fpu_sqrt.c revision 92986
1/* 2 * Copyright (c) 1992, 1993 3 * The Regents of the University of California. 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 * All advertising materials mentioning features or use of this software 10 * must display the following acknowledgement: 11 * This product includes software developed by the University of 12 * California, Lawrence Berkeley Laboratory. 13 * 14 * Redistribution and use in source and binary forms, with or without 15 * modification, are permitted provided that the following conditions 16 * are met: 17 * 1. Redistributions of source code must retain the above copyright 18 * notice, this list of conditions and the following disclaimer. 19 * 2. Redistributions in binary form must reproduce the above copyright 20 * notice, this list of conditions and the following disclaimer in the 21 * documentation and/or other materials provided with the distribution. 22 * 3. All advertising materials mentioning features or use of this software 23 * must display the following acknowledgement: 24 * This product includes software developed by the University of 25 * California, Berkeley and its contributors. 26 * 4. 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 * $NetBSD: fpu_sqrt.c,v 1.2 1994/11/20 20:52:46 deraadt Exp $ 44 */ 45 46#include <sys/cdefs.h> 47__FBSDID("$FreeBSD: head/lib/libc/sparc64/fpu/fpu_sqrt.c 92986 2002-03-22 21:53:29Z obrien $"); 48 49/* 50 * Perform an FPU square root (return sqrt(x)). 51 */ 52 53#include <sys/types.h> 54 55#include <machine/frame.h> 56#include <machine/fp.h> 57 58#include "fpu_arith.h" 59#include "fpu_emu.h" 60#include "fpu_extern.h" 61 62/* 63 * Our task is to calculate the square root of a floating point number x0. 64 * This number x normally has the form: 65 * 66 * exp 67 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 68 * 69 * This can be left as it stands, or the mantissa can be doubled and the 70 * exponent decremented: 71 * 72 * exp-1 73 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 74 * 75 * If the exponent `exp' is even, the square root of the number is best 76 * handled using the first form, and is by definition equal to: 77 * 78 * exp/2 79 * sqrt(x) = sqrt(mant) * 2 80 * 81 * If exp is odd, on the other hand, it is convenient to use the second 82 * form, giving: 83 * 84 * (exp-1)/2 85 * sqrt(x) = sqrt(2 * mant) * 2 86 * 87 * In the first case, we have 88 * 89 * 1 <= mant < 2 90 * 91 * and therefore 92 * 93 * sqrt(1) <= sqrt(mant) < sqrt(2) 94 * 95 * while in the second case we have 96 * 97 * 2 <= 2*mant < 4 98 * 99 * and therefore 100 * 101 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 102 * 103 * so that in any case, we are sure that 104 * 105 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 106 * 107 * or 108 * 109 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 110 * 111 * This root is therefore a properly formed mantissa for a floating 112 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 113 * as above. This leaves us with the problem of finding the square root 114 * of a fixed-point number in the range [1..4). 115 * 116 * Though it may not be instantly obvious, the following square root 117 * algorithm works for any integer x of an even number of bits, provided 118 * that no overflows occur: 119 * 120 * let q = 0 121 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 122 * x *= 2 -- multiply by radix, for next digit 123 * if x >= 2q + 2^k then -- if adding 2^k does not 124 * x -= 2q + 2^k -- exceed the correct root, 125 * q += 2^k -- add 2^k and adjust x 126 * fi 127 * done 128 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 129 * 130 * If NBITS is odd (so that k is initially even), we can just add another 131 * zero bit at the top of x. Doing so means that q is not going to acquire 132 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 133 * final value in x is not needed, or can be off by a factor of 2, this is 134 * equivalant to moving the `x *= 2' step to the bottom of the loop: 135 * 136 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 137 * 138 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 139 * (Since the algorithm is destructive on x, we will call x's initial 140 * value, for which q is some power of two times its square root, x0.) 