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 #include <sys/types.h> 51 #include <sys/systm.h> 52 53 #include <machine/fpu.h> 54 55 #include <powerpc/fpu/fpu_arith.h> 56 #include <powerpc/fpu/fpu_emu.h> 57 58 /* 59 * Our task is to calculate the square root of a floating point number x0. 60 * This number x normally has the form: 61 * 62 * exp 63 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 64 * 65 * This can be left as it stands, or the mantissa can be doubled and the 66 * exponent decremented: 67 * 68 * exp-1 69 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 70 * 71 * If the exponent `exp' is even, the square root of the number is best 72 * handled using the first form, and is by definition equal to: 73 * 74 * exp/2 75 * sqrt(x) = sqrt(mant) * 2 76 * 77 * If exp is odd, on the other hand, it is convenient to use the second 78 * form, giving: 79 * 80 * (exp-1)/2 81 * sqrt(x) = sqrt(2 * mant) * 2 82 * 83 * In the first case, we have 84 * 85 * 1 <= mant < 2 86 * 87 * and therefore 88 * 89 * sqrt(1) <= sqrt(mant) < sqrt(2) 90 * 91 * while in the second case we have 92 * 93 * 2 <= 2*mant < 4 94 * 95 * and therefore 96 * 97 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 98 * 99 * so that in any case, we are sure that 100 * 101 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 102 * 103 * or 104 * 105 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 106 * 107 * This root is therefore a properly formed mantissa for a floating 108 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 109 * as above. This leaves us with the problem of finding the square root 110 * of a fixed-point number in the range [1..4). 111 * 112 * Though it may not be instantly obvious, the following square root 113 * algorithm works for any integer x of an even number of bits, provided 114 * that no overflows occur: 115 * 116 * let q = 0 117 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 118 * x *= 2 -- multiply by radix, for next digit 119 * if x >= 2q + 2^k then -- if adding 2^k does not 120 * x -= 2q + 2^k -- exceed the correct root, 121 * q += 2^k -- add 2^k and adjust x 122 * fi 123 * done 124 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 125 * 126 * If NBITS is odd (so that k is initially even), we can just add another 127 * zero bit at the top of x. Doing so means that q is not going to acquire 128 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 129 * final value in x is not needed, or can be off by a factor of 2, this is 130 * equivalant to moving the `x *= 2' step to the bottom of the loop: 131 * 132 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 133 * 134 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 135 * (Since the algorithm is destructive on x, we will call x's initial 136 * value, for which q is some power of two times its square root, x0.) 137 * 138 * If we insert a loop invariant y = 2q, we can then rewrite this using 139 * C notation as: 140 * 141 * q = y = 0; x = x0; 142 * for (k = NBITS; --k >= 0;) { 143 * #if (NBITS is even) 144 * x *= 2; 145 * #endif 146 * t = y + (1 << k); 147 * if (x >= t) { 148 * x -= t; 149 * q += 1 << k; 150 * y += 1 << (k + 1); 151 * } 152 * #if (NBITS is odd) 153 * x *= 2; 154 * #endif 155 * } 156 * 157 * If x0 is fixed point, rather than an integer, we can simply alter the 158 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 159 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 160 * 161 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 162 * integers, which adds some complication. But note that q is built one 163 * bit at a time, from the top down, and is not used itself in the loop 164 * (we use 2q as held in y instead). This means we can build our answer 165 * in an integer, one word at a time, which saves a bit of work. Also, 166 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 167 * `new' bits in y and we can set them with an `or' operation rather than 168 * a full-blown multiword add. 169 * 170 * We are almost done, except for one snag. We must prove that none of our 171 * intermediate calculations can overflow. We know that x0 is in [1..4) 172 * and therefore the square root in q will be in [1..2), but what about x, 173 * y, and t? 174 * 175 * We know that y = 2q at the beginning of each loop. (The relation only 176 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 177 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 178 * Furthermore, we can prove with a bit of work that x never exceeds y by 179 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 180 * an exercise to the reader, mostly because I have become tired of working 181 * on this comment.) 