Revert (leaving the complex math part alone). Some stuff is depending

on this historical behavior;  so we're stuck in this stupid situation.
No cookie for me.

1 files changed, 104 insertions, 0 deletions

diff --git a/lib/libc/gen/modf.c b/lib/libc/gen/modf.c
new file mode 100644
index 00000000000..24effd38045
--- /dev/null
+++ b/lib/libc/gen/modf.c
@@ -0,0 +1,104 @@
+/*	$OpenBSD: modf.c,v 1.3 2011/07/08 22:28:33 martynas 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(double val, double *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);
+}