141 * 142 * If we insert a loop invariant y = 2q, we can then rewrite this using 143 * C notation as: 144 * 145 * q = y = 0; x = x0; 146 * for (k = NBITS; --k >= 0;) { 147 * #if (NBITS is even) 148 * x *= 2; 149 * #endif 150 * t = y + (1 << k); 151 * if (x >= t) { 152 * x -= t; 153 * q += 1 << k; 154 * y += 1 << (k + 1); 155 * } 156 * #if (NBITS is odd) 157 * x *= 2; 158 * #endif 159 * } 160 * 161 * If x0 is fixed point, rather than an integer, we can simply alter the 162 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 163 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 164 * 165 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 166 * integers, which adds some complication. But note that q is built one 167 * bit at a time, from the top down, and is not used itself in the loop 168 * (we use 2q as held in y instead). This means we can build our answer 169 * in an integer, one word at a time, which saves a bit of work. Also, 170 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 171 * `new' bits in y and we can set them with an `or' operation rather than 172 * a full-blown multiword add. 173 * 174 * We are almost done, except for one snag. We must prove that none of our 175 * intermediate calculations can overflow. We know that x0 is in [1..4) 176 * and therefore the square root in q will be in [1..2), but what about x, 177 * y, and t? 178 * 179 * We know that y = 2q at the beginning of each loop. (The relation only 180 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 181 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 182 * Furthermore, we can prove with a bit of work that x never exceeds y by 183 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 184 * an exercise to the reader, mostly because I have become tired of working 185 * on this comment.) 186 * 187 * If our floating point mantissas (which are of the form 1.frac) occupy 188 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 189 * In fact, we want even one more bit (for a carry, to avoid compares), or 190 * three extra. There is a comment in fpu_emu.h reminding maintainers of 191 * this, so we have some justification in assuming it. 192 */ 193struct fpn * 194__fpu_sqrt(fe) 195 struct fpemu *fe; 196{ 197 struct fpn *x = &fe->fe_f1; 198 u_int bit, q, tt; 199 u_int x0, x1, x2, x3; 200 u_int y0, y1, y2, y3; 201 u_int d0, d1, d2, d3; 202 int e; 203 204 /* 205 * Take care of special cases first. In order: 206 * 207 * sqrt(NaN) = NaN 208 * sqrt(+0) = +0 209 * sqrt(-0) = -0 210 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 211 * sqrt(+Inf) = +Inf 212 * 213 * Then all that remains are numbers with mantissas in [1..2). 214 */ 215 if (ISNAN(x) || ISZERO(x)) 216 return (x); 217 if (x->fp_sign) 218 return (__fpu_newnan(fe)); 219 if (ISINF(x)) 220 return (x); 221 222 /* 223 * Calculate result exponent. As noted above, this may involve 224 * doubling the mantissa. We will also need to double x each 225 * time around the loop, so we define a macro for this here, and 226 * we break out the multiword mantissa. 