1/* $OpenBSD: fpu_sqrt.c,v 1.7 2024/03/29 21:07:11 miod Exp $ */ 2/* $NetBSD: fpu_sqrt.c,v 1.2 1994/11/20 20:52:46 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_sqrt.c 8.1 (Berkeley) 6/11/93 42 */ 43 44/* 45 * Perform an FPU square root (return sqrt(x)). 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 * Our task is to calculate the square root of a floating point number x0. 57 * This number x normally has the form: 58 * 59 * exp 60 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 61 * 62 * This can be left as it stands, or the mantissa can be doubled and the 63 * exponent decremented: 64 * 65 * exp-1 66 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 67 * 68 * If the exponent `exp' is even, the square root of the number is best 69 * handled using the first form, and is by definition equal to: 70 * 71 * exp/2 72 * sqrt(x) = sqrt(mant) * 2 73 * 74 * If exp is odd, on the other hand, it is convenient to use the second 75 * form, giving: 76 * 77 * (exp-1)/2 78 * sqrt(x) = sqrt(2 * mant) * 2 79 * 80 * In the first case, we have 81 * 82 * 1 <= mant < 2 83 * 84 * and therefore 85 * 86 * sqrt(1) <= sqrt(mant) < sqrt(2) 87 * 88 * while in the second case we have 89 * 90 * 2 <= 2*mant < 4 91 * 92 * and therefore 93 * 94 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 95 * 96 * so that in any case, we are sure that 97 * 98 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 99 * 100 * or 101 * 102 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 103 * 104 * This root is therefore a properly formed mantissa for a floating 105 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 106 * as above. This leaves us with the problem of finding the square root 107 * of a fixed-point number in the range [1..4). 108 * 109 * Though it may not be instantly obvious, the following square root 110 * algorithm works for any integer x of an even number of bits, provided 111 * that no overflows occur: 112 * 113 * let q = 0 114 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 115 * x *= 2 -- multiply by radix, for next digit 116 * if x >= 2q + 2^k then -- if adding 2^k does not 117 * x -= 2q + 2^k -- exceed the correct root, 118 * q += 2^k -- add 2^k and adjust x 119 * fi 120 * done 121 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 122 * 123 * If NBITS is odd (so that k is initially even), we can just add another 124 * zero bit at the top of x. Doing so means that q is not going to acquire 125 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 126 * final value in x is not needed, or can be off by a factor of 2, this is 127 * equivalent to moving the `x *= 2' step to the bottom of the loop: 128 * 129 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 130 * 131 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 132 * (Since the algorithm is destructive on x, we will call x's initial 133 * value, for which q is some power of two times its square root, x0.) 134 * 135 * If we insert a loop invariant y = 2q, we can then rewrite this using 136 * C notation as: 137 * 138 * q = y = 0; x = x0; 139 * for (k = NBITS; --k >= 0;) { 140 * #if (NBITS is even) 141 * x *= 2; 142 * #endif 143 * t = y + (1 << k); 144 * if (x >= t) { 145 * x -= t; 146 * q += 1 << k; 147 * y += 1 << (k + 1); 148 * } 149 * #if (NBITS is odd) 150 * x *= 2; 151 * #endif 152 * } 153 * 154 * If x0 is fixed point, rather than an integer, we can simply alter the 155 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 156 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 157 * 158 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 159 * integers, which adds some complication. But note that q is built one 160 * bit at a time, from the top down, and is not used itself in the loop 161 * (we use 2q as held in y instead). This means we can build our answer 162 * in an integer, one word at a time, which saves a bit of work. Also, 163 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 164 * `new' bits in y and we can set them with an `or' operation rather than 165 * a full-blown multiword add. 166 * 167 * We are almost done, except for one snag. We must prove that none of our 168 * intermediate calculations can overflow. We know that x0 is in [1..4) 169 * and therefore the square root in q will be in [1..2), but what about x, 170 * y, and t? 171 * 172 * We know that y = 2q at the beginning of each loop. (The relation only 173 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 174 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 175 * Furthermore, we can prove with a bit of work that x never exceeds y by 176 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 177 * an exercise to the reader, mostly because I have become tired of working 178 * on this comment.) 179 * 180 * If our floating point mantissas (which are of the form 1.frac) occupy 181 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 182 * In fact, we want even one more bit (for a carry, to avoid compares), or 183 * three extra. There is a comment in fpu_emu.h reminding maintainers of 184 * this, so we have some justification in assuming it. 185 */ 186struct fpn * 187fpu_sqrt(struct fpemu *fe) 188{ 189 struct fpn *x = &fe->fe_f1; 190 u_int bit, q, tt; 191 u_int x0, x1, x2, x3; 192 u_int y0, y1, y2, y3; 193 u_int d0, d1, d2, d3; 194 int e; 195 196 /* 197 * Take care of special cases first. In order: 198 * 199 * sqrt(NaN) = NaN 200 * sqrt(+0) = +0 201 * sqrt(-0) = -0 202 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 203 * sqrt(+Inf) = +Inf 204 * 205 * Then all that remains are numbers with mantissas in [1..2). 