fpu_div.c revision 1.4
1/* $OpenBSD: fpu_div.c,v 1.4 2022/10/16 01:22:39 jsg Exp $ */ 2/* $NetBSD: fpu_div.c,v 1.2 1994/11/20 20:52:38 deraadt Exp $ */ 3 4/* 5 * Copyright (c) 1992, 1993 6 * The Regents of the University of California. All rights reserved. 7 * 8 * This software was developed by the Computer Systems Engineering group 9 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 10 * contributed to Berkeley. 11 * 12 * All advertising materials mentioning features or use of this software 13 * must display the following acknowledgement: 14 * This product includes software developed by the University of 15 * California, Lawrence Berkeley Laboratory. 16 * 17 * Redistribution and use in source and binary forms, with or without 18 * modification, are permitted provided that the following conditions 19 * are met: 20 * 1. Redistributions of source code must retain the above copyright 21 * notice, this list of conditions and the following disclaimer. 22 * 2. Redistributions in binary form must reproduce the above copyright 23 * notice, this list of conditions and the following disclaimer in the 24 * documentation and/or other materials provided with the distribution. 25 * 3. Neither the name of the University nor the names of its contributors 26 * may be used to endorse or promote products derived from this software 27 * without specific prior written permission. 28 * 29 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 30 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 31 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 32 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 33 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 34 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 35 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 36 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 37 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 38 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 39 * SUCH DAMAGE. 40 * 41 * @(#)fpu_div.c 8.1 (Berkeley) 6/11/93 42 */ 43 44/* 45 * Perform an FPU divide (return x / y). 46 */ 47 48#include <sys/types.h> 49 50#include <machine/reg.h> 51 52#include <sparc64/fpu/fpu_arith.h> 53#include <sparc64/fpu/fpu_emu.h> 54 55/* 56 * Division of normal numbers is done as follows: 57 * 58 * x and y are floating point numbers, i.e., in the form 1.bbbb * 2^e. 59 * If X and Y are the mantissas (1.bbbb's), the quotient is then: 60 * 61 * q = (X / Y) * 2^((x exponent) - (y exponent)) 62 * 63 * Since X and Y are both in [1.0,2.0), the quotient's mantissa (X / Y) 64 * will be in [0.5,2.0). Moreover, it will be less than 1.0 if and only 65 * if X < Y. In that case, it will have to be shifted left one bit to 66 * become a normal number, and the exponent decremented. Thus, the 67 * desired exponent is: 68 * 69 * left_shift = x->fp_mant < y->fp_mant; 70 * result_exp = x->fp_exp - y->fp_exp - left_shift; 71 * 72 * The quotient mantissa X/Y can then be computed one bit at a time 73 * using the following algorithm: 74 * 75 * Q = 0; -- Initial quotient. 76 * R = X; -- Initial remainder, 77 * if (left_shift) -- but fixed up in advance. 78 * R *= 2; 79 * for (bit = FP_NMANT; --bit >= 0; R *= 2) { 80 * if (R >= Y) { 81 * Q |= 1 << bit; 82 * R -= Y; 83 * } 84 * } 85 * 86 * The subtraction R -= Y always removes the uppermost bit from R (and 87 * can sometimes remove additional lower-order 1 bits); this proof is 88 * left to the reader. 89 * 90 * This loop correctly calculates the guard and round bits since they are 91 * included in the expanded internal representation. The sticky bit 92 * is to be set if and only if any other bits beyond guard and round 93 * would be set. From the above it is obvious that this is true if and 94 * only if the remainder R is nonzero when the loop terminates. 95 * 96 * Examining the loop above, we can see that the quotient Q is built 97 * one bit at a time ``from the top down''. This means that we can 98 * dispense with the multi-word arithmetic and just build it one word 99 * at a time, writing each result word when it is done. 100 * 101 * Furthermore, since X and Y are both in [1.0,2.0), we know that, 102 * initially, R >= Y. (Recall that, if X < Y, R is set to X * 2 and 103 * is therefore at in [2.0,4.0).) Thus Q is sure to have bit FP_NMANT-1 104 * set, and R can be set initially to either X - Y (when X >= Y) or 105 * 2X - Y (when X < Y). In addition, comparing R and Y is difficult, 106 * so we will simply calculate R - Y and see if that underflows. 