227 */ 228#ifdef FPU_SHL1_BY_ADD 229#define DOUBLE_X { \ 230 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 231 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 232} 233#else 234#define DOUBLE_X { \ 235 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 236 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 237} 238#endif 239#if (FP_NMANT & 1) != 0 240# define ODD_DOUBLE DOUBLE_X 241# define EVEN_DOUBLE /* nothing */ 242#else 243# define ODD_DOUBLE /* nothing */ 244# define EVEN_DOUBLE DOUBLE_X 245#endif 246 x0 = x->fp_mant[0]; 247 x1 = x->fp_mant[1]; 248 x2 = x->fp_mant[2]; 249 x3 = x->fp_mant[3]; 250 e = x->fp_exp; 251 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 252 DOUBLE_X; 253 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 254 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 255 256 /* 257 * Now calculate the mantissa root. Since x is now in [1..4), 258 * we know that the first trip around the loop will definitely 259 * set the top bit in q, so we can do that manually and start 260 * the loop at the next bit down instead. We must be sure to 261 * double x correctly while doing the `known q=1.0'. 262 * 263 * We do this one mantissa-word at a time, as noted above, to 264 * save work. To avoid `(1 << 31) << 1', we also do the top bit 265 * outside of each per-word loop. 266 * 267 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 268 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 269 * is always a `new' one, this means that three of the `t?'s are 270 * just the corresponding `y?'; we use `#define's here for this. 271 * The variable `tt' holds the actual `t?' variable. 272 */ 273 274 /* calculate q0 */ 275#define t0 tt 276 bit = FP_1; 277 EVEN_DOUBLE; 278 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 279 q = bit; 280 x0 -= bit; 281 y0 = bit << 1; 282 /* } */ 283 ODD_DOUBLE; 284 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 285 EVEN_DOUBLE; 286 t0 = y0 | bit; /* t = y + bit */ 287 if (x0 >= t0) { /* if x >= t then */ 288 x0 -= t0; /* x -= t */ 289 q |= bit; /* q += bit */ 290 y0 |= bit << 1; /* y += bit << 1 */ 291 } 292 ODD_DOUBLE; 293 } 294 x->fp_mant[0] = q; 295#undef t0 296 297 /* calculate q1. note (y0&1)==0. */ 298#define t0 y0 299#define t1 tt 300 q = 0; 301 y1 = 0; 302 bit = 1 << 31; 303 EVEN_DOUBLE; 304 t1 = bit; 305 FPU_SUBS(d1, x1, t1); 306 FPU_SUBC(d0, x0, t0); /* d = x - t */ 307 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 308 x0 = d0, x1 = d1; /* x -= t */ 309 q = bit; /* q += bit */ 310 y0 |= 1; /* y += bit << 1 */ 311 } 312 ODD_DOUBLE; 313 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 314 EVEN_DOUBLE; /* as before */ 315 t1 = y1 | bit; 316 FPU_SUBS(d1, x1, t1); 317 FPU_SUBC(d0, x0, t0); 318 if ((int)d0 >= 0) { 319 x0 = d0, x1 = d1; 320 q |= bit; 321 y1 |= bit << 1; 322 } 323 ODD_DOUBLE; 324 } 325 x->fp_mant[1] = q; 326#undef t1 327 328 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 329#define t1 y1 330#define t2 tt 331 q = 0; 332 y2 = 0; 333 bit = 1 << 31; 334 EVEN_DOUBLE; 335 t2 = bit; 336 FPU_SUBS(d2, x2, t2); 337 FPU_SUBCS(d1, x1, t1); 338 FPU_SUBC(d0, x0, t0); 339 if ((int)d0 >= 0) { 340 x0 = d0, x1 = d1, x2 = d2; 341 q |= bit; 342 y1 |= 1; /* now t1, y1 are set in concrete */ 343 } 344 ODD_DOUBLE; 345 while ((bit >>= 1) != 0) { 346 EVEN_DOUBLE; 347 t2 = y2 | bit; 348 FPU_SUBS(d2, x2, t2); 349 FPU_SUBCS(d1, x1, t1); 350 FPU_SUBC(d0, x0, t0); 351 if ((int)d0 >= 0) { 352 x0 = d0, x1 = d1, x2 = d2; 353 q |= bit; 354 y2 |= bit << 1; 355 } 356 ODD_DOUBLE; 357 } 358 x->fp_mant[2] = q; 359#undef t2 360 361 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 362#define t2 y2 363#define t3 tt 364 q = 0; 365 y3 = 0; 366 bit = 1 << 31; 367 EVEN_DOUBLE; 368 t3 = bit; 369 FPU_SUBS(d3, x3, t3); 370 FPU_SUBCS(d2, x2, t2); 371 FPU_SUBCS(d1, x1, t1); 372 FPU_SUBC(d0, x0, t0); 373 ODD_DOUBLE; 374 if ((int)d0 >= 0) { 375 x0 = d0, x1 = d1, x2 = d2; 376 q |= bit; 377 y2 |= 1; 378 } 379 while ((bit >>= 1) != 0) { 380 EVEN_DOUBLE; 381 t3 = y3 | bit; 382 FPU_SUBS(d3, x3, t3); 383 FPU_SUBCS(d2, x2, t2); 384 FPU_SUBCS(d1, x1, t1); 385 FPU_SUBC(d0, x0, t0); 386 if ((int)d0 >= 0) { 387 x0 = d0, x1 = d1, x2 = d2; 388 q |= bit; 389 y3 |= bit << 1; 390 } 391 ODD_DOUBLE; 392 } 393 x->fp_mant[3] = q; 394 395 /* 396 * The result, which includes guard and round bits, is exact iff 397 * x is now zero; any nonzero bits in x represent sticky bits. 398 */ 399 x->fp_sticky = x0 | x1 | x2 | x3; 400 return (x); 401} 402