/* $OpenBSD: fpu_add.c,v 1.3 2003/06/02 23:27:48 millert Exp $ */ /* $NetBSD: fpu_add.c,v 1.2 1996/04/30 11:52:09 briggs Exp $ */ /* * Copyright (c) 1992, 1993 * 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 Laboratory. * * Redistribution and use in source and binary forms, with or without * modification, are permitted provided that the following conditions * are met: * 1. Redistributions of source code must retain the above copyright * notice, this list of conditions and the following disclaimer. * 2. Redistributions in binary form must reproduce the above copyright * notice, this list of conditions and the following disclaimer in the * documentation and/or other materials provided with the distribution. * 3. Neither the name of the University nor the names of its contributors * may be used to endorse or promote products derived from this software * without specific prior written permission. * * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF * SUCH DAMAGE. * * @(#)fpu_add.c 8.1 (Berkeley) 6/11/93 */ /* * Perform an FPU add (return x + y). * * To subtract, negate y and call add. */ #include #include #include #include "fpu_arith.h" #include "fpu_emulate.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_fpcr & FPCR_ROUND); if (ISZERO(y)) { if (rd != FPCR_MINF) /* 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 == FPCR_MINF); 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); }