1 /* $OpenBSD: fpu_sqrt.c,v 1.6 2022/12/27 17:10:06 jmc 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 /* 49 * Perform an FPU square root (return sqrt(x)). 50 */ 51 52 #include <sys/types.h> 53 54 #include <machine/frame.h> 55 56 #include "fpu_arith.h" 57 #include "fpu_emu.h" 58 #include "fpu_extern.h" 59 60 /* 61 * Our task is to calculate the square root of a floating point number x0. 62 * This number x normally has the form: 63 * 64 * exp 65 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 66 * 67 * This can be left as it stands, or the mantissa can be doubled and the 68 * exponent decremented: 69 * 70 * exp-1 71 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 72 * 73 * If the exponent `exp' is even, the square root of the number is best 74 * handled using the first form, and is by definition equal to: 75 * 76 * exp/2 77 * sqrt(x) = sqrt(mant) * 2 78 * 79 * If exp is odd, on the other hand, it is convenient to use the second 80 * form, giving: 81 * 82 * (exp-1)/2 83 * sqrt(x) = sqrt(2 * mant) * 2 84 * 85 * In the first case, we have 86 * 87 * 1 <= mant < 2 88 * 89 * and therefore 90 * 91 * sqrt(1) <= sqrt(mant) < sqrt(2) 92 * 93 * while in the second case we have 94 * 95 * 2 <= 2*mant < 4 96 * 97 * and therefore 98 * 99 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 100 * 101 * so that in any case, we are sure that 102 * 103 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 104 * 105 * or 106 * 107 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 108 * 109 * This root is therefore a properly formed mantissa for a floating 110 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 111 * as above. This leaves us with the problem of finding the square root 112 * of a fixed-point number in the range [1..4). 113 * 114 * Though it may not be instantly obvious, the following square root 115 * algorithm works for any integer x of an even number of bits, provided 116 * that no overflows occur: 117 * 118 * let q = 0 119 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 120 * x *= 2 -- multiply by radix, for next digit 121 * if x >= 2q + 2^k then -- if adding 2^k does not 122 * x -= 2q + 2^k -- exceed the correct root, 123 * q += 2^k -- add 2^k and adjust x 124 * fi 125 * done 126 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 127 * 128 * If NBITS is odd (so that k is initially even), we can just add another 129 * zero bit at the top of x. Doing so means that q is not going to acquire 130 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 131 * final value in x is not needed, or can be off by a factor of 2, this is 132 * equivalent to moving the `x *= 2' step to the bottom of the loop: 133 * 134 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 135 * 136 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 137 * (Since the algorithm is destructive on x, we will call x's initial 138 * value, for which q is some power of two times its square root, x0.) 139 * 140 * If we insert a loop invariant y = 2q, we can then rewrite this using 141 * C notation as: 142 * 143 * q = y = 0; x = x0; 144 * for (k = NBITS; --k >= 0;) { 145 * #if (NBITS is even) 146 * x *= 2; 147 * #endif 148 * t = y + (1 << k); 149 * if (x >= t) { 150 * x -= t; 151 * q += 1 << k; 152 * y += 1 << (k + 1); 153 * } 154 * #if (NBITS is odd) 155 * x *= 2; 156 * #endif 157 * } 158 * 159 * If x0 is fixed point, rather than an integer, we can simply alter the 160 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 161 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 162 * 163 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 164 * integers, which adds some complication. But note that q is built one 165 * bit at a time, from the top down, and is not used itself in the loop 166 * (we use 2q as held in y instead). This means we can build our answer 167 * in an integer, one word at a time, which saves a bit of work. Also, 168 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 169 * `new' bits in y and we can set them with an `or' operation rather than 170 * a full-blown multiword add. 171 * 172 * We are almost done, except for one snag. We must prove that none of our 173 * intermediate calculations can overflow. We know that x0 is in [1..4) 174 * and therefore the square root in q will be in [1..2), but what about x, 175 * y, and t? 176 * 177 * We know that y = 2q at the beginning of each loop. (The relation only 178 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 179 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 180 * Furthermore, we can prove with a bit of work that x never exceeds y by 181 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 182 * an exercise to the reader, mostly because I have become tired of working 183 * on this comment.) 184 * 185 * If our floating point mantissas (which are of the form 1.frac) occupy 186 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 187 * In fact, we want even one more bit (for a carry, to avoid compares), or 188 * three extra. There is a comment in fpu_emu.h reminding maintainers of 189 * this, so we have some justification in assuming it. 190 */ 191 struct fpn * 192 __fpu_sqrt(fe) 193 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; 201 202 /* 203 * Take care of special cases first. In order: 204 * 205 * sqrt(NaN) = NaN 206 * sqrt(+0) = +0 207 * sqrt(-0) = -0 208 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 209 * sqrt(+Inf) = +Inf 210 * 211 * Then all that remains are numbers with mantissas in [1..2). 