1 /* $NetBSD: fpu_add.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */
4 * SPDX-License-Identifier: BSD-3-Clause
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42 * @(#)fpu_add.c 8.1 (Berkeley) 6/11/93
46 * Perform an FPU add (return x + y).
48 * To subtract, negate y and call add.
51 #include <sys/cdefs.h>
52 __FBSDID("$FreeBSD$");
54 #include <sys/types.h>
55 #include <sys/systm.h>
57 #include <machine/fpu.h>
58 #include <machine/ieeefp.h>
59 #include <machine/reg.h>
61 #include <powerpc/fpu/fpu_arith.h>
62 #include <powerpc/fpu/fpu_emu.h>
65 fpu_add(struct fpemu *fe)
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.
91 DPRINTF(FPE_REG, ("fpu_add:\n"));
94 DPRINTF(FPE_REG, ("=>\n"));
97 fe->fe_cx |= FPSCR_VXSNAN;
102 if (ISINF(x) && x->fp_sign != y->fp_sign) {
103 fe->fe_cx |= FPSCR_VXISI;
104 return (fpu_newnan(fe));
109 rd = ((fe->fe_fpscr) & FPSCR_RN);
111 if (rd != FP_RM) /* only -0 + -0 gives -0 */
112 y->fp_sign &= x->fp_sign;
113 else /* any -0 operand gives -0 */
114 y->fp_sign |= x->fp_sign;
123 * We really have two numbers to add, although their signs may
124 * differ. Make the exponents match, by shifting the smaller
125 * number right (e.g., 1.011 => 0.1011) and increasing its
126 * exponent (2^3 => 2^4). Note that we do not alter the exponents
130 r->fp_class = FPC_NUM;
131 if (x->fp_exp == y->fp_exp) {
132 r->fp_exp = x->fp_exp;
135 if (x->fp_exp < y->fp_exp) {
137 * Try to avoid subtract case iii (see below).
138 * This also guarantees that x->fp_sticky = 0.
142 /* now x->fp_exp > y->fp_exp */
143 r->fp_exp = x->fp_exp;
144 r->fp_sticky = fpu_shr(y, x->fp_exp - y->fp_exp);
146 r->fp_sign = x->fp_sign;
147 if (x->fp_sign == y->fp_sign) {
151 * The signs match, so we simply add the numbers. The result
152 * may be `supernormal' (as big as 1.111...1 + 1.111...1, or
153 * 11.111...0). If so, a single bit shift-right will fix it
154 * (but remember to adjust the exponent).
156 /* r->fp_mant = x->fp_mant + y->fp_mant */
157 FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]);
158 FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]);
159 FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]);
160 FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]);
161 if ((r->fp_mant[0] = r0) >= FP_2) {
162 (void) fpu_shr(r, 1);
169 * The signs differ, so things are rather more difficult.
170 * H&P would have us negate the negative operand and add;
171 * this is the same as subtracting the negative operand.
172 * This is quite a headache. Instead, we will subtract
173 * y from x, regardless of whether y itself is the negative
174 * operand. When this is done one of three conditions will
175 * hold, depending on the magnitudes of x and y:
176 * case i) |x| > |y|. The result is just x - y,
177 * with x's sign, but it may need to be normalized.
178 * case ii) |x| = |y|. The result is 0 (maybe -0)
179 * so must be fixed up.
180 * case iii) |x| < |y|. We goofed; the result should
181 * be (y - x), with the same sign as y.
182 * We could compare |x| and |y| here and avoid case iii,
183 * but that would take just as much work as the subtract.
184 * We can tell case iii has occurred by an overflow.
186 * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
188 /* r->fp_mant = x->fp_mant - y->fp_mant */
189 FPU_SET_CARRY(y->fp_sticky);
190 FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]);
191 FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]);
192 FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]);
193 FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]);
196 if ((r0 | r1 | r2 | r3) == 0) {
198 r->fp_class = FPC_ZERO;
199 r->fp_sign = rd == FP_RM;
204 * Oops, case iii. This can only occur when the
205 * exponents were equal, in which case neither
206 * x nor y have sticky bits set. Flip the sign
207 * (to y's sign) and negate the result to get y - x.
210 if (x->fp_exp != y->fp_exp || r->fp_sticky)
213 r->fp_sign = y->fp_sign;
215 FPU_SUBCS(r2, 0, r2);
216 FPU_SUBCS(r1, 0, r1);