1 /* $OpenBSD: fpu_sqrt.c,v 1.1 2003/07/21 18:41:30 jason Exp $ */ 2 3 /* 4 * Copyright (c) 1992, 1993 5 * The Regents of the University of California. All rights reserved. 6 * 7 * This software was developed by the Computer Systems Engineering group 8 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 9 * contributed to Berkeley. 10 * 11 * All advertising materials mentioning features or use of this software 12 * must display the following acknowledgement: 13 * This product includes software developed by the University of 14 * California, Lawrence Berkeley Laboratory. 15 * 16 * Redistribution and use in source and binary forms, with or without 17 * modification, are permitted provided that the following conditions 18 * are met: 19 * 1. Redistributions of source code must retain the above copyright 20 * notice, this list of conditions and the following disclaimer. 21 * 2. Redistributions in binary form must reproduce the above copyright 22 * notice, this list of conditions and the following disclaimer in the 23 * documentation and/or other materials provided with the distribution. 24 * 3. All advertising materials mentioning features or use of this software 25 * must display the following acknowledgement: 26 * This product includes software developed by the University of 27 * California, Berkeley and its contributors. 28 * 4. Neither the name of the University nor the names of its contributors 29 * may be used to endorse or promote products derived from this software 30 * without specific prior written permission. 31 * 32 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 33 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 34 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 35 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 36 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 37 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 38 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 39 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 40 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 41 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 42 * SUCH DAMAGE. 43 * 44 * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93 45 * $NetBSD: fpu_sqrt.c,v 1.2 1994/11/20 20:52:46 deraadt Exp $ 46 */ 47 48 #include <sys/cdefs.h> 49 #if 0 50 __FBSDID("$FreeBSD: src/lib/libc/sparc64/fpu/fpu_sqrt.c,v 1.3 2002/03/22 21:52:58 obrien Exp $"); 51 #endif 52 53 /* 54 * Perform an FPU square root (return sqrt(x)). 55 */ 56 57 #include <sys/types.h> 58 59 #include <machine/frame.h> 60 61 #include "fpu_arith.h" 62 #include "fpu_emu.h" 63 #include "fpu_extern.h" 64 65 /* 66 * Our task is to calculate the square root of a floating point number x0. 67 * This number x normally has the form: 68 * 69 * exp 70 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 71 * 72 * This can be left as it stands, or the mantissa can be doubled and the 73 * exponent decremented: 74 * 75 * exp-1 76 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 77 * 78 * If the exponent `exp' is even, the square root of the number is best 79 * handled using the first form, and is by definition equal to: 80 * 81 * exp/2 82 * sqrt(x) = sqrt(mant) * 2 83 * 84 * If exp is odd, on the other hand, it is convenient to use the second 85 * form, giving: 86 * 87 * (exp-1)/2 88 * sqrt(x) = sqrt(2 * mant) * 2 89 * 90 * In the first case, we have 91 * 92 * 1 <= mant < 2 93 * 94 * and therefore 95 * 96 * sqrt(1) <= sqrt(mant) < sqrt(2) 97 * 98 * while in the second case we have 99 * 100 * 2 <= 2*mant < 4 101 * 102 * and therefore 103 * 104 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 105 * 106 * so that in any case, we are sure that 107 * 108 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 109 * 110 * or 111 * 112 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 113 * 114 * This root is therefore a properly formed mantissa for a floating 115 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 116 * as above. This leaves us with the problem of finding the square root 117 * of a fixed-point number in the range [1..4). 