1 /* 2 * Copyright (c) 1992 The Regents of the University of California. 3 * 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 * %sccs.include.redist.c% 10 * 11 * @(#)fpu_add.c 7.1 (Berkeley) 07/13/92 12 * 13 * from: $Header: fpu_add.c,v 1.3 92/06/17 18:11:43 mccanne Exp $ 14 */ 15 16 /* 17 * Perform an FPU add (return x + y). 18 * 19 * To subtract, negate y and call add. 20 */ 21 22 #include "sys/types.h" 23 24 #include "machine/reg.h" 25 26 #include "fpu_arith.h" 27 #include "fpu_emu.h" 28 29 struct fpn * 30 fpu_add(fe) 31 register struct fpemu *fe; 32 { 33 register struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2, *r; 34 register u_int r0, r1, r2, r3; 35 register int rd; 36 37 /* 38 * Put the `heavier' operand on the right (see fpu_emu.h). 39 * Then we will have one of the following cases, taken in the 40 * following order: 41 * 42 * - y = NaN. Implied: if only one is a signalling NaN, y is. 43 * The result is y. 44 * - y = Inf. Implied: x != NaN (is 0, number, or Inf: the NaN 45 * case was taken care of earlier). 46 * If x = -y, the result is NaN. Otherwise the result 47 * is y (an Inf of whichever sign). 48 * - y is 0. Implied: x = 0. 49 * If x and y differ in sign (one positive, one negative), 50 * the result is +0 except when rounding to -Inf. If same: 51 * +0 + +0 = +0; -0 + -0 = -0. 52 * - x is 0. Implied: y != 0. 53 * Result is y. 54 * - other. Implied: both x and y are numbers. 55 * Do addition a la Hennessey & Patterson. 56 */ 57 ORDER(x, y); 58 if (ISNAN(y)) 59 return (y); 60 if (ISINF(y)) { 61 if (ISINF(x) && x->fp_sign != y->fp_sign) 62 return (fpu_newnan(fe)); 63 return (y); 64 } 65 rd = ((fe->fe_fsr >> FSR_RD_SHIFT) & FSR_RD_MASK); 66 if (ISZERO(y)) { 67 if (rd != FSR_RD_RM) /* only -0 + -0 gives -0 */ 68 y->fp_sign &= x->fp_sign; 69 else /* any -0 operand gives -0 */ 70 y->fp_sign |= x->fp_sign; 71 return (y); 72 } 73 if (ISZERO(x)) 74 return (y); 75 /* 76 * We really have two numbers to add, although their signs may 77 * differ. Make the exponents match, by shifting the smaller 78 * number right (e.g., 1.011 => 0.1011) and increasing its 79 * exponent (2^3 => 2^4). Note that we do not alter the exponents 80 * of x and y here. 81 */ 82 r = &fe->fe_f3; 83 r->fp_class = FPC_NUM; 84 if (x->fp_exp == y->fp_exp) { 85 r->fp_exp = x->fp_exp; 86 r->fp_sticky = 0; 87 } else { 88 if (x->fp_exp < y->fp_exp) { 89 /* 90 * Try to avoid subtract case iii (see below). 91 * This also guarantees that x->fp_sticky = 0. 92 */ 93 SWAP(x, y); 94 } 95 /* now x->fp_exp > y->fp_exp */ 96 r->fp_exp = x->fp_exp; 97 r->fp_sticky = fpu_shr(y, x->fp_exp - y->fp_exp); 98 } 99 r->fp_sign = x->fp_sign; 100 if (x->fp_sign == y->fp_sign) { 101 FPU_DECL_CARRY 102 103 /* 104 * The signs match, so we simply add the numbers. The result 105 * may be `supernormal' (as big as 1.111...1 + 1.111...1, or 106 * 11.111...0). If so, a single bit shift-right will fix it 107 * (but remember to adjust the exponent). 108 */ 109 /* r->fp_mant = x->fp_mant + y->fp_mant */ 110 FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]); 111 FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]); 112 FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]); 113 FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]); 114 if ((r->fp_mant[0] = r0) >= FP_2) { 115 (void) fpu_shr(r, 1); 116 r->fp_exp++; 117 } 118 } else { 119 FPU_DECL_CARRY 120 121 /* 122 * The signs differ, so things are rather more difficult. 123 * H&P would have us negate the negative operand and add; 124 * this is the same as subtracting the negative operand. 125 * This is quite a headache. Instead, we will subtract 126 * y from x, regardless of whether y itself is the negative 127 * operand. When this is done one of three conditions will 128 * hold, depending on the magnitudes of x and y: 129 * case i) |x| > |y|. The result is just x - y, 130 * with x's sign, but it may need to be normalized. 131 * case ii) |x| = |y|. The result is 0 (maybe -0) 132 * so must be fixed up. 133 * case iii) |x| < |y|. We goofed; the result should 134 * be (y - x), with the same sign as y. 135 * We could compare |x| and |y| here and avoid case iii, 136 * but that would take just as much work as the subtract. 137 * We can tell case iii has occurred by an overflow. 138 * 139 * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0. 140 */ 141 /* r->fp_mant = x->fp_mant - y->fp_mant */ 142 FPU_SET_CARRY(y->fp_sticky); 143 FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]); 144 FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]); 145 FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]); 146 FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]); 147 if (r0 < FP_2) { 148 /* cases i and ii */ 149 if ((r0 | r1 | r2 | r3) == 0) { 150 /* case ii */ 151 r->fp_class = FPC_ZERO; 152 r->fp_sign = rd == FSR_RD_RM; 153 return (r); 154 } 155 } else { 156 /* 157 * Oops, case iii. This can only occur when the 158 * exponents were equal, in which case neither 159 * x nor y have sticky bits set. Flip the sign 160 * (to y's sign) and negate the result to get y - x. 161 */ 162 #ifdef DIAGNOSTIC 163 if (x->fp_exp != y->fp_exp || r->fp_sticky) 164 panic("fpu_add"); 165 #endif 166 r->fp_sign = y->fp_sign; 167 FPU_SUBS(r3, 0, r3); 168 FPU_SUBCS(r2, 0, r2); 169 FPU_SUBCS(r1, 0, r1); 170 FPU_SUBC(r0, 0, r0); 171 } 172 r->fp_mant[3] = r3; 173 r->fp_mant[2] = r2; 174 r->fp_mant[1] = r1; 175 r->fp_mant[0] = r0; 176 if (r0 < FP_1) 177 fpu_norm(r); 178 } 179 return (r); 180 } 181