1/* $NetBSD: fpu_sqrt.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */ 2 3/*- 4 * SPDX-License-Identifier: BSD-3-Clause 5 * 6 * Copyright (c) 1992, 1993 7 * The Regents of the University of California. All rights reserved. 8 * 9 * This software was developed by the Computer Systems Engineering group 10 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 11 * contributed to Berkeley. 12 * 13 * All advertising materials mentioning features or use of this software 14 * must display the following acknowledgement: 15 * This product includes software developed by the University of 16 * California, Lawrence Berkeley Laboratory. 17 * 18 * Redistribution and use in source and binary forms, with or without 19 * modification, are permitted provided that the following conditions 20 * are met: 21 * 1. Redistributions of source code must retain the above copyright 22 * notice, this list of conditions and the following disclaimer. 23 * 2. Redistributions in binary form must reproduce the above copyright 24 * notice, this list of conditions and the following disclaimer in the 25 * documentation and/or other materials provided with the distribution. 26 * 3. Neither the name of the University nor the names of its contributors 27 * may be used to endorse or promote products derived from this software 28 * without specific prior written permission. 29 * 30 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 31 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 32 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 33 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 34 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 35 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 36 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 37 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 38 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 39 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 40 * SUCH DAMAGE. 41 * 42 * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93 43 */ 44 45/* 46 * Perform an FPU square root (return sqrt(x)). 47 */ 48 49#include <sys/cdefs.h> 50__FBSDID("$FreeBSD$"); 51 52#include <sys/types.h> 53#include <sys/systm.h> 54 55#include <machine/fpu.h> 56#include <machine/reg.h> 57 58#include <powerpc/fpu/fpu_arith.h> 59#include <powerpc/fpu/fpu_emu.h> 60 61/* 62 * Our task is to calculate the square root of a floating point number x0. 63 * This number x normally has the form: 64 * 65 * exp 66 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 67 * 68 * This can be left as it stands, or the mantissa can be doubled and the 69 * exponent decremented: 70 * 71 * exp-1 72 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 73 * 74 * If the exponent `exp' is even, the square root of the number is best 75 * handled using the first form, and is by definition equal to: 76 * 77 * exp/2 78 * sqrt(x) = sqrt(mant) * 2 79 * 80 * If exp is odd, on the other hand, it is convenient to use the second 81 * form, giving: 82 * 83 * (exp-1)/2 84 * sqrt(x) = sqrt(2 * mant) * 2 85 * 86 * In the first case, we have 87 * 88 * 1 <= mant < 2 89 * 90 * and therefore 91 * 92 * sqrt(1) <= sqrt(mant) < sqrt(2) 93 * 94 * while in the second case we have 95 * 96 * 2 <= 2*mant < 4 97 * 98 * and therefore 99 * 100 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 101 * 102 * so that in any case, we are sure that 103 * 104 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 105 * 106 * or 107 * 108 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 109 * 110 * This root is therefore a properly formed mantissa for a floating 111 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 112 * as above. This leaves us with the problem of finding the square root 113 * of a fixed-point number in the range [1..4). 