NetBSD/lib/libc/arch/alpha/gen/modf.c

105 lines
2.9 KiB
C

/* $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 <sys/types.h>
#include <machine/ieee.h>
#include <errno.h>
#include <math.h>
/*
* 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 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);
}