xref: /illumos-gate/usr/src/lib/libc/i386/fp/_X_cplx_mul.c (revision 7c478bd9)
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 2004 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  * _X_cplx_mul(z, w) returns z * w with infinities handled according
31  * to C99.
32  *
33  * If z and w are both finite, _X_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 _X_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, _X_cplx_mul delivers an infinite result.  If one factor
44  * is infinite and the other is zero, _X_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(i386) && !defined(__i386) && !defined(__amd64)
56 #error This code is for x86 only
57 #endif
58 
59 static union {
60 	int	i;
61 	float	f;
62 } inf = {
63 	0x7f800000
64 };
65 
66 /*
67  * Return +1 if x is +Inf, -1 if x is -Inf, and 0 otherwise
68  */
69 static int
70 testinfl(long double x)
71 {
72 	union {
73 		int		i[3];
74 		long double	e;
75 	} xx;
76 
77 	xx.e = x;
78 	if ((xx.i[2] & 0x7fff) != 0x7fff || ((xx.i[1] << 1) | xx.i[0]) != 0)
79 		return (0);
80 	return (1 | ((xx.i[2] << 16) >> 31));
81 }
82 
83 long double _Complex
84 _X_cplx_mul(long double _Complex z, long double _Complex w)
85 {
86 	long double _Complex	v;
87 	long double		a, b, c, d, x, y;
88 	int			recalc, i, j;
89 
90 	/*
91 	 * The following is equivalent to
92 	 *
93 	 *  a = creall(z); b = cimagl(z);
94 	 *  c = creall(w); d = cimagl(w);
95 	 */
96 	a = ((long double *)&z)[0];
97 	b = ((long double *)&z)[1];
98 	c = ((long double *)&w)[0];
99 	d = ((long double *)&w)[1];
100 
101 	x = a * c - b * d;
102 	y = a * d + b * c;
103 
104 	if (x != x && y != y) {
105 		/*
106 		 * Both x and y are NaN, so z and w can't both be finite.
107 		 * If at least one of z or w is a complex NaN, and neither
108 		 * is infinite, then we might as well deliver NaN + I * NaN.
109 		 * So the only cases to check are when one of z or w is
110 		 * infinite.
111 		 */
112 		recalc = 0;
113 		i = testinfl(a);
114 		j = testinfl(b);
115 		if (i | j) { /* z is infinite */
116 			/* "factor out" infinity */
117 			a = i;
118 			b = j;
119 			recalc = 1;
120 		}
121 		i = testinfl(c);
122 		j = testinfl(d);
123 		if (i | j) { /* w is infinite */
124 			/* "factor out" infinity */
125 			c = i;
126 			d = j;
127 			recalc = 1;
128 		}
129 		if (recalc) {
130 			x = inf.f * (a * c - b * d);
131 			y = inf.f * (a * d + b * c);
132 		}
133 	}
134 
135 	/*
136 	 * The following is equivalent to
137 	 *
138 	 *  return x + I * y;
139 	 */
140 	((long double *)&v)[0] = x;
141 	((long double *)&v)[1] = y;
142 	return (v);
143 }
144