114 * 115 * Though it may not be instantly obvious, the following square root 116 * algorithm works for any integer x of an even number of bits, provided 117 * that no overflows occur: 118 * 119 * let q = 0 120 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 121 * x *= 2 -- multiply by radix, for next digit 122 * if x >= 2q + 2^k then -- if adding 2^k does not 123 * x -= 2q + 2^k -- exceed the correct root, 124 * q += 2^k -- add 2^k and adjust x 125 * fi 126 * done 127 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 128 * 129 * If NBITS is odd (so that k is initially even), we can just add another 130 * zero bit at the top of x. Doing so means that q is not going to acquire 131 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 132 * final value in x is not needed, or can be off by a factor of 2, this is 133 * equivalant to moving the `x *= 2' step to the bottom of the loop: 134 * 135 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 136 * 137 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 138 * (Since the algorithm is destructive on x, we will call x's initial 139 * value, for which q is some power of two times its square root, x0.) 140 * 141 * If we insert a loop invariant y = 2q, we can then rewrite this using 142 * C notation as: 143 * 144 * q = y = 0; x = x0; 145 * for (k = NBITS; --k >= 0;) { 146 * #if (NBITS is even) 147 * x *= 2; 148 * #endif 149 * t = y + (1 << k); 150 * if (x >= t) { 151 * x -= t; 152 * q += 1 << k; 153 * y += 1 << (k + 1); 154 * } 155 * #if (NBITS is odd) 156 * x *= 2; 157 * #endif 158 * } 159 * 160 * If x0 is fixed point, rather than an integer, we can simply alter the 161 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 162 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 163 * 164 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 165 * integers, which adds some complication. But note that q is built one 166 * bit at a time, from the top down, and is not used itself in the loop 167 * (we use 2q as held in y instead). This means we can build our answer 168 * in an integer, one word at a time, which saves a bit of work. Also, 169 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 170 * `new' bits in y and we can set them with an `or' operation rather than 171 * a full-blown multiword add. 172 * 173 * We are almost done, except for one snag. We must prove that none of our 174 * intermediate calculations can overflow. We know that x0 is in [1..4) 175 * and therefore the square root in q will be in [1..2), but what about x, 176 * y, and t? 177 * 178 * We know that y = 2q at the beginning of each loop. (The relation only 179 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 180 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 181 * Furthermore, we can prove with a bit of work that x never exceeds y by 182 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 183 * an exercise to the reader, mostly because I have become tired of working 184 * on this comment.) 185 * 186 * If our floating point mantissas (which are of the form 1.frac) occupy 187 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 188 * In fact, we want even one more bit (for a carry, to avoid compares), or 189 * three extra. There is a comment in fpu_emu.h reminding maintainers of 190 * this, so we have some justification in assuming it. 191 */ 192struct fpn * 193fpu_sqrt(struct fpemu *fe) 194{ 195 struct fpn *x = &fe->fe_f1; 196 u_int bit, q, tt; 197 u_int x0, x1, x2, x3; 198 u_int y0, y1, y2, y3; 199 u_int d0, d1, d2, d3; 200 int e; 201 FPU_DECL_CARRY; 202 203 /* 204 * Take care of special cases first. In order: 205 * 206 * sqrt(NaN) = NaN 207 * sqrt(+0) = +0 208 * sqrt(-0) = -0 209 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 210 * sqrt(+Inf) = +Inf 211 * 212 * Then all that remains are numbers with mantissas in [1..2). 213 */ 214 DPRINTF(FPE_REG, ("fpu_sqer:\n")); 215 DUMPFPN(FPE_REG, x); 216 DPRINTF(FPE_REG, ("=>\n")); 217 if (ISNAN(x)) { 218 fe->fe_cx |= FPSCR_VXSNAN; 219 DUMPFPN(FPE_REG, x); 220 return (x); 221 } 222 if (ISZERO(x)) { 223 fe->fe_cx |= FPSCR_ZX; 224 x->fp_class = FPC_INF; 225 DUMPFPN(FPE_REG, x); 226 return (x); 227 } 228 if (x->fp_sign) { 229 fe->fe_cx |= FPSCR_VXSQRT; 230 return (fpu_newnan(fe)); 231 } 232 if (ISINF(x)) { 233 DUMPFPN(FPE_REG, x); 234 return (x); 235 } 236 237 /* 238 * Calculate result exponent. As noted above, this may involve 239 * doubling the mantissa. We will also need to double x each 240 * time around the loop, so we define a macro for this here, and 241 * we break out the multiword mantissa. 