1 /* $NetBSD: fpu_add.c,v 1.5 2003/08/07 16:29:36 agc 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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17 * modification, are permitted provided that the following conditions
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25 * may be used to endorse or promote products derived from this software
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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 1.5 2003/08/07 16:29:36 agc Exp $");
52 #include <sys/types.h>
54 #include <sys/systm.h>
57 #include <machine/reg.h>
58 #include <machine/instr.h>
60 #include <sparc/fpu/fpu_arith.h>
61 #include <sparc/fpu/fpu_emu.h>
62 #include <sparc/fpu/fpu_extern.h>
65 fpu_add(struct fpemu
*fe
)
67 register struct fpn
*x
= &fe
->fe_f1
, *y
= &fe
->fe_f2
, *r
;
68 register u_int r0
, r1
, r2
, r3
;
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
= ((fe
->fe_fsr
>> FSR_RD_SHIFT
) & FSR_RD_MASK
);
101 if (rd
!= FSR_RD_RM
) /* 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_RM
;
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
)
200 r
->fp_sign
= y
->fp_sign
;
202 FPU_SUBCS(r2
, 0, r2
);
203 FPU_SUBCS(r1
, 0, r1
);