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40 * @(#)fpu_add.c 8.1 (Berkeley) 6/11/93
41 * $NetBSD: fpu_add.c,v 1.3 1996/03/14 19:41:52 christos Exp $
44 #include <sys/cdefs.h>
45 __FBSDID("$FreeBSD$");
48 * Perform an FPU add (return x + y).
50 * To subtract, negate y and call add.
53 #include <sys/param.h>
55 #include <machine/frame.h>
56 #include <machine/fp.h>
57 #include <machine/fsr.h>
58 #include <machine/instr.h>
60 #include "fpu_arith.h"
62 #include "fpu_extern.h"
63 #include "__sparc_utrap_private.h"
69 struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2, *r;
74 * Put the `heavier' operand on the right (see fpu_emu.h).
75 * Then we will have one of the following cases, taken in the
78 * - y = NaN. Implied: if only one is a signalling NaN, y is.
80 * - y = Inf. Implied: x != NaN (is 0, number, or Inf: the NaN
81 * case was taken care of earlier).
82 * If x = -y, the result is NaN. Otherwise the result
83 * is y (an Inf of whichever sign).
84 * - y is 0. Implied: x = 0.
85 * If x and y differ in sign (one positive, one negative),
86 * the result is +0 except when rounding to -Inf. If same:
87 * +0 + +0 = +0; -0 + -0 = -0.
88 * - x is 0. Implied: y != 0.
90 * - other. Implied: both x and y are numbers.
91 * Do addition a la Hennessey & Patterson.
97 if (ISINF(x) && x->fp_sign != y->fp_sign)
98 return (__fpu_newnan(fe));
101 rd = FSR_GET_RD(fe->fe_fsr);
103 if (rd != FSR_RD_NINF) /* only -0 + -0 gives -0 */
104 y->fp_sign &= x->fp_sign;
105 else /* any -0 operand gives -0 */
106 y->fp_sign |= x->fp_sign;
112 * We really have two numbers to add, although their signs may
113 * differ. Make the exponents match, by shifting the smaller
114 * number right (e.g., 1.011 => 0.1011) and increasing its
115 * exponent (2^3 => 2^4). Note that we do not alter the exponents
119 r->fp_class = FPC_NUM;
120 if (x->fp_exp == y->fp_exp) {
121 r->fp_exp = x->fp_exp;
124 if (x->fp_exp < y->fp_exp) {
126 * Try to avoid subtract case iii (see below).
127 * This also guarantees that x->fp_sticky = 0.
131 /* now x->fp_exp > y->fp_exp */
132 r->fp_exp = x->fp_exp;
133 r->fp_sticky = __fpu_shr(y, x->fp_exp - y->fp_exp);
135 r->fp_sign = x->fp_sign;
136 if (x->fp_sign == y->fp_sign) {
140 * The signs match, so we simply add the numbers. The result
141 * may be `supernormal' (as big as 1.111...1 + 1.111...1, or
142 * 11.111...0). If so, a single bit shift-right will fix it
143 * (but remember to adjust the exponent).
145 /* r->fp_mant = x->fp_mant + y->fp_mant */
146 FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]);
147 FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]);
148 FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]);
149 FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]);
150 if ((r->fp_mant[0] = r0) >= FP_2) {
151 (void) __fpu_shr(r, 1);
158 * The signs differ, so things are rather more difficult.
159 * H&P would have us negate the negative operand and add;
160 * this is the same as subtracting the negative operand.
161 * This is quite a headache. Instead, we will subtract
162 * y from x, regardless of whether y itself is the negative
163 * operand. When this is done one of three conditions will
164 * hold, depending on the magnitudes of x and y:
165 * case i) |x| > |y|. The result is just x - y,
166 * with x's sign, but it may need to be normalized.
167 * case ii) |x| = |y|. The result is 0 (maybe -0)
168 * so must be fixed up.
169 * case iii) |x| < |y|. We goofed; the result should
170 * be (y - x), with the same sign as y.
171 * We could compare |x| and |y| here and avoid case iii,
172 * but that would take just as much work as the subtract.
173 * We can tell case iii has occurred by an overflow.
175 * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
177 /* r->fp_mant = x->fp_mant - y->fp_mant */
178 FPU_SET_CARRY(y->fp_sticky);
179 FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]);
180 FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]);
181 FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]);
182 FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]);
185 if ((r0 | r1 | r2 | r3) == 0) {
187 r->fp_class = FPC_ZERO;
188 r->fp_sign = rd == FSR_RD_NINF;
193 * Oops, case iii. This can only occur when the
194 * exponents were equal, in which case neither
195 * x nor y have sticky bits set. Flip the sign
196 * (to y's sign) and negate the result to get y - x.
199 if (x->fp_exp != y->fp_exp || r->fp_sticky)
200 __utrap_panic("fpu_add");
202 r->fp_sign = y->fp_sign;
204 FPU_SUBCS(r2, 0, r2);
205 FPU_SUBCS(r1, 0, r1);