/*
 * 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
 * or http://www.opensolaris.org/os/licensing.
 * 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 2003 Sun Microsystems, Inc.  All rights reserved.
 * Use is subject to license terms.
 */

/*
 * _F_cplx_mul(z, w) returns z * w with infinities handled according
 * to C99.
 *
 * If z and w are both finite, _F_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 _F_cplx_mul(z, w) delivers x + I * y
 * where x = a * c - b * d and y = a * d + b * c.  This implementation
 * uses double precision to form these expressions, so none of the
 * intermediate products can overflow.
 *
 * If one of z or w is infinite and the other is either finite nonzero
 * or infinite, _F_cplx_mul delivers an infinite result.  If one factor
 * is infinite and the other is zero, _F_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 invalid operation and inexact
 * exceptions.  C99 allows this.
 */

#if !defined(sparc) && !defined(__sparc)
#error This code is for SPARC only
#endif

static union {
        int     i[2];
        double  d;
} inf = {
        0x7ff00000, 0
};

/*
 * Return +1 if x is +Inf, -1 if x is -Inf, and 0 otherwise
 */
static int
testinff(float x)
{
        union {
                int     i;
                float   f;
        } xx;

        xx.f = x;
        return ((((xx.i << 1) - 0xff000000) == 0)? (1 | (xx.i >> 31)) : 0);
}

float _Complex
_F_cplx_mul(float _Complex z, float _Complex w)
{
        float _Complex  v = 0;
        float           a, b, c, d;
        double          x, y;
        int             recalc, i, j;

        /*
         * The following is equivalent to
         *
         *  a = crealf(z); b = cimagf(z);
         *  c = crealf(w); d = cimagf(w);
         */
        a = ((float *)&z)[0];
        b = ((float *)&z)[1];
        c = ((float *)&w)[0];
        d = ((float *)&w)[1];

        x = (double)a * c - (double)b * d;
        y = (double)a * d + (double)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 = testinff(a);
                j = testinff(b);
                if (i | j) { /* z is infinite */
                        /* "factor out" infinity */
                        a = i;
                        b = j;
                        recalc = 1;
                }
                i = testinff(c);
                j = testinff(d);
                if (i | j) { /* w is infinite */
                        /* "factor out" infinity */
                        c = i;
                        d = j;
                        recalc = 1;
                }
                if (recalc) {
                        x = inf.d * ((double)a * c - (double)b * d);
                        y = inf.d * ((double)a * d + (double)b * c);
                }
        }

        /*
         * The following is equivalent to
         *
         *  return x + I * y;
         */
        ((float *)&v)[0] = (float)x;
        ((float *)&v)[1] = (float)y;
        return (v);
}