xref: /netbsd-src/lib/libc/compat/gen/compat_modf_ieee754.c (revision d48f14661dda8638fee055ba15d35bdfb29b9fa8)
1 /* $NetBSD: compat_modf_ieee754.c,v 1.1 2006/06/27 18:16:47 drochner Exp $ */
2 
3 /*
4  * Copyright (c) 1994, 1995 Carnegie-Mellon University.
5  * All rights reserved.
6  *
7  * Author: Chris G. Demetriou
8  *
9  * Permission to use, copy, modify and distribute this software and
10  * its documentation is hereby granted, provided that both the copyright
11  * notice and this permission notice appear in all copies of the
12  * software, derivative works or modified versions, and any portions
13  * thereof, and that both notices appear in supporting documentation.
14  *
15  * CARNEGIE MELLON ALLOWS FREE USE OF THIS SOFTWARE IN ITS "AS IS"
16  * CONDITION.  CARNEGIE MELLON DISCLAIMS ANY LIABILITY OF ANY KIND
17  * FOR ANY DAMAGES WHATSOEVER RESULTING FROM THE USE OF THIS SOFTWARE.
18  *
19  * Carnegie Mellon requests users of this software to return to
20  *
21  *  Software Distribution Coordinator  or  Software.Distribution@CS.CMU.EDU
22  *  School of Computer Science
23  *  Carnegie Mellon University
24  *  Pittsburgh PA 15213-3890
25  *
26  * any improvements or extensions that they make and grant Carnegie the
27  * rights to redistribute these changes.
28  */
29 
30 #include <sys/types.h>
31 #include <machine/ieee.h>
32 #include <errno.h>
33 #include <math.h>
34 
35 /*
36  * double modf(double val, double *iptr)
37  * returns: f and i such that |f| < 1.0, (f + i) = val, and
38  *	sign(f) == sign(i) == sign(val).
39  *
40  * Beware signedness when doing subtraction, and also operand size!
41  */
42 double
43 modf(double val, double *iptr)
44 {
45 	union ieee_double_u u, v;
46 	u_int64_t frac;
47 
48 	/*
49 	 * If input is Inf or NaN, return it and leave i alone.
50 	 */
51 	u.dblu_d = val;
52 	if (u.dblu_dbl.dbl_exp == DBL_EXP_INFNAN)
53 		return (u.dblu_d);
54 
55 	/*
56 	 * If input can't have a fractional part, return
57 	 * (appropriately signed) zero, and make i be the input.
58 	 */
59 	if ((int)u.dblu_dbl.dbl_exp - DBL_EXP_BIAS > DBL_FRACBITS - 1) {
60 		*iptr = u.dblu_d;
61 		v.dblu_d = 0.0;
62 		v.dblu_dbl.dbl_sign = u.dblu_dbl.dbl_sign;
63 		return (v.dblu_d);
64 	}
65 
66 	/*
67 	 * If |input| < 1.0, return it, and set i to the appropriately
68 	 * signed zero.
69 	 */
70 	if (u.dblu_dbl.dbl_exp < DBL_EXP_BIAS) {
71 		v.dblu_d = 0.0;
72 		v.dblu_dbl.dbl_sign = u.dblu_dbl.dbl_sign;
73 		*iptr = v.dblu_d;
74 		return (u.dblu_d);
75 	}
76 
77 	/*
78 	 * There can be a fractional part of the input.
79 	 * If you look at the math involved for a few seconds, it's
80 	 * plain to see that the integral part is the input, with the
81 	 * low (DBL_FRACBITS - (exponent - DBL_EXP_BIAS)) bits zeroed,
82 	 * the fractional part is the part with the rest of the
83 	 * bits zeroed.  Just zeroing the high bits to get the
84 	 * fractional part would yield a fraction in need of
85 	 * normalization.  Therefore, we take the easy way out, and
86 	 * just use subtraction to get the fractional part.
87 	 */
88 	v.dblu_d = u.dblu_d;
89 	/* Zero the low bits of the fraction, the sleazy way. */
90 	frac = ((u_int64_t)v.dblu_dbl.dbl_frach << 32) + v.dblu_dbl.dbl_fracl;
91 	frac >>= DBL_FRACBITS - (u.dblu_dbl.dbl_exp - DBL_EXP_BIAS);
92 	frac <<= DBL_FRACBITS - (u.dblu_dbl.dbl_exp - DBL_EXP_BIAS);
93 	v.dblu_dbl.dbl_fracl = (unsigned int)frac & 0xffffffff;
94 	v.dblu_dbl.dbl_frach = (unsigned int)(frac >> 32);
95 	*iptr = v.dblu_d;
96 
97 	u.dblu_d -= v.dblu_d;
98 	u.dblu_dbl.dbl_sign = v.dblu_dbl.dbl_sign;
99 	return (u.dblu_d);
100 }
101