118 * 119 * Though it may not be instantly obvious, the following square root 120 * algorithm works for any integer x of an even number of bits, provided 121 * that no overflows occur: 122 * 123 * let q = 0 124 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 125 * x *= 2 -- multiply by radix, for next digit 126 * if x >= 2q + 2^k then -- if adding 2^k does not 127 * x -= 2q + 2^k -- exceed the correct root, 128 * q += 2^k -- add 2^k and adjust x 129 * fi 130 * done 131 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 132 * 133 * If NBITS is odd (so that k is initially even), we can just add another 134 * zero bit at the top of x. Doing so means that q is not going to acquire 135 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 136 * final value in x is not needed, or can be off by a factor of 2, this is 137 * equivalant to moving the `x *= 2' step to the bottom of the loop: 138 * 139 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 140 * 141 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 142 * (Since the algorithm is destructive on x, we will call x's initial 143 * value, for which q is some power of two times its square root, x0.) 144 * 145 * If we insert a loop invariant y = 2q, we can then rewrite this using 146 * C notation as: 147 * 148 * q = y = 0; x = x0; 149 * for (k = NBITS; --k >= 0;) { 150 * #if (NBITS is even) 151 * x *= 2; 152 * #endif 153 * t = y + (1 << k); 154 * if (x >= t) { 155 * x -= t; 156 * q += 1 << k; 157 * y += 1 << (k + 1); 158 * } 159 * #if (NBITS is odd) 160 * x *= 2; 161 * #endif 162 * } 163 * 164 * If x0 is fixed point, rather than an integer, we can simply alter the 165 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 166 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 167 * 168 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 169 * integers, which adds some complication. But note that q is built one 170 * bit at a time, from the top down, and is not used itself in the loop 171 * (we use 2q as held in y instead). This means we can build our answer 172 * in an integer, one word at a time, which saves a bit of work. Also, 173 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 174 * `new' bits in y and we can set them with an `or' operation rather than 175 * a full-blown multiword add. 176 * 177 * We are almost done, except for one snag. We must prove that none of our 178 * intermediate calculations can overflow. We know that x0 is in [1..4) 179 * and therefore the square root in q will be in [1..2), but what about x, 180 * y, and t? 181 * 182 * We know that y = 2q at the beginning of each loop. (The relation only 183 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 184 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 185 * Furthermore, we can prove with a bit of work that x never exceeds y by 186 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 187 * an exercise to the reader, mostly because I have become tired of working 188 * on this comment.) 189 * 190 * If our floating point mantissas (which are of the form 1.frac) occupy 191 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 192 * In fact, we want even one more bit (for a carry, to avoid compares), or 193 * three extra. There is a comment in fpu_emu.h reminding maintainers of 194 * this, so we have some justification in assuming it. 195 */ 196 struct fpn * 197 __fpu_sqrt(fe) 198 struct fpemu *fe; 199 { 200 struct fpn *x = &fe->fe_f1; 201 u_int bit, q, tt; 202 u_int x0, x1, x2, x3; 203 u_int y0, y1, y2, y3; 204 u_int d0, d1, d2, d3; 205 int e; 206 207 /* 208 * Take care of special cases first. In order: 209 * 210 * sqrt(NaN) = NaN 211 * sqrt(+0) = +0 212 * sqrt(-0) = -0 213 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 214 * sqrt(+Inf) = +Inf 215 * 216 * Then all that remains are numbers with mantissas in [1..2). 217 */ 218 if (ISNAN(x) || ISZERO(x)) 219 return (x); 220 if (x->fp_sign) 221 return (__fpu_newnan(fe)); 222 if (ISINF(x)) 223 return (x); 224 225 /* 226 * Calculate result exponent. As noted above, this may involve 227 * doubling the mantissa. We will also need to double x each 228 * time around the loop, so we define a macro for this here, and 229 * we break out the multiword mantissa. 