2 * Copyright (c) 1992, 1993
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5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
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38 * @(#)fpu_add.c 8.1 (Berkeley) 6/11/93
39 * $NetBSD: fpu_add.c,v 1.3 1996/03/14 19:41:52 christos Exp $
42 #include <sys/cdefs.h>
43 __FBSDID("$FreeBSD$");
46 * Perform an FPU add (return x + y).
48 * To subtract, negate y and call add.
51 #include <sys/param.h>
53 #include <machine/frame.h>
54 #include <machine/fp.h>
55 #include <machine/fsr.h>
56 #include <machine/instr.h>
58 #include "fpu_arith.h"
60 #include "fpu_extern.h"
61 #include "__sparc_utrap_private.h"
67 struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2, *r;
72 * Put the `heavier' operand on the right (see fpu_emu.h).
73 * Then we will have one of the following cases, taken in the
76 * - y = NaN. Implied: if only one is a signalling NaN, y is.
78 * - y = Inf. Implied: x != NaN (is 0, number, or Inf: the NaN
79 * case was taken care of earlier).
80 * If x = -y, the result is NaN. Otherwise the result
81 * is y (an Inf of whichever sign).
82 * - y is 0. Implied: x = 0.
83 * If x and y differ in sign (one positive, one negative),
84 * the result is +0 except when rounding to -Inf. If same:
85 * +0 + +0 = +0; -0 + -0 = -0.
86 * - x is 0. Implied: y != 0.
88 * - other. Implied: both x and y are numbers.
89 * Do addition a la Hennessey & Patterson.
95 if (ISINF(x) && x->fp_sign != y->fp_sign)
96 return (__fpu_newnan(fe));
99 rd = FSR_GET_RD(fe->fe_fsr);
101 if (rd != FSR_RD_NINF) /* only -0 + -0 gives -0 */
102 y->fp_sign &= x->fp_sign;
103 else /* any -0 operand gives -0 */
104 y->fp_sign |= x->fp_sign;
110 * We really have two numbers to add, although their signs may
111 * differ. Make the exponents match, by shifting the smaller
112 * number right (e.g., 1.011 => 0.1011) and increasing its
113 * exponent (2^3 => 2^4). Note that we do not alter the exponents
117 r->fp_class = FPC_NUM;
118 if (x->fp_exp == y->fp_exp) {
119 r->fp_exp = x->fp_exp;
122 if (x->fp_exp < y->fp_exp) {
124 * Try to avoid subtract case iii (see below).
125 * This also guarantees that x->fp_sticky = 0.
129 /* now x->fp_exp > y->fp_exp */
130 r->fp_exp = x->fp_exp;
131 r->fp_sticky = __fpu_shr(y, x->fp_exp - y->fp_exp);
133 r->fp_sign = x->fp_sign;
134 if (x->fp_sign == y->fp_sign) {
138 * The signs match, so we simply add the numbers. The result
139 * may be `supernormal' (as big as 1.111...1 + 1.111...1, or
140 * 11.111...0). If so, a single bit shift-right will fix it
141 * (but remember to adjust the exponent).
143 /* r->fp_mant = x->fp_mant + y->fp_mant */
144 FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]);
145 FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]);
146 FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]);
147 FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]);
148 if ((r->fp_mant[0] = r0) >= FP_2) {
149 (void) __fpu_shr(r, 1);
156 * The signs differ, so things are rather more difficult.
157 * H&P would have us negate the negative operand and add;
158 * this is the same as subtracting the negative operand.
159 * This is quite a headache. Instead, we will subtract
160 * y from x, regardless of whether y itself is the negative
161 * operand. When this is done one of three conditions will
162 * hold, depending on the magnitudes of x and y:
163 * case i) |x| > |y|. The result is just x - y,
164 * with x's sign, but it may need to be normalized.
165 * case ii) |x| = |y|. The result is 0 (maybe -0)
166 * so must be fixed up.
167 * case iii) |x| < |y|. We goofed; the result should
168 * be (y - x), with the same sign as y.
169 * We could compare |x| and |y| here and avoid case iii,
170 * but that would take just as much work as the subtract.
171 * We can tell case iii has occurred by an overflow.
173 * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
175 /* r->fp_mant = x->fp_mant - y->fp_mant */
176 FPU_SET_CARRY(y->fp_sticky);
177 FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]);
178 FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]);
179 FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]);
180 FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]);
183 if ((r0 | r1 | r2 | r3) == 0) {
185 r->fp_class = FPC_ZERO;
186 r->fp_sign = rd == FSR_RD_NINF;
191 * Oops, case iii. This can only occur when the
192 * exponents were equal, in which case neither
193 * x nor y have sticky bits set. Flip the sign
194 * (to y's sign) and negate the result to get y - x.
197 if (x->fp_exp != y->fp_exp || r->fp_sticky)
198 __utrap_panic("fpu_add");
200 r->fp_sign = y->fp_sign;
202 FPU_SUBCS(r2, 0, r2);
203 FPU_SUBCS(r1, 0, r1);