xref: /titanic_44/usr/src/lib/libc/i386/fp/_D_cplx_mul.c (revision 7c478bd95313f5f23a4c958a745db2134aa03244)
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  * _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.  This implementation
37  * uses extended precision to form these expressions, so none of the
38  * intermediate products can overflow.
39  *
40  * If one of z or w is infinite and the other is either finite nonzero
41  * or infinite, _D_cplx_mul delivers an infinite result.  If one factor
42  * is infinite and the other is zero, _D_cplx_mul delivers NaN + I * NaN.
43  * C99 doesn't specify the latter case.
44  *
45  * C99 also doesn't specify what should happen if either z or w is a
46  * complex NaN (i.e., neither finite nor infinite).  This implementation
47  * delivers NaN + I * NaN in this case.
48  *
49  * This implementation can raise spurious invalid operation and inexact
50  * exceptions.  C99 allows this.
51  */
52 
53 #if !defined(i386) && !defined(__i386) && !defined(__amd64)
54 #error This code is for x86 only
55 #endif
56 
57 static union {
58 	int	i;
59 	float	f;
60 } inf = {
61 	0x7f800000
62 };
63 
64 /*
65  * Return +1 if x is +Inf, -1 if x is -Inf, and 0 otherwise
66  */
67 static int
testinf(double x)68 testinf(double x)
69 {
70 	union {
71 		int	i[2];
72 		double	d;
73 	} xx;
74 
75 	xx.d = x;
76 	return (((((xx.i[1] << 1) - 0xffe00000) | xx.i[0]) == 0)?
77 		(1 | (xx.i[1] >> 31)) : 0);
78 }
79 
80 double _Complex
_D_cplx_mul(double _Complex z,double _Complex w)81 _D_cplx_mul(double _Complex z, double _Complex w)
82 {
83 	double _Complex	v;
84 	double		a, b, c, d;
85 	long double	x, y;
86 	int		recalc, i, j;
87 
88 	/*
89 	 * The following is equivalent to
90 	 *
91 	 *  a = creal(z); b = cimag(z);
92 	 *  c = creal(w); d = cimag(w);
93 	 */
94 	/* LINTED alignment */
95 	a = ((double *)&z)[0];
96 	/* LINTED alignment */
97 	b = ((double *)&z)[1];
98 	/* LINTED alignment */
99 	c = ((double *)&w)[0];
100 	/* LINTED alignment */
101 	d = ((double *)&w)[1];
102 
103 	x = (long double)a * c - (long double)b * d;
104 	y = (long double)a * d + (long double)b * c;
105 
106 	if (x != x && y != y) {
107 		/*
108 		 * Both x and y are NaN, so z and w can't both be finite.
109 		 * If at least one of z or w is a complex NaN, and neither
110 		 * is infinite, then we might as well deliver NaN + I * NaN.
111 		 * So the only cases to check are when one of z or w is
112 		 * infinite.
113 		 */
114 		recalc = 0;
115 		i = testinf(a);
116 		j = testinf(b);
117 		if (i | j) { /* z is infinite */
118 			/* "factor out" infinity */
119 			a = i;
120 			b = j;
121 			recalc = 1;
122 		}
123 		i = testinf(c);
124 		j = testinf(d);
125 		if (i | j) { /* w is infinite */
126 			/* "factor out" infinity */
127 			c = i;
128 			d = j;
129 			recalc = 1;
130 		}
131 		if (recalc) {
132 			x = inf.f * ((long double)a * c - (long double)b * d);
133 			y = inf.f * ((long double)a * d + (long double)b * c);
134 		}
135 	}
136 
137 	/*
138 	 * The following is equivalent to
139 	 *
140 	 *  return x + I * y;
141 	 */
142 	/* LINTED alignment */
143 	((double *)&v)[0] = (double)x;
144 	/* LINTED alignment */
145 	((double *)&v)[1] = (double)y;
146 	return (v);
147 }
148