242 */ 243#ifdef FPU_SHL1_BY_ADD 244#define DOUBLE_X { \ 245 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 246 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 247} 248#else 249#define DOUBLE_X { \ 250 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 251 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 252} 253#endif 254#if (FP_NMANT & 1) != 0 255# define ODD_DOUBLE DOUBLE_X 256# define EVEN_DOUBLE /* nothing */ 257#else 258# define ODD_DOUBLE /* nothing */ 259# define EVEN_DOUBLE DOUBLE_X 260#endif 261 x0 = x->fp_mant[0]; 262 x1 = x->fp_mant[1]; 263 x2 = x->fp_mant[2]; 264 x3 = x->fp_mant[3]; 265 e = x->fp_exp; 266 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 267 DOUBLE_X; 268 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 269 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 270 271 /* 272 * Now calculate the mantissa root. Since x is now in [1..4), 273 * we know that the first trip around the loop will definitely 274 * set the top bit in q, so we can do that manually and start 275 * the loop at the next bit down instead. We must be sure to 276 * double x correctly while doing the `known q=1.0'. 277 * 278 * We do this one mantissa-word at a time, as noted above, to 279 * save work. To avoid `(1U << 31) << 1', we also do the top bit 280 * outside of each per-word loop. 281 * 282 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 283 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 284 * is always a `new' one, this means that three of the `t?'s are 285 * just the corresponding `y?'; we use `#define's here for this. 286 * The variable `tt' holds the actual `t?' variable. 287 */ 288 289 /* calculate q0 */ 290#define t0 tt 291 bit = FP_1; 292 EVEN_DOUBLE; 293 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 294 q = bit; 295 x0 -= bit; 296 y0 = bit << 1; 297 /* } */ 298 ODD_DOUBLE; 299 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 300 EVEN_DOUBLE; 301 t0 = y0 | bit; /* t = y + bit */ 302 if (x0 >= t0) { /* if x >= t then */ 303 x0 -= t0; /* x -= t */ 304 q |= bit; /* q += bit */ 305 y0 |= bit << 1; /* y += bit << 1 */ 306 } 307 ODD_DOUBLE; 308 } 309 x->fp_mant[0] = q; 310#undef t0 311 312 /* calculate q1. note (y0&1)==0. */ 313#define t0 y0 314#define t1 tt 315 q = 0; 316 y1 = 0; 317 bit = 1 << 31; 318 EVEN_DOUBLE; 319 t1 = bit; 320 FPU_SUBS(d1, x1, t1); 321 FPU_SUBC(d0, x0, t0); /* d = x - t */ 322 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 323 x0 = d0, x1 = d1; /* x -= t */ 324 q = bit; /* q += bit */ 325 y0 |= 1; /* y += bit << 1 */ 326 } 327 ODD_DOUBLE; 328 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 329 EVEN_DOUBLE; /* as before */ 330 t1 = y1 | bit; 331 FPU_SUBS(d1, x1, t1); 332 FPU_SUBC(d0, x0, t0); 333 if ((int)d0 >= 0) { 334 x0 = d0, x1 = d1; 335 q |= bit; 336 y1 |= bit << 1; 337 } 338 ODD_DOUBLE; 339 } 340 x->fp_mant[1] = q; 341#undef t1 342 343 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 344#define t1 y1 345#define t2 tt 346 q = 0; 347 y2 = 0; 348 bit = 1 << 31; 349 EVEN_DOUBLE; 350 t2 = bit; 351 FPU_SUBS(d2, x2, t2); 352 FPU_SUBCS(d1, x1, t1); 353 FPU_SUBC(d0, x0, t0); 354 if ((int)d0 >= 0) { 355 x0 = d0, x1 = d1, x2 = d2; 356 q = bit; 357 y1 |= 1; /* now t1, y1 are set in concrete */ 358 } 359 ODD_DOUBLE; 360 while ((bit >>= 1) != 0) { 361 EVEN_DOUBLE; 362 t2 = y2 | bit; 363 FPU_SUBS(d2, x2, t2); 364 FPU_SUBCS(d1, x1, t1); 365 FPU_SUBC(d0, x0, t0); 366 if ((int)d0 >= 0) { 367 x0 = d0, x1 = d1, x2 = d2; 368 q |= bit; 369 y2 |= bit << 1; 370 } 371 ODD_DOUBLE; 372 } 373 x->fp_mant[2] = q; 374#undef t2 375 376 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 377#define t2 y2 378#define t3 tt 379 q = 0; 380 y3 = 0; 381 bit = 1 << 31; 382 EVEN_DOUBLE; 383 t3 = bit; 384 FPU_SUBS(d3, x3, t3); 385 FPU_SUBCS(d2, x2, t2); 386 FPU_SUBCS(d1, x1, t1); 387 FPU_SUBC(d0, x0, t0); 388 if ((int)d0 >= 0) { 389 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 390 q = bit; 391 y2 |= 1; 392 } 393 ODD_DOUBLE; 394 while ((bit >>= 1) != 0) { 395 EVEN_DOUBLE; 396 t3 = y3 | bit; 397 FPU_SUBS(d3, x3, t3); 398 FPU_SUBCS(d2, x2, t2); 399 FPU_SUBCS(d1, x1, t1); 400 FPU_SUBC(d0, x0, t0); 401 if ((int)d0 >= 0) { 402 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 403 q |= bit; 404 y3 |= bit << 1; 405 } 406 ODD_DOUBLE; 407 } 408 x->fp_mant[3] = q; 409 410 /* 411 * The result, which includes guard and round bits, is exact iff 412 * x is now zero; any nonzero bits in x represent sticky bits. 413 */ 414 x->fp_sticky = x0 | x1 | x2 | x3; 415 DUMPFPN(FPE_REG, x); 416 return (x); 417} 418