freebsd-dev/lib/libc/ia64/gen/ldexp.c
2002-03-21 18:49:23 +00:00

136 lines
3.7 KiB
C

/* $NetBSD: ldexp.c,v 1.1 1995/02/10 17:50:24 cgd Exp $ */
/* $FreeBSD$ */
/*
* 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 ldexp(double val, int exp)
* returns: val * (2**exp)
*/
double
ldexp(val, exp)
double val;
int exp;
{
int oldexp, newexp, mulexp;
union doub {
double v;
struct ieee_double s;
} u, mul;
/*
* If input is zero, or no change, just return input.
* Likewise, if input is Inf or NaN, just return it.
*/
u.v = val;
oldexp = u.s.dbl_exp;
if (val == 0 || exp == 0 || oldexp == DBL_EXP_INFNAN)
return (val);
/*
* Compute new exponent and check for over/under flow.
* Underflow, unfortunately, could mean switching to denormal.
* If result out of range, set ERANGE and return 0 if too small
* or Inf if too big, with the same sign as the input value.
*/
newexp = oldexp + exp;
if (newexp >= DBL_EXP_INFNAN) {
/* u.s.dbl_sign = val < 0; -- already set */
u.s.dbl_exp = DBL_EXP_INFNAN;
u.s.dbl_frach = u.s.dbl_fracl = 0;
errno = ERANGE;
return (u.v); /* Inf */
}
if (newexp <= 0) {
/*
* The output number is either a denormal or underflows
* (see comments in machine/ieee.h).
*/
if (newexp <= -DBL_FRACBITS) {
/* u.s.dbl_sign = val < 0; -- already set */
u.s.dbl_exp = 0;
u.s.dbl_frach = u.s.dbl_fracl = 0;
errno = ERANGE;
return (u.v); /* zero */
}
/*
* We are going to produce a denorm. Our `exp' argument
* might be as small as -2097, and we cannot compute
* 2^-2097, so we may have to do this as many as three
* steps (not just two, as for positive `exp's below).
*/
mul.v = 0;
while (exp <= -DBL_EXP_BIAS) {
mul.s.dbl_exp = 1;
val *= mul.v;
exp += DBL_EXP_BIAS - 1;
}
mul.s.dbl_exp = exp + DBL_EXP_BIAS;
val *= mul.v;
return (val);
}
/*
* Newexp is positive.
*
* If oldexp is zero, we are starting with a denorm, and simply
* adjusting the exponent will produce bogus answers. We need
* to fix that first.
*/
if (oldexp == 0) {
/*
* Multiply by 2^mulexp to make the number normalizable.
* We cannot multiply by more than 2^1023, but `exp'
* argument might be as large as 2046. A single
* adjustment, however, will normalize the number even
* for huge `exp's, and then we can use exponent
* arithmetic just as for normal `double's.
*/
mulexp = exp <= DBL_EXP_BIAS ? exp : DBL_EXP_BIAS;
mul.v = 0;
mul.s.dbl_exp = mulexp + DBL_EXP_BIAS;
val *= mul.v;
if (mulexp == exp)
return (val);
u.v = val;
newexp -= mulexp;
}
/*
* Both oldexp and newexp are positive; just replace the
* old exponent with the new one.
*/
u.s.dbl_exp = newexp;
return (u.v);
}