230 */ 231 #ifdef FPU_SHL1_BY_ADD 232 #define DOUBLE_X { \ 233 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 234 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 235 } 236 #else 237 #define DOUBLE_X { \ 238 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 239 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 240 } 241 #endif 242 #if (FP_NMANT & 1) != 0 243 # define ODD_DOUBLE DOUBLE_X 244 # define EVEN_DOUBLE /* nothing */ 245 #else 246 # define ODD_DOUBLE /* nothing */ 247 # define EVEN_DOUBLE DOUBLE_X 248 #endif 249 x0 = x->fp_mant[0]; 250 x1 = x->fp_mant[1]; 251 x2 = x->fp_mant[2]; 252 x3 = x->fp_mant[3]; 253 e = x->fp_exp; 254 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 255 DOUBLE_X; 256 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 257 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 258 259 /* 260 * Now calculate the mantissa root. Since x is now in [1..4), 261 * we know that the first trip around the loop will definitely 262 * set the top bit in q, so we can do that manually and start 263 * the loop at the next bit down instead. We must be sure to 264 * double x correctly while doing the `known q=1.0'. 265 * 266 * We do this one mantissa-word at a time, as noted above, to 267 * save work. To avoid `(1 << 31) << 1', we also do the top bit 268 * outside of each per-word loop. 269 * 270 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 271 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 272 * is always a `new' one, this means that three of the `t?'s are 273 * just the corresponding `y?'; we use `#define's here for this. 274 * The variable `tt' holds the actual `t?' variable. 275 */ 276 277 /* calculate q0 */ 278 #define t0 tt 279 bit = FP_1; 280 EVEN_DOUBLE; 281 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 282 q = bit; 283 x0 -= bit; 284 y0 = bit << 1; 285 /* } */ 286 ODD_DOUBLE; 287 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 288 EVEN_DOUBLE; 289 t0 = y0 | bit; /* t = y + bit */ 290 if (x0 >= t0) { /* if x >= t then */ 291 x0 -= t0; /* x -= t */ 292 q |= bit; /* q += bit */ 293 y0 |= bit << 1; /* y += bit << 1 */ 294 } 295 ODD_DOUBLE; 296 } 297 x->fp_mant[0] = q; 298 #undef t0 299 300 /* calculate q1. note (y0&1)==0. */ 301 #define t0 y0 302 #define t1 tt 303 q = 0; 304 y1 = 0; 305 bit = 1 << 31; 306 EVEN_DOUBLE; 307 t1 = bit; 308 FPU_SUBS(d1, x1, t1); 309 FPU_SUBC(d0, x0, t0); /* d = x - t */ 310 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 311 x0 = d0, x1 = d1; /* x -= t */ 312 q = bit; /* q += bit */ 313 y0 |= 1; /* y += bit << 1 */ 314 } 315 ODD_DOUBLE; 316 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 317 EVEN_DOUBLE; /* as before */ 318 t1 = y1 | bit; 319 FPU_SUBS(d1, x1, t1); 320 FPU_SUBC(d0, x0, t0); 321 if ((int)d0 >= 0) { 322 x0 = d0, x1 = d1; 323 q |= bit; 324 y1 |= bit << 1; 325 } 326 ODD_DOUBLE; 327 } 328 x->fp_mant[1] = q; 329 #undef t1 330 331 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 332 #define t1 y1 333 #define t2 tt 334 q = 0; 335 y2 = 0; 336 bit = 1 << 31; 337 EVEN_DOUBLE; 338 t2 = bit; 339 FPU_SUBS(d2, x2, t2); 340 FPU_SUBCS(d1, x1, t1); 341 FPU_SUBC(d0, x0, t0); 342 if ((int)d0 >= 0) { 343 x0 = d0, x1 = d1, x2 = d2; 344 q |= bit; 345 y1 |= 1; /* now t1, y1 are set in concrete */ 346 } 347 ODD_DOUBLE; 348 while ((bit >>= 1) != 0) { 349 EVEN_DOUBLE; 350 t2 = y2 | 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 y2 |= bit << 1; 358 } 359 ODD_DOUBLE; 360 } 361 x->fp_mant[2] = q; 362 #undef t2 363 364 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 365 #define t2 y2 366 #define t3 tt 367 q = 0; 368 y3 = 0; 369 bit = 1 << 31; 370 EVEN_DOUBLE; 371 t3 = bit; 372 FPU_SUBS(d3, x3, t3); 373 FPU_SUBCS(d2, x2, t2); 374 FPU_SUBCS(d1, x1, t1); 375 FPU_SUBC(d0, x0, t0); 376 ODD_DOUBLE; 377 if ((int)d0 >= 0) { 378 x0 = d0, x1 = d1, x2 = d2; 379 q |= bit; 380 y2 |= 1; 381 } 382 while ((bit >>= 1) != 0) { 383 EVEN_DOUBLE; 384 t3 = y3 | bit; 385 FPU_SUBS(d3, x3, t3); 386 FPU_SUBCS(d2, x2, t2); 387 FPU_SUBCS(d1, x1, t1); 388 FPU_SUBC(d0, x0, t0); 389 if ((int)d0 >= 0) { 390 x0 = d0, x1 = d1, x2 = d2; 391 q |= bit; 392 y3 |= bit << 1; 393 } 394 ODD_DOUBLE; 395 } 396 x->fp_mant[3] = q; 397 398 /* 399 * The result, which includes guard and round bits, is exact iff 400 * x is now zero; any nonzero bits in x represent sticky bits. 401 */ 402 x->fp_sticky = x0 | x1 | x2 | x3; 403 return (x); 404 } 405