/* $OpenBSD: modf.c,v 1.2 2005/08/07 16:40:14 espie Exp $ */ /* $NetBSD: modf.c,v 1.1 1995/02/10 17:50:25 cgd Exp $ */ /* * Copyright (c) 1994, 1995 Carnegie-Mellon University. * All rights reserved. * * Author: Chris G. Demetriou * * Permission to use, copy, modify and distribute this software and * its documentation is hereby granted, provided that both the copyright * notice and this permission notice appear in all copies of the * software, derivative works or modified versions, and any portions * thereof, and that both notices appear in supporting documentation. * * CARNEGIE MELLON ALLOWS FREE USE OF THIS SOFTWARE IN ITS "AS IS" * CONDITION. CARNEGIE MELLON DISCLAIMS ANY LIABILITY OF ANY KIND * FOR ANY DAMAGES WHATSOEVER RESULTING FROM THE USE OF THIS SOFTWARE. * * Carnegie Mellon requests users of this software to return to * * Software Distribution Coordinator or Software.Distribution@CS.CMU.EDU * School of Computer Science * Carnegie Mellon University * Pittsburgh PA 15213-3890 * * any improvements or extensions that they make and grant Carnegie the * rights to redistribute these changes. */ #include #include #include #include /* * double modf(double val, double *iptr) * returns: f and i such that |f| < 1.0, (f + i) = val, and * sign(f) == sign(i) == sign(val). * * Beware signedness when doing subtraction, and also operand size! */ double modf(val, iptr) double val, *iptr; { union doub { double v; struct ieee_double s; } u, v; u_int64_t frac; /* * If input is Inf or NaN, return it and leave i alone. */ u.v = val; if (u.s.dbl_exp == DBL_EXP_INFNAN) return (u.v); /* * If input can't have a fractional part, return * (appropriately signed) zero, and make i be the input. */ if ((int)u.s.dbl_exp - DBL_EXP_BIAS > DBL_FRACBITS - 1) { *iptr = u.v; v.v = 0.0; v.s.dbl_sign = u.s.dbl_sign; return (v.v); } /* * If |input| < 1.0, return it, and set i to the appropriately * signed zero. */ if (u.s.dbl_exp < DBL_EXP_BIAS) { v.v = 0.0; v.s.dbl_sign = u.s.dbl_sign; *iptr = v.v; return (u.v); } /* * There can be a fractional part of the input. * If you look at the math involved for a few seconds, it's * plain to see that the integral part is the input, with the * low (DBL_FRACBITS - (exponent - DBL_EXP_BIAS)) bits zeroed, * the fractional part is the part with the rest of the * bits zeroed. Just zeroing the high bits to get the * fractional part would yield a fraction in need of * normalization. Therefore, we take the easy way out, and * just use subtraction to get the fractional part. */ v.v = u.v; /* Zero the low bits of the fraction, the sleazy way. */ frac = ((u_int64_t)v.s.dbl_frach << 32) + v.s.dbl_fracl; frac >>= DBL_FRACBITS - (u.s.dbl_exp - DBL_EXP_BIAS); frac <<= DBL_FRACBITS - (u.s.dbl_exp - DBL_EXP_BIAS); v.s.dbl_fracl = frac & 0xffffffff; v.s.dbl_frach = frac >> 32; *iptr = v.v; u.v -= v.v; u.s.dbl_sign = v.s.dbl_sign; return (u.v); }