206 */ 207 if (ISNAN(x) || ISZERO(x)) 208 return (x); 209 if (x->fp_sign) 210 return (fpu_newnan(fe)); 211 if (ISINF(x)) 212 return (x); 213 214 /* 215 * Calculate result exponent. As noted above, this may involve 216 * doubling the mantissa. We will also need to double x each 217 * time around the loop, so we define a macro for this here, and 218 * we break out the multiword mantissa. 219 */ 220#ifdef FPU_SHL1_BY_ADD 221#define DOUBLE_X { \ 222 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 223 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 224} 225#else 226#define DOUBLE_X { \ 227 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 228 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 229} 230#endif 231#if (FP_NMANT & 1) != 0 232# define ODD_DOUBLE DOUBLE_X 233# define EVEN_DOUBLE /* nothing */ 234#else 235# define ODD_DOUBLE /* nothing */ 236# define EVEN_DOUBLE DOUBLE_X 237#endif 238 x0 = x->fp_mant[0]; 239 x1 = x->fp_mant[1]; 240 x2 = x->fp_mant[2]; 241 x3 = x->fp_mant[3]; 242 e = x->fp_exp; 243 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 244 DOUBLE_X; 245 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 246 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 247 248 /* 249 * Now calculate the mantissa root. Since x is now in [1..4), 250 * we know that the first trip around the loop will definitely 251 * set the top bit in q, so we can do that manually and start 252 * the loop at the next bit down instead. We must be sure to 253 * double x correctly while doing the `known q=1.0'. 254 * 255 * We do this one mantissa-word at a time, as noted above, to 256 * save work. To avoid `(1U << 31) << 1', we also do the top bit 257 * outside of each per-word loop. 258 * 259 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 260 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 261 * is always a `new' one, this means that three of the `t?'s are 262 * just the corresponding `y?'; we use `#define's here for this. 263 * The variable `tt' holds the actual `t?' variable. 264 */ 265 266 /* calculate q0 */ 267#define t0 tt 268 bit = FP_1; 269 EVEN_DOUBLE; 270 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 271 q = bit; 272 x0 -= bit; 273 y0 = bit << 1; 274 /* } */ 275 ODD_DOUBLE; 276 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 277 EVEN_DOUBLE; 278 t0 = y0 | bit; /* t = y + bit */ 279 if (x0 >= t0) { /* if x >= t then */ 280 x0 -= t0; /* x -= t */ 281 q |= bit; /* q += bit */ 282 y0 |= bit << 1; /* y += bit << 1 */ 283 } 284 ODD_DOUBLE; 285 } 286 x->fp_mant[0] = q; 287#undef t0 288 289 /* calculate q1. note (y0&1)==0. */ 290#define t0 y0 291#define t1 tt 292 q = 0; 293 y1 = 0; 294 bit = 1U << 31; 295 EVEN_DOUBLE; 296 t1 = bit; 297 FPU_SUBS(d1, x1, t1); 298 FPU_SUBC(d0, x0, t0); /* d = x - t */ 299 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 300 x0 = d0, x1 = d1; /* x -= t */ 301 q = bit; /* q += bit */ 302 y0 |= 1; /* y += bit << 1 */ 303 } 304 ODD_DOUBLE; 305 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 306 EVEN_DOUBLE; /* as before */ 307 t1 = y1 | bit; 308 FPU_SUBS(d1, x1, t1); 309 FPU_SUBC(d0, x0, t0); 310 if ((int)d0 >= 0) { 311 x0 = d0, x1 = d1; 312 q |= bit; 313 y1 |= bit << 1; 314 } 315 ODD_DOUBLE; 316 } 317 x->fp_mant[1] = q; 318#undef t1 319 320 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 321#define t1 y1 322#define t2 tt 323 q = 0; 324 y2 = 0; 325 bit = 1U << 31; 326 EVEN_DOUBLE; 327 t2 = bit; 328 FPU_SUBS(d2, x2, t2); 329 FPU_SUBCS(d1, x1, t1); 330 FPU_SUBC(d0, x0, t0); 331 if ((int)d0 >= 0) { 332 x0 = d0, x1 = d1, x2 = d2; 333 q |= bit; 334 y1 |= 1; /* now t1, y1 are set in concrete */ 335 } 336 ODD_DOUBLE; 337 while ((bit >>= 1) != 0) { 338 EVEN_DOUBLE; 339 t2 = y2 | bit; 340 FPU_SUBS(d2, x2, t2); 341 FPU_SUBCS(d1, x1, t1); 342 FPU_SUBC(d0, x0, t0); 343 if ((int)d0 >= 0) { 344 x0 = d0, x1 = d1, x2 = d2; 345 q |= bit; 346 y2 |= bit << 1; 347 } 348 ODD_DOUBLE; 349 } 350 x->fp_mant[2] = q; 351#undef t2 352 353 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 354#define t2 y2 355#define t3 tt 356 q = 0; 357 y3 = 0; 358 bit = 1U << 31; 359 EVEN_DOUBLE; 360 t3 = bit; 361 FPU_SUBS(d3, x3, t3); 362 FPU_SUBCS(d2, x2, t2); 363 FPU_SUBCS(d1, x1, t1); 364 FPU_SUBC(d0, x0, t0); 365 if ((int)d0 >= 0) { 366 x0 = d0, x1 = d1, x2 = d2, x3 = d3; 367 q |= bit; 368 y2 |= 1; 369 } 370 ODD_DOUBLE; 371 while ((bit >>= 1) != 0) { 372 EVEN_DOUBLE; 373 t3 = y3 | bit; 374 FPU_SUBS(d3, x3, t3); 375 FPU_SUBCS(d2, x2, t2); 376 FPU_SUBCS(d1, x1, t1); 377 FPU_SUBC(d0, x0, t0); 378 if ((int)d0 >= 0) { 379 x0 = d0, x1 = d1, x2 = d2, x3 = d3; 380 q |= bit; 381 y3 |= bit << 1; 382 } 383 ODD_DOUBLE; 384 } 385 x->fp_mant[3] = q; 386 387 /* 388 * The result, which includes guard and round bits, is exact iff 389 * x is now zero; any nonzero bits in x represent sticky bits. 390 */ 391 x->fp_sticky = x0 | x1 | x2 | x3; 392 return (x); 393} 394