1 /*
2  * CDDL HEADER START
3  *
4  * The contents of this file are subject to the terms of the
5  * Common Development and Distribution License, Version 1.0 only
6  * (the "License").  You may not use this file except in compliance
7  * with the License.
8  *
9  * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
10  * or http://www.opensolaris.org/os/licensing.
11  * See the License for the specific language governing permissions
12  * and limitations under the License.
13  *
14  * When distributing Covered Code, include this CDDL HEADER in each
15  * file and include the License file at usr/src/OPENSOLARIS.LICENSE.
16  * If applicable, add the following below this CDDL HEADER, with the
17  * fields enclosed by brackets "[]" replaced with your own identifying
18  * information: Portions Copyright [yyyy] [name of copyright owner]
19  *
20  * CDDL HEADER END
21  */
22 /*
23  * Copyright 2003 Sun Microsystems, Inc.  All rights reserved.
24  * Use is subject to license terms.
25  */
26 
27 #pragma ident	"%Z%%M%	%I%	%E% SMI"
28 
29 /*
30  * _D_cplx_mul(z, w) returns z * w with infinities handled according
31  * to C99.
32  *
33  * If z and w are both finite, _D_cplx_mul(z, w) delivers the complex
34  * product according to the usual formula: let a = Re(z), b = Im(z),
35  * c = Re(w), and d = Im(w); then _D_cplx_mul(z, w) delivers x + I * y
36  * where x = a * c - b * d and y = a * d + b * c.  Note that if both
37  * ac and bd overflow, then at least one of ad or bc must also over-
38  * flow, and vice versa, so that if one component of the product is
39  * NaN, the other is infinite.  (Such a value is considered infinite
40  * according to C99.)
41  *
42  * If one of z or w is infinite and the other is either finite nonzero
43  * or infinite, _D_cplx_mul delivers an infinite result.  If one factor
44  * is infinite and the other is zero, _D_cplx_mul delivers NaN + I * NaN.
45  * C99 doesn't specify the latter case.
46  *
47  * C99 also doesn't specify what should happen if either z or w is a
48  * complex NaN (i.e., neither finite nor infinite).  This implementation
49  * delivers NaN + I * NaN in this case.
50  *
51  * This implementation can raise spurious underflow, overflow, invalid
52  * operation, and inexact exceptions.  C99 allows this.
53  */
54 
55 #if !defined(sparc) && !defined(__sparc)
56 #error This code is for SPARC only
57 #endif
58 
59 static union {
60 	int	i[2];
61 	double	d;
62 } inf = {
63 	0x7ff00000, 0
64 };
65 
66 /*
67  * Return +1 if x is +Inf, -1 if x is -Inf, and 0 otherwise
68  */
69 static int
70 testinf(double x)
71 {
72 	union {
73 		int	i[2];
74 		double	d;
75 	} xx;
76 
77 	xx.d = x;
78 	return (((((xx.i[0] << 1) - 0xffe00000) | xx.i[1]) == 0)?
79 		(1 | (xx.i[0] >> 31)) : 0);
80 }
81 
82 double _Complex
83 _D_cplx_mul(double _Complex z, double _Complex w)
84 {
85 	double _Complex	v;
86 	double		a, b, c, d, x, y;
87 	int		recalc, i, j;
88 
89 	/*
90 	 * The following is equivalent to
91 	 *
92 	 *  a = creal(z); b = cimag(z);
93 	 *  c = creal(w); d = cimag(w);
94 	 */
95 	a = ((double *)&z)[0];
96 	b = ((double *)&z)[1];
97 	c = ((double *)&w)[0];
98 	d = ((double *)&w)[1];
99 
100 	x = a * c - b * d;
101 	y = a * d + b * c;
102 
103 	if (x != x && y != y) {
104 		/*
105 		 * Both x and y are NaN, so z and w can't both be finite.
106 		 * If at least one of z or w is a complex NaN, and neither
107 		 * is infinite, then we might as well deliver NaN + I * NaN.
108 		 * So the only cases to check are when one of z or w is
109 		 * infinite.
110 		 */
111 		recalc = 0;
112 		i = testinf(a);
113 		j = testinf(b);
114 		if (i | j) { /* z is infinite */
115 			/* "factor out" infinity */
116 			a = i;
117 			b = j;
118 			recalc = 1;
119 		}
120 		i = testinf(c);
121 		j = testinf(d);
122 		if (i | j) { /* w is infinite */
123 			/* "factor out" infinity */
124 			c = i;
125 			d = j;
126 			recalc = 1;
127 		}
128 		if (recalc) {
129 			x = inf.d * (a * c - b * d);
130 			y = inf.d * (a * d + b * c);
131 		}
132 	}
133 
134 	/*
135 	 * The following is equivalent to
136 	 *
137 	 *  return x + I * y;
138 	 */
139 	((double *)&v)[0] = x;
140 	((double *)&v)[1] = y;
141 	return (v);
142 }
143