/*
* CDDL HEADER START
*
* The contents of this file are subject to the terms of the
* Common Development and Distribution License, Version 1.0 only
* (the "License"). You may not use this file except in compliance
* with the License.
*
* You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE
* See the License for the specific language governing permissions
* and limitations under the License.
*
* When distributing Covered Code, include this CDDL HEADER in each
* file and include the License file at usr/src/OPENSOLARIS.LICENSE.
* If applicable, add the following below this CDDL HEADER, with the
* fields enclosed by brackets "[]" replaced with your own identifying
* information: Portions Copyright [yyyy] [name of copyright owner]
*
* CDDL HEADER END
*/
/*
* Copyright 2004 Sun Microsystems, Inc. All rights reserved.
* Use is subject to license terms.
*/
#pragma ident "%Z%%M% %I% %E% SMI"
/*
* _X_cplx_mul(z, w) returns z * w with infinities handled according
* to C99.
*
* If z and w are both finite, _X_cplx_mul(z, w) delivers the complex
* product according to the usual formula: let a = Re(z), b = Im(z),
* c = Re(w), and d = Im(w); then _X_cplx_mul(z, w) delivers x + I * y
* where x = a * c - b * d and y = a * d + b * c. Note that if both
* ac and bd overflow, then at least one of ad or bc must also over-
* flow, and vice versa, so that if one component of the product is
* NaN, the other is infinite. (Such a value is considered infinite
* according to C99.)
*
* If one of z or w is infinite and the other is either finite nonzero
* or infinite, _X_cplx_mul delivers an infinite result. If one factor
* is infinite and the other is zero, _X_cplx_mul delivers NaN + I * NaN.
* C99 doesn't specify the latter case.
*
* C99 also doesn't specify what should happen if either z or w is a
* complex NaN (i.e., neither finite nor infinite). This implementation
* delivers NaN + I * NaN in this case.
*
* This implementation can raise spurious underflow, overflow, invalid
* operation, and inexact exceptions. C99 allows this.
*/
#endif
static union {
int i;
float f;
} inf = {
0x7f800000
};
/*
* Return +1 if x is +Inf, -1 if x is -Inf, and 0 otherwise
*/
static int
testinfl(long double x)
{
union {
int i[3];
long double e;
} xx;
xx.e = x;
return (0);
}
long double _Complex
_X_cplx_mul(long double _Complex z, long double _Complex w)
{
long double _Complex v;
long double a, b, c, d, x, y;
int recalc, i, j;
/*
* The following is equivalent to
*
* a = creall(z); b = cimagl(z);
* c = creall(w); d = cimagl(w);
*/
a = ((long double *)&z)[0];
b = ((long double *)&z)[1];
c = ((long double *)&w)[0];
d = ((long double *)&w)[1];
x = a * c - b * d;
y = a * d + b * c;
if (x != x && y != y) {
/*
* Both x and y are NaN, so z and w can't both be finite.
* If at least one of z or w is a complex NaN, and neither
* is infinite, then we might as well deliver NaN + I * NaN.
* So the only cases to check are when one of z or w is
* infinite.
*/
recalc = 0;
i = testinfl(a);
j = testinfl(b);
if (i | j) { /* z is infinite */
/* "factor out" infinity */
a = i;
b = j;
recalc = 1;
}
i = testinfl(c);
j = testinfl(d);
if (i | j) { /* w is infinite */
/* "factor out" infinity */
c = i;
d = j;
recalc = 1;
}
if (recalc) {
x = inf.f * (a * c - b * d);
y = inf.f * (a * d + b * c);
}
}
/*
* The following is equivalent to
*
* return x + I * y;
*/
((long double *)&v)[0] = x;
((long double *)&v)[1] = y;
return (v);
}