- ldexp implementation has issues. switch to the one from libm

- remove frexp in hppa64, cloned from hppa
- move generic ieee754 implementations of modf and ldexp to gen
ok kettenis@, "looks good" millert@

1 files changed, 0 insertions, 105 deletions

diff --git a/lib/libc/arch/alpha/gen/modf.c b/lib/libc/arch/alpha/gen/modf.c
deleted file mode 100644
index 2315ef19034..00000000000
--- a/lib/libc/arch/alpha/gen/modf.c
+++ /dev/null
@@ -1,105 +0,0 @@
-/*	$OpenBSD: modf.c,v 1.5 2005/08/07 16:40:13 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 <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 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);
-}