *
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|  * Beware signedness when doing subtraction, and also operand size!
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|  */
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| double
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| modf(double val, double *iptr)
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| {
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| 	union ieee_double_u u, v;
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| 	u_int64_t frac;
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| 
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| 	/*
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| 	 * If input is +/-Inf or NaN, return +/-0 or NaN.
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| 	 */
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| 	u.dblu_d = val;
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| 	if (u.dblu_dbl.dbl_exp == DBL_EXP_INFNAN) {
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| 		*iptr = u.dblu_d;
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| 		return (0.0 / u.dblu_d);
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| 	}
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| 
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| 	/*
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| 	 * If input can't have a fractional part, return
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| 	 * (appropriately signed) zero, and make i be the input.
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| 	 */
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| 	if ((int)u.dblu_dbl.dbl_exp - DBL_EXP_BIAS > DBL_FRACBITS - 1) {
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| 		*iptr = u.dblu_d;
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| 		v.dblu_d = 0.0;
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| 		v.dblu_dbl.dbl_sign = u.dblu_dbl.dbl_sign;
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| 		return (v.dblu_d);
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| 	}
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| 
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| 	/*
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| 	 * If |input| < 1.0, return it, and set i to the appropriately
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| 	 * signed zero.
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| 	 */
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| 	if (u.dblu_dbl.dbl_exp < DBL_EXP_BIAS) {
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| 		v.dblu_d = 0.0;
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| 		v.dblu_dbl.dbl_sign = u.dblu_dbl.dbl_sign;
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| 		*iptr = v.dblu_d;
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| 		return (u.dblu_d);
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| 	}
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| 
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| 	/*
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| 	 * There can be a fractional part of the input.
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| 	 * If you look at the math involved for a few seconds, it's
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| 	 * plain to see that the integral part is the input, with the
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| 	 * low (DBL_FRACBITS - (exponent - DBL_EXP_BIAS)) bits zeroed,
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| 	 * the fractional part is the part with the rest of the
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| 	 * bits zeroed.  Just zeroing the high bits to get the
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| 	 * fractional part would yield a fraction in need of
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| 	 * normalization.  Therefore, we take the easy way out, and
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| 	 * just use subtraction to get the fractional part.
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| 	 */
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| 	v.dblu_d = u.dblu_d;
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| 	/* Zero the low bits of the fraction, the sleazy way. */
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| 	frac = ((u_int64_t)v.dblu_dbl.dbl_frach << 32) + v.dblu_dbl.dbl_fracl;
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| 	frac >>= DBL_FRACBITS - (u.dblu_dbl.dbl_exp - DBL_EXP_BIAS);
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| 	frac <<= DBL_FRACBITS - (u.dblu_dbl.dbl_exp - DBL_EXP_BIAS);
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| 	v.dblu_dbl.dbl_fracl = (unsigned int)frac & 0xffffffff;
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| 	v.dblu_dbl.dbl_frach = (unsigned int)(frac >> 32);
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| 	*iptr = v.dblu_d;
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| 
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| 	u.dblu_d -= v.dblu_d;
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| 	u.dblu_dbl.dbl_sign = v.dblu_dbl.dbl_sign;
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| 	return (u.dblu_d);
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| }
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