107 * This leads to the following revised version of the algorithm: 108 * 109 * R = X; 110 * bit = FP_1; 111 * D = R - Y; 112 * if (D >= 0) { 113 * result_exp = x->fp_exp - y->fp_exp; 114 * R = D; 115 * q = bit; 116 * bit >>= 1; 117 * } else { 118 * result_exp = x->fp_exp - y->fp_exp - 1; 119 * q = 0; 120 * } 121 * R <<= 1; 122 * do { 123 * D = R - Y; 124 * if (D >= 0) { 125 * q |= bit; 126 * R = D; 127 * } 128 * R <<= 1; 129 * } while ((bit >>= 1) != 0); 130 * Q[0] = q; 131 * for (i = 1; i < 4; i++) { 132 * q = 0, bit = 1U << 31; 133 * do { 134 * D = R - Y; 135 * if (D >= 0) { 136 * q |= bit; 137 * R = D; 138 * } 139 * R <<= 1; 140 * } while ((bit >>= 1) != 0); 141 * Q[i] = q; 142 * } 143 * 144 * This can be refined just a bit further by moving the `R <<= 1' 145 * calculations to the front of the do-loops and eliding the first one. 146 * The process can be terminated immediately whenever R becomes 0, but 147 * this is relatively rare, and we do not bother. 148 */ 149 150struct fpn * 151fpu_div(register struct fpemu *fe) 152{ 153 register struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2; 154 register u_int q, bit; 155 register u_int r0, r1, r2, r3, d0, d1, d2, d3, y0, y1, y2, y3; 156 FPU_DECL_CARRY 157 158 /* 159 * Since divide is not commutative, we cannot just use ORDER. 160 * Check either operand for NaN first; if there is at least one, 161 * order the signalling one (if only one) onto the right, then 162 * return it. Otherwise we have the following cases: 163 * 164 * Inf / Inf = NaN, plus NV exception 165 * Inf / num = Inf [i.e., return x] 166 * Inf / 0 = Inf [i.e., return x] 167 * 0 / Inf = 0 [i.e., return x] 168 * 0 / num = 0 [i.e., return x] 169 * 0 / 0 = NaN, plus NV exception 170 * num / Inf = 0 171 * num / num = num (do the divide) 172 * num / 0 = Inf, plus DZ exception 173 */ 174 if (ISNAN(x) || ISNAN(y)) { 175 ORDER(x, y); 176 return (y); 177 } 178 if (ISINF(x) || ISZERO(x)) { 179 if (x->fp_class == y->fp_class) 180 return (fpu_newnan(fe)); 181 return (x); 182 } 183 184 /* all results at this point use XOR of operand signs */ 185 x->fp_sign ^= y->fp_sign; 186 if (ISINF(y)) { 187 x->fp_class = FPC_ZERO; 188 return (x); 189 } 190 if (ISZERO(y)) { 191 fe->fe_cx = FSR_DZ; 192 x->fp_class = FPC_INF; 193 return (x); 194 } 195 196 /* 197 * Macros for the divide. See comments at top for algorithm. 198 * Note that we expand R, D, and Y here. 199 */ 200 201#define SUBTRACT /* D = R - Y */ \ 202 FPU_SUBS(d3, r3, y3); FPU_SUBCS(d2, r2, y2); \ 203 FPU_SUBCS(d1, r1, y1); FPU_SUBC(d0, r0, y0) 204 205#define NONNEGATIVE /* D >= 0 */ \ 206 ((int)d0 >= 0) 207 208#ifdef FPU_SHL1_BY_ADD 209#define SHL1 /* R <<= 1 */ \ 210 FPU_ADDS(r3, r3, r3); FPU_ADDCS(r2, r2, r2); \ 211 FPU_ADDCS(r1, r1, r1); FPU_ADDC(r0, r0, r0) 212#else 213#define SHL1 \ 214 r0 = (r0 << 1) | (r1 >> 31), r1 = (r1 << 1) | (r2 >> 31), \ 215 r2 = (r2 << 1) | (r3 >> 31), r3 <<= 1 216#endif 217 218#define LOOP /* do ... while (bit >>= 1) */ \ 219 do { \ 220 SHL1; \ 221 SUBTRACT; \ 222 if (NONNEGATIVE) { \ 223 q |= bit; \ 224 r0 = d0, r1 = d1, r2 = d2, r3 = d3; \ 225 } \ 226 } while ((bit >>= 1) != 0) 227 228#define WORD(r, i) /* calculate r->fp_mant[i] */ \ 229 q = 0; \ 230 bit = 1U << 31; \ 231 LOOP; \ 232 (x)->fp_mant[i] = q 233 234 /* Setup. Note that we put our result in x. */ 235 r0 = x->fp_mant[0]; 236 r1 = x->fp_mant[1]; 237 r2 = x->fp_mant[2]; 238 r3 = x->fp_mant[3]; 239 y0 = y->fp_mant[0]; 240 y1 = y->fp_mant[1]; 241 y2 = y->fp_mant[2]; 242 y3 = y->fp_mant[3]; 243 244 bit = FP_1; 245 SUBTRACT; 246 if (NONNEGATIVE) { 247 x->fp_exp -= y->fp_exp; 248 r0 = d0, r1 = d1, r2 = d2, r3 = d3; 249 q = bit; 250 bit >>= 1; 251 } else { 252 x->fp_exp -= y->fp_exp + 1; 253 q = 0; 254 } 255 LOOP; 256 x->fp_mant[0] = q; 257 WORD(x, 1); 258 WORD(x, 2); 259 WORD(x, 3); 260 x->fp_sticky = r0 | r1 | r2 | r3; 261 262 return (x); 263} 264