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 57 #include <powerpc/fpu/fpu_arith.h> 58 #include <powerpc/fpu/fpu_emu.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 * equivalant 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(struct fpemu *fe) 193 { 194 struct fpn *x = &fe->fe_f1; 195 u_int bit, q, tt; 196 u_int x0, x1, x2, x3; 197 u_int y0, y1, y2, y3; 198 u_int d0, d1, d2, d3; 199 int e; 200 FPU_DECL_CARRY; 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 DPRINTF(FPE_REG, ("fpu_sqer:\n")); 214 DUMPFPN(FPE_REG, x); 215 DPRINTF(FPE_REG, ("=>\n")); 216 if (ISNAN(x)) { 217 fe->fe_cx |= FPSCR_VXSNAN; 218 DUMPFPN(FPE_REG, x); 219 return (x); 220 } 221 if (ISZERO(x)) { 222 fe->fe_cx |= FPSCR_ZX; 223 x->fp_class = FPC_INF; 224 DUMPFPN(FPE_REG, x); 225 return (x); 226 } 227 if (x->fp_sign) { 228 fe->fe_cx |= FPSCR_VXSQRT; 229 return (fpu_newnan(fe)); 230 } 231 if (ISINF(x)) { 232 DUMPFPN(FPE_REG, x); 233 return (x); 234 } 235 236 /* 237 * Calculate result exponent. As noted above, this may involve 238 * doubling the mantissa. We will also need to double x each 239 * time around the loop, so we define a macro for this here, and 240 * we break out the multiword mantissa. 241 */ 242 #ifdef FPU_SHL1_BY_ADD 243 #define DOUBLE_X { \ 244 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 245 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 246 } 247 #else 248 #define DOUBLE_X { \ 249 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 250 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 251 } 252 #endif 253 #if (FP_NMANT & 1) != 0 254 # define ODD_DOUBLE DOUBLE_X 255 # define EVEN_DOUBLE /* nothing */ 256 #else 257 # define ODD_DOUBLE /* nothing */ 258 # define EVEN_DOUBLE DOUBLE_X 259 #endif 260 x0 = x->fp_mant[0]; 261 x1 = x->fp_mant[1]; 262 x2 = x->fp_mant[2]; 263 x3 = x->fp_mant[3]; 264 e = x->fp_exp; 265 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 266 DOUBLE_X; 267 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 268 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 269 270 /* 271 * Now calculate the mantissa root. Since x is now in [1..4), 272 * we know that the first trip around the loop will definitely 273 * set the top bit in q, so we can do that manually and start 274 * the loop at the next bit down instead. We must be sure to 275 * double x correctly while doing the `known q=1.0'. 276 * 277 * We do this one mantissa-word at a time, as noted above, to 278 * save work. To avoid `(1U << 31) << 1', we also do the top bit 279 * outside of each per-word loop. 280 * 281 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 282 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 283 * is always a `new' one, this means that three of the `t?'s are 284 * just the corresponding `y?'; we use `#define's here for this. 285 * The variable `tt' holds the actual `t?' variable. 286 */ 287 288 /* calculate q0 */ 289 #define t0 tt 290 bit = FP_1; 291 EVEN_DOUBLE; 292 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 293 q = bit; 294 x0 -= bit; 295 y0 = bit << 1; 296 /* } */ 297 ODD_DOUBLE; 298 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 299 EVEN_DOUBLE; 300 t0 = y0 | bit; /* t = y + bit */ 301 if (x0 >= t0) { /* if x >= t then */ 302 x0 -= t0; /* x -= t */ 303 q |= bit; /* q += bit */ 304 y0 |= bit << 1; /* y += bit << 1 */ 305 } 306 ODD_DOUBLE; 307 } 308 x->fp_mant[0] = q; 309 #undef t0 310 311 /* calculate q1. note (y0&1)==0. */ 312 #define t0 y0 313 #define t1 tt 314 q = 0; 315 y1 = 0; 316 bit = 1 << 31; 317 EVEN_DOUBLE; 318 t1 = bit; 319 FPU_SUBS(d1, x1, t1); 320 FPU_SUBC(d0, x0, t0); /* d = x - t */ 321 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 322 x0 = d0, x1 = d1; /* x -= t */ 323 q = bit; /* q += bit */ 324 y0 |= 1; /* y += bit << 1 */ 325 } 326 ODD_DOUBLE; 327 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 328 EVEN_DOUBLE; /* as before */ 329 t1 = y1 | bit; 330 FPU_SUBS(d1, x1, t1); 331 FPU_SUBC(d0, x0, t0); 332 if ((int)d0 >= 0) { 333 x0 = d0, x1 = d1; 334 q |= bit; 335 y1 |= bit << 1; 336 } 337 ODD_DOUBLE; 338 } 339 x->fp_mant[1] = q; 340 #undef t1 341 342 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 343 #define t1 y1 344 #define t2 tt 345 q = 0; 346 y2 = 0; 347 bit = 1 << 31; 348 EVEN_DOUBLE; 349 t2 = bit; 350 FPU_SUBS(d2, x2, t2); 351 FPU_SUBCS(d1, x1, t1); 352 FPU_SUBC(d0, x0, t0); 353 if ((int)d0 >= 0) { 354 x0 = d0, x1 = d1, x2 = d2; 355 q = bit; 356 y1 |= 1; /* now t1, y1 are set in concrete */ 357 } 358 ODD_DOUBLE; 359 while ((bit >>= 1) != 0) { 360 EVEN_DOUBLE; 361 t2 = y2 | bit; 362 FPU_SUBS(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; 367 q |= bit; 368 y2 |= bit << 1; 369 } 370 ODD_DOUBLE; 371 } 372 x->fp_mant[2] = q; 373 #undef t2 374 375 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 376 #define t2 y2 377 #define t3 tt 378 q = 0; 379 y3 = 0; 380 bit = 1 << 31; 381 EVEN_DOUBLE; 382 t3 = bit; 383 FPU_SUBS(d3, x3, t3); 384 FPU_SUBCS(d2, x2, t2); 385 FPU_SUBCS(d1, x1, t1); 386 FPU_SUBC(d0, x0, t0); 387 if ((int)d0 >= 0) { 388 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 389 q = bit; 390 y2 |= 1; 391 } 392 ODD_DOUBLE; 393 while ((bit >>= 1) != 0) { 394 EVEN_DOUBLE; 395 t3 = y3 | bit; 396 FPU_SUBS(d3, x3, t3); 397 FPU_SUBCS(d2, x2, t2); 398 FPU_SUBCS(d1, x1, t1); 399 FPU_SUBC(d0, x0, t0); 400 if ((int)d0 >= 0) { 401 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 402 q |= bit; 403 y3 |= bit << 1; 404 } 405 ODD_DOUBLE; 406 } 407 x->fp_mant[3] = q; 408 409 /* 410 * The result, which includes guard and round bits, is exact iff 411 * x is now zero; any nonzero bits in x represent sticky bits. 412 */ 413 x->fp_sticky = x0 | x1 | x2 | x3; 414 DUMPFPN(FPE_REG, x); 415 return (x); 416 } 417