1 /*
2 * Copyright (c) 1992, 1993
3 * The Regents of the University of California. All rights reserved.
4 *
5 * This software was developed by the Computer Systems Engineering group
6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
7 * contributed to Berkeley.
8 *
9 * All advertising materials mentioning features or use of this software
10 * must display the following acknowledgement:
11 * This product includes software developed by the University of
12 * California, Lawrence Berkeley Laboratory.
13 *
14 * %sccs.include.redist.c%
15 *
16 * @(#)fpu_add.c 8.1 (Berkeley) 06/11/93
17 *
18 * from: $Header: fpu_add.c,v 1.4 92/11/26 01:39:46 torek Exp $
19 */
20
21 /*
22 * Perform an FPU add (return x + y).
23 *
24 * To subtract, negate y and call add.
25 */
26
27 #include <sys/types.h>
28
29 #include <machine/reg.h>
30
31 #include <sparc/fpu/fpu_arith.h>
32 #include <sparc/fpu/fpu_emu.h>
33
34 struct fpn *
fpu_add(fe)35 fpu_add(fe)
36 register struct fpemu *fe;
37 {
38 register struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2, *r;
39 register u_int r0, r1, r2, r3;
40 register int rd;
41
42 /*
43 * Put the `heavier' operand on the right (see fpu_emu.h).
44 * Then we will have one of the following cases, taken in the
45 * following order:
46 *
47 * - y = NaN. Implied: if only one is a signalling NaN, y is.
48 * The result is y.
49 * - y = Inf. Implied: x != NaN (is 0, number, or Inf: the NaN
50 * case was taken care of earlier).
51 * If x = -y, the result is NaN. Otherwise the result
52 * is y (an Inf of whichever sign).
53 * - y is 0. Implied: x = 0.
54 * If x and y differ in sign (one positive, one negative),
55 * the result is +0 except when rounding to -Inf. If same:
56 * +0 + +0 = +0; -0 + -0 = -0.
57 * - x is 0. Implied: y != 0.
58 * Result is y.
59 * - other. Implied: both x and y are numbers.
60 * Do addition a la Hennessey & Patterson.
61 */
62 ORDER(x, y);
63 if (ISNAN(y))
64 return (y);
65 if (ISINF(y)) {
66 if (ISINF(x) && x->fp_sign != y->fp_sign)
67 return (fpu_newnan(fe));
68 return (y);
69 }
70 rd = ((fe->fe_fsr >> FSR_RD_SHIFT) & FSR_RD_MASK);
71 if (ISZERO(y)) {
72 if (rd != FSR_RD_RM) /* only -0 + -0 gives -0 */
73 y->fp_sign &= x->fp_sign;
74 else /* any -0 operand gives -0 */
75 y->fp_sign |= x->fp_sign;
76 return (y);
77 }
78 if (ISZERO(x))
79 return (y);
80 /*
81 * We really have two numbers to add, although their signs may
82 * differ. Make the exponents match, by shifting the smaller
83 * number right (e.g., 1.011 => 0.1011) and increasing its
84 * exponent (2^3 => 2^4). Note that we do not alter the exponents
85 * of x and y here.
86 */
87 r = &fe->fe_f3;
88 r->fp_class = FPC_NUM;
89 if (x->fp_exp == y->fp_exp) {
90 r->fp_exp = x->fp_exp;
91 r->fp_sticky = 0;
92 } else {
93 if (x->fp_exp < y->fp_exp) {
94 /*
95 * Try to avoid subtract case iii (see below).
96 * This also guarantees that x->fp_sticky = 0.
97 */
98 SWAP(x, y);
99 }
100 /* now x->fp_exp > y->fp_exp */
101 r->fp_exp = x->fp_exp;
102 r->fp_sticky = fpu_shr(y, x->fp_exp - y->fp_exp);
103 }
104 r->fp_sign = x->fp_sign;
105 if (x->fp_sign == y->fp_sign) {
106 FPU_DECL_CARRY
107
108 /*
109 * The signs match, so we simply add the numbers. The result
110 * may be `supernormal' (as big as 1.111...1 + 1.111...1, or
111 * 11.111...0). If so, a single bit shift-right will fix it
112 * (but remember to adjust the exponent).
113 */
114 /* r->fp_mant = x->fp_mant + y->fp_mant */
115 FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]);
116 FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]);
117 FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]);
118 FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]);
119 if ((r->fp_mant[0] = r0) >= FP_2) {
120 (void) fpu_shr(r, 1);
121 r->fp_exp++;
122 }
123 } else {
124 FPU_DECL_CARRY
125
126 /*
127 * The signs differ, so things are rather more difficult.
128 * H&P would have us negate the negative operand and add;
129 * this is the same as subtracting the negative operand.
130 * This is quite a headache. Instead, we will subtract
131 * y from x, regardless of whether y itself is the negative
132 * operand. When this is done one of three conditions will
133 * hold, depending on the magnitudes of x and y:
134 * case i) |x| > |y|. The result is just x - y,
135 * with x's sign, but it may need to be normalized.
136 * case ii) |x| = |y|. The result is 0 (maybe -0)
137 * so must be fixed up.
138 * case iii) |x| < |y|. We goofed; the result should
139 * be (y - x), with the same sign as y.
140 * We could compare |x| and |y| here and avoid case iii,
141 * but that would take just as much work as the subtract.
142 * We can tell case iii has occurred by an overflow.
143 *
144 * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
145 */
146 /* r->fp_mant = x->fp_mant - y->fp_mant */
147 FPU_SET_CARRY(y->fp_sticky);
148 FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]);
149 FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]);
150 FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]);
151 FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]);
152 if (r0 < FP_2) {
153 /* cases i and ii */
154 if ((r0 | r1 | r2 | r3) == 0) {
155 /* case ii */
156 r->fp_class = FPC_ZERO;
157 r->fp_sign = rd == FSR_RD_RM;
158 return (r);
159 }
160 } else {
161 /*
162 * Oops, case iii. This can only occur when the
163 * exponents were equal, in which case neither
164 * x nor y have sticky bits set. Flip the sign
165 * (to y's sign) and negate the result to get y - x.
166 */
167 #ifdef DIAGNOSTIC
168 if (x->fp_exp != y->fp_exp || r->fp_sticky)
169 panic("fpu_add");
170 #endif
171 r->fp_sign = y->fp_sign;
172 FPU_SUBS(r3, 0, r3);
173 FPU_SUBCS(r2, 0, r2);
174 FPU_SUBCS(r1, 0, r1);
175 FPU_SUBC(r0, 0, r0);
176 }
177 r->fp_mant[3] = r3;
178 r->fp_mant[2] = r2;
179 r->fp_mant[1] = r1;
180 r->fp_mant[0] = r0;
181 if (r0 < FP_1)
182 fpu_norm(r);
183 }
184 return (r);
185 }
186