1 /* $NetBSD: fpu_add.c,v 1.6 2005/12/11 12:17:52 christos Exp $ */
4 * Copyright (c) 1992, 1993
5 * The Regents of the University of California. All rights reserved.
7 * This software was developed by the Computer Systems Engineering group
8 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
9 * contributed to Berkeley.
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12 * must display the following acknowledgement:
13 * This product includes software developed by the University of
14 * California, Lawrence Berkeley Laboratory.
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37 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
40 * @(#)fpu_add.c 8.1 (Berkeley) 6/11/93
44 * Perform an FPU add (return x + y).
46 * To subtract, negate y and call add.
49 #include <sys/cdefs.h>
50 __KERNEL_RCSID(0, "$NetBSD: fpu_add.c,v 1.6 2005/12/11 12:17:52 christos Exp $");
52 #include <sys/types.h>
53 #include <sys/systm.h>
55 #include <machine/reg.h>
57 #include "fpu_arith.h"
58 #include "fpu_emulate.h"
61 fpu_add(register struct fpemu
*fe
)
63 register struct fpn
*x
= &fe
->fe_f1
, *y
= &fe
->fe_f2
, *r
;
64 register u_int r0
, r1
, r2
;
68 * Put the `heavier' operand on the right (see fpu_emu.h).
69 * Then we will have one of the following cases, taken in the
72 * - y = NaN. Implied: if only one is a signalling NaN, y is.
74 * - y = Inf. Implied: x != NaN (is 0, number, or Inf: the NaN
75 * case was taken care of earlier).
76 * If x = -y, the result is NaN. Otherwise the result
77 * is y (an Inf of whichever sign).
78 * - y is 0. Implied: x = 0.
79 * If x and y differ in sign (one positive, one negative),
80 * the result is +0 except when rounding to -Inf. If same:
81 * +0 + +0 = +0; -0 + -0 = -0.
82 * - x is 0. Implied: y != 0.
84 * - other. Implied: both x and y are numbers.
85 * Do addition a la Hennessey & Patterson.
91 if (ISINF(x
) && x
->fp_sign
!= y
->fp_sign
)
92 return (fpu_newnan(fe
));
95 rd
= (fe
->fe_fpcr
& FPCR_ROUND
);
97 if (rd
!= FPCR_MINF
) /* only -0 + -0 gives -0 */
98 y
->fp_sign
&= x
->fp_sign
;
99 else /* any -0 operand gives -0 */
100 y
->fp_sign
|= x
->fp_sign
;
106 * We really have two numbers to add, although their signs may
107 * differ. Make the exponents match, by shifting the smaller
108 * number right (e.g., 1.011 => 0.1011) and increasing its
109 * exponent (2^3 => 2^4). Note that we do not alter the exponents
113 r
->fp_class
= FPC_NUM
;
114 if (x
->fp_exp
== y
->fp_exp
) {
115 r
->fp_exp
= x
->fp_exp
;
118 if (x
->fp_exp
< y
->fp_exp
) {
120 * Try to avoid subtract case iii (see below).
121 * This also guarantees that x->fp_sticky = 0.
125 /* now x->fp_exp > y->fp_exp */
126 r
->fp_exp
= x
->fp_exp
;
127 r
->fp_sticky
= fpu_shr(y
, x
->fp_exp
- y
->fp_exp
);
129 r
->fp_sign
= x
->fp_sign
;
130 if (x
->fp_sign
== y
->fp_sign
) {
134 * The signs match, so we simply add the numbers. The result
135 * may be `supernormal' (as big as 1.111...1 + 1.111...1, or
136 * 11.111...0). If so, a single bit shift-right will fix it
137 * (but remember to adjust the exponent).
139 /* r->fp_mant = x->fp_mant + y->fp_mant */
140 FPU_ADDS(r
->fp_mant
[2], x
->fp_mant
[2], y
->fp_mant
[2]);
141 FPU_ADDCS(r
->fp_mant
[1], x
->fp_mant
[1], y
->fp_mant
[1]);
142 FPU_ADDC(r0
, x
->fp_mant
[0], y
->fp_mant
[0]);
143 if ((r
->fp_mant
[0] = r0
) >= FP_2
) {
144 (void) fpu_shr(r
, 1);
151 * The signs differ, so things are rather more difficult.
152 * H&P would have us negate the negative operand and add;
153 * this is the same as subtracting the negative operand.
154 * This is quite a headache. Instead, we will subtract
155 * y from x, regardless of whether y itself is the negative
156 * operand. When this is done one of three conditions will
157 * hold, depending on the magnitudes of x and y:
158 * case i) |x| > |y|. The result is just x - y,
159 * with x's sign, but it may need to be normalized.
160 * case ii) |x| = |y|. The result is 0 (maybe -0)
161 * so must be fixed up.
162 * case iii) |x| < |y|. We goofed; the result should
163 * be (y - x), with the same sign as y.
164 * We could compare |x| and |y| here and avoid case iii,
165 * but that would take just as much work as the subtract.
166 * We can tell case iii has occurred by an overflow.
168 * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
170 /* r->fp_mant = x->fp_mant - y->fp_mant */
171 FPU_SET_CARRY(y
->fp_sticky
);
172 FPU_SUBCS(r2
, x
->fp_mant
[2], y
->fp_mant
[2]);
173 FPU_SUBCS(r1
, x
->fp_mant
[1], y
->fp_mant
[1]);
174 FPU_SUBC(r0
, x
->fp_mant
[0], y
->fp_mant
[0]);
177 if ((r0
| r1
| r2
) == 0) {
179 r
->fp_class
= FPC_ZERO
;
180 r
->fp_sign
= (rd
== FPCR_MINF
);
185 * Oops, case iii. This can only occur when the
186 * exponents were equal, in which case neither
187 * x nor y have sticky bits set. Flip the sign
188 * (to y's sign) and negate the result to get y - x.
191 if (x
->fp_exp
!= y
->fp_exp
|| r
->fp_sticky
)
194 r
->fp_sign
= y
->fp_sign
;
196 FPU_SUBCS(r1
, 0, r1
);