usr/src/sys/sparc/fpu/fpu_add.c

/*
 * Copyright (c) 1992 The Regents of the University of California.
 * All rights reserved.
 *
 * This software was developed by the Computer Systems Engineering group
 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
 * contributed to Berkeley.
 *
 * All advertising materials mentioning features or use of this software
 * must display the following acknowledgement:
 *      This product includes software developed by the University of
 *      California, Lawrence Berkeley Laboratories.
 *
 * %sccs.include.redist.c%
 *
 *      @(#)fpu_add.c   7.3 (Berkeley) %G%
 *
 * from: $Header: fpu_add.c,v 1.3 92/06/17 18:11:43 mccanne Exp $
 */

/*
 * Perform an FPU add (return x + y).
 *
 * To subtract, negate y and call add.
 */

#include <sys/types.h>

#include <machine/reg.h>

#include <sparc/fpu/fpu_arith.h>
#include <sparc/fpu/fpu_emu.h>

struct fpn *
fpu_add(fe)
        register struct fpemu *fe;
{
        register struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2, *r;
        register u_int r0, r1, r2, r3;
        register int rd;

        /*
         * Put the `heavier' operand on the right (see fpu_emu.h).
         * Then we will have one of the following cases, taken in the
         * following order:
         *
         *  - y = NaN.  Implied: if only one is a signalling NaN, y is.
         *      The result is y.
         *  - y = Inf.  Implied: x != NaN (is 0, number, or Inf: the NaN
         *    case was taken care of earlier).
         *      If x = -y, the result is NaN.  Otherwise the result
         *      is y (an Inf of whichever sign).
         *  - y is 0.  Implied: x = 0.
         *      If x and y differ in sign (one positive, one negative),
         *      the result is +0 except when rounding to -Inf.  If same:
         *      +0 + +0 = +0; -0 + -0 = -0.
         *  - x is 0.  Implied: y != 0.
         *      Result is y.
         *  - other.  Implied: both x and y are numbers.
         *      Do addition a la Hennessey & Patterson.
         */
        ORDER(x, y);
        if (ISNAN(y))
                return (y);
        if (ISINF(y)) {
                if (ISINF(x) && x->fp_sign != y->fp_sign)
                        return (fpu_newnan(fe));
                return (y);
        }
        rd = ((fe->fe_fsr >> FSR_RD_SHIFT) & FSR_RD_MASK);
        if (ISZERO(y)) {
                if (rd != FSR_RD_RM)    /* only -0 + -0 gives -0 */
                        y->fp_sign &= x->fp_sign;
                else                    /* any -0 operand gives -0 */
                        y->fp_sign |= x->fp_sign;
                return (y);
        }
        if (ISZERO(x))
                return (y);
        /*
         * We really have two numbers to add, although their signs may
         * differ.  Make the exponents match, by shifting the smaller
         * number right (e.g., 1.011 => 0.1011) and increasing its
         * exponent (2^3 => 2^4).  Note that we do not alter the exponents
         * of x and y here.
         */
        r = &fe->fe_f3;
        r->fp_class = FPC_NUM;
        if (x->fp_exp == y->fp_exp) {
                r->fp_exp = x->fp_exp;
                r->fp_sticky = 0;
        } else {
                if (x->fp_exp < y->fp_exp) {
                        /*
                         * Try to avoid subtract case iii (see below).
                         * This also guarantees that x->fp_sticky = 0.
                         */
                        SWAP(x, y);
                }
                /* now x->fp_exp > y->fp_exp */
                r->fp_exp = x->fp_exp;
                r->fp_sticky = fpu_shr(y, x->fp_exp - y->fp_exp);
        }
        r->fp_sign = x->fp_sign;
        if (x->fp_sign == y->fp_sign) {
                FPU_DECL_CARRY

                /*
                 * The signs match, so we simply add the numbers.  The result
                 * may be `supernormal' (as big as 1.111...1 + 1.111...1, or
                 * 11.111...0).  If so, a single bit shift-right will fix it
                 * (but remember to adjust the exponent).
                 */
                /* r->fp_mant = x->fp_mant + y->fp_mant */
                FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]);
                FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]);
                FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]);
                FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]);
                if ((r->fp_mant[0] = r0) >= FP_2) {
                        (void) fpu_shr(r, 1);
                        r->fp_exp++;
                }
        } else {
                FPU_DECL_CARRY

                /*
                 * The signs differ, so things are rather more difficult.
                 * H&P would have us negate the negative operand and add;
                 * this is the same as subtracting the negative operand.
                 * This is quite a headache.  Instead, we will subtract
                 * y from x, regardless of whether y itself is the negative
                 * operand.  When this is done one of three conditions will
                 * hold, depending on the magnitudes of x and y:
                 *   case i)   |x| > |y|.  The result is just x - y,
                 *      with x's sign, but it may need to be normalized.
                 *   case ii)  |x| = |y|.  The result is 0 (maybe -0)
                 *      so must be fixed up.
                 *   case iii) |x| < |y|.  We goofed; the result should
                 *      be (y - x), with the same sign as y.
                 * We could compare |x| and |y| here and avoid case iii,
                 * but that would take just as much work as the subtract.
                 * We can tell case iii has occurred by an overflow.
                 *
                 * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
                 */
                /* r->fp_mant = x->fp_mant - y->fp_mant */
                FPU_SET_CARRY(y->fp_sticky);
                FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]);
                FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]);
                FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]);
                FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]);
                if (r0 < FP_2) {
                        /* cases i and ii */
                        if ((r0 | r1 | r2 | r3) == 0) {
                                /* case ii */
                                r->fp_class = FPC_ZERO;
                                r->fp_sign = rd == FSR_RD_RM;
                                return (r);
                        }
                } else {
                        /*
                         * Oops, case iii.  This can only occur when the
                         * exponents were equal, in which case neither
                         * x nor y have sticky bits set.  Flip the sign
                         * (to y's sign) and negate the result to get y - x.
                         */
#ifdef DIAGNOSTIC
                        if (x->fp_exp != y->fp_exp || r->fp_sticky)
                                panic("fpu_add");
#endif
                        r->fp_sign = y->fp_sign;
                        FPU_SUBS(r3, 0, r3);
                        FPU_SUBCS(r2, 0, r2);
                        FPU_SUBCS(r1, 0, r1);
                        FPU_SUBC(r0, 0, r0);
                }
                r->fp_mant[3] = r3;
                r->fp_mant[2] = r2;
                r->fp_mant[1] = r1;
                r->fp_mant[0] = r0;
                if (r0 < FP_1)
                        fpu_norm(r);
        }
        return (r);
}