212 */ 213 if (ISNAN(x) || ISZERO(x)) 214 return (x); 215 if (x->fp_sign) 216 return (__fpu_newnan(fe)); 217 if (ISINF(x)) 218 return (x); 219 220 /* 221 * Calculate result exponent. As noted above, this may involve 222 * doubling the mantissa. We will also need to double x each 223 * time around the loop, so we define a macro for this here, and 224 * we break out the multiword mantissa. 225 */ 226 #ifdef FPU_SHL1_BY_ADD 227 #define DOUBLE_X { \ 228 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 229 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 230 } 231 #else 232 #define DOUBLE_X { \ 233 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 234 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 235 } 236 #endif 237 #if (FP_NMANT & 1) != 0 238 # define ODD_DOUBLE DOUBLE_X 239 # define EVEN_DOUBLE /* nothing */ 240 #else 241 # define ODD_DOUBLE /* nothing */ 242 # define EVEN_DOUBLE DOUBLE_X 243 #endif 244 x0 = x->fp_mant[0]; 245 x1 = x->fp_mant[1]; 246 x2 = x->fp_mant[2]; 247 x3 = x->fp_mant[3]; 248 e = x->fp_exp; 249 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 250 DOUBLE_X; 251 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 252 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 253 254 /* 255 * Now calculate the mantissa root. Since x is now in [1..4), 256 * we know that the first trip around the loop will definitely 257 * set the top bit in q, so we can do that manually and start 258 * the loop at the next bit down instead. We must be sure to 259 * double x correctly while doing the `known q=1.0'. 260 * 261 * We do this one mantissa-word at a time, as noted above, to 262 * save work. To avoid `(1U << 31) << 1', we also do the top bit 263 * outside of each per-word loop. 264 * 265 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 266 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 267 * is always a `new' one, this means that three of the `t?'s are 268 * just the corresponding `y?'; we use `#define's here for this. 269 * The variable `tt' holds the actual `t?' variable. 270 */ 271 272 /* calculate q0 */ 273 #define t0 tt 274 bit = FP_1; 275 EVEN_DOUBLE; 276 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 277 q = bit; 278 x0 -= bit; 279 y0 = bit << 1; 280 /* } */ 281 ODD_DOUBLE; 282 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 283 EVEN_DOUBLE; 284 t0 = y0 | bit; /* t = y + bit */ 285 if (x0 >= t0) { /* if x >= t then */ 286 x0 -= t0; /* x -= t */ 287 q |= bit; /* q += bit */ 288 y0 |= bit << 1; /* y += bit << 1 */ 289 } 290 ODD_DOUBLE; 291 } 292 x->fp_mant[0] = q; 293 #undef t0 294 295 /* calculate q1. note (y0&1)==0. */ 296 #define t0 y0 297 #define t1 tt 298 q = 0; 299 y1 = 0; 300 bit = 1U << 31; 301 EVEN_DOUBLE; 302 t1 = bit; 303 FPU_SUBS(d1, x1, t1); 304 FPU_SUBC(d0, x0, t0); /* d = x - t */ 305 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 306 x0 = d0, x1 = d1; /* x -= t */ 307 q = bit; /* q += bit */ 308 y0 |= 1; /* y += bit << 1 */ 309 } 310 ODD_DOUBLE; 311 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 312 EVEN_DOUBLE; /* as before */ 313 t1 = y1 | bit; 314 FPU_SUBS(d1, x1, t1); 315 FPU_SUBC(d0, x0, t0); 316 if ((int)d0 >= 0) { 317 x0 = d0, x1 = d1; 318 q |= bit; 319 y1 |= bit << 1; 320 } 321 ODD_DOUBLE; 322 } 323 x->fp_mant[1] = q; 324 #undef t1 325 326 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 327 #define t1 y1 328 #define t2 tt 329 q = 0; 330 y2 = 0; 331 bit = 1U << 31; 332 EVEN_DOUBLE; 333 t2 = bit; 334 FPU_SUBS(d2, x2, t2); 335 FPU_SUBCS(d1, x1, t1); 336 FPU_SUBC(d0, x0, t0); 337 if ((int)d0 >= 0) { 338 x0 = d0, x1 = d1, x2 = d2; 339 q |= bit; 340 y1 |= 1; /* now t1, y1 are set in concrete */ 341 } 342 ODD_DOUBLE; 343 while ((bit >>= 1) != 0) { 344 EVEN_DOUBLE; 345 t2 = y2 | bit; 346 FPU_SUBS(d2, x2, t2); 347 FPU_SUBCS(d1, x1, t1); 348 FPU_SUBC(d0, x0, t0); 349 if ((int)d0 >= 0) { 350 x0 = d0, x1 = d1, x2 = d2; 351 q |= bit; 352 y2 |= bit << 1; 353 } 354 ODD_DOUBLE; 355 } 356 x->fp_mant[2] = q; 357 #undef t2 358 359 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 360 #define t2 y2 361 #define t3 tt 362 q = 0; 363 y3 = 0; 364 bit = 1U << 31; 365 EVEN_DOUBLE; 366 t3 = bit; 367 FPU_SUBS(d3, x3, t3); 368 FPU_SUBCS(d2, x2, t2); 369 FPU_SUBCS(d1, x1, t1); 370 FPU_SUBC(d0, x0, t0); 371 if ((int)d0 >= 0) { 372 x0 = d0, x1 = d1, x2 = d2, x3 = d3; 373 q |= bit; 374 y2 |= 1; 375 } 376 ODD_DOUBLE; 377 while ((bit >>= 1) != 0) { 378 EVEN_DOUBLE; 379 t3 = y3 | bit; 380 FPU_SUBS(d3, x3, t3); 381 FPU_SUBCS(d2, x2, t2); 382 FPU_SUBCS(d1, x1, t1); 383 FPU_SUBC(d0, x0, t0); 384 if ((int)d0 >= 0) { 385 x0 = d0, x1 = d1, x2 = d2, x3 = d3; 386 q |= bit; 387 y3 |= bit << 1; 388 } 389 ODD_DOUBLE; 390 } 391 x->fp_mant[3] = q; 392 393 /* 394 * The result, which includes guard and round bits, is exact iff 395 * x is now zero; any nonzero bits in x represent sticky bits. 396 */ 397 x->fp_sticky = x0 | x1 | x2 | x3; 398 return (x); 399 } 400