182 * 183 * If our floating point mantissas (which are of the form 1.frac) occupy 184 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 185 * In fact, we want even one more bit (for a carry, to avoid compares), or 186 * three extra. There is a comment in fpu_emu.h reminding maintainers of 187 * this, so we have some justification in assuming it. 188 */ 189 struct fpn * 190 fpu_sqrt(struct fpemu *fe) 191 { 192 struct fpn *x = &fe->fe_f1; 193 u_int bit, q, tt; 194 u_int x0, x1, x2, x3; 195 u_int y0, y1, y2, y3; 196 u_int d0, d1, d2, d3; 197 int e; 198 FPU_DECL_CARRY; 199 200 /* 201 * Take care of special cases first. In order: 202 * 203 * sqrt(NaN) = NaN 204 * sqrt(+0) = +0 205 * sqrt(-0) = -0 206 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 207 * sqrt(+Inf) = +Inf 208 * 209 * Then all that remains are numbers with mantissas in [1..2). 210 */ 211 DPRINTF(FPE_REG, ("fpu_sqer:\n")); 212 DUMPFPN(FPE_REG, x); 213 DPRINTF(FPE_REG, ("=>\n")); 214 if (ISNAN(x)) { 215 fe->fe_cx |= FPSCR_VXSNAN; 216 DUMPFPN(FPE_REG, x); 217 return (x); 218 } 219 if (ISZERO(x)) { 220 fe->fe_cx |= FPSCR_ZX; 221 x->fp_class = FPC_INF; 222 DUMPFPN(FPE_REG, x); 223 return (x); 224 } 225 if (x->fp_sign) { 226 fe->fe_cx |= FPSCR_VXSQRT; 227 return (fpu_newnan(fe)); 228 } 229 if (ISINF(x)) { 230 DUMPFPN(FPE_REG, x); 231 return (x); 232 } 233 234 /* 235 * Calculate result exponent. As noted above, this may involve 236 * doubling the mantissa. We will also need to double x each 237 * time around the loop, so we define a macro for this here, and 238 * we break out the multiword mantissa. 239 */ 240 #ifdef FPU_SHL1_BY_ADD 241 #define DOUBLE_X { \ 242 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 243 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 244 } 245 #else 246 #define DOUBLE_X { \ 247 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 248 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 249 } 250 #endif 251 #if (FP_NMANT & 1) != 0 252 # define ODD_DOUBLE DOUBLE_X 253 # define EVEN_DOUBLE /* nothing */ 254 #else 255 # define ODD_DOUBLE /* nothing */ 256 # define EVEN_DOUBLE DOUBLE_X 257 #endif 258 x0 = x->fp_mant[0]; 259 x1 = x->fp_mant[1]; 260 x2 = x->fp_mant[2]; 261 x3 = x->fp_mant[3]; 262 e = x->fp_exp; 263 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 264 DOUBLE_X; 265 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 266 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 267 268 /* 269 * Now calculate the mantissa root. Since x is now in [1..4), 270 * we know that the first trip around the loop will definitely 271 * set the top bit in q, so we can do that manually and start 272 * the loop at the next bit down instead. We must be sure to 273 * double x correctly while doing the `known q=1.0'. 274 * 275 * We do this one mantissa-word at a time, as noted above, to 276 * save work. To avoid `(1U << 31) << 1', we also do the top bit 277 * outside of each per-word loop. 278 * 279 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 280 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 281 * is always a `new' one, this means that three of the `t?'s are 282 * just the corresponding `y?'; we use `#define's here for this. 283 * The variable `tt' holds the actual `t?' variable. 284 */ 285 286 /* calculate q0 */ 287 #define t0 tt 288 bit = FP_1; 289 EVEN_DOUBLE; 290 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 291 q = bit; 292 x0 -= bit; 293 y0 = bit << 1; 294 /* } */ 295 ODD_DOUBLE; 296 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 297 EVEN_DOUBLE; 298 t0 = y0 | bit; /* t = y + bit */ 299 if (x0 >= t0) { /* if x >= t then */ 300 x0 -= t0; /* x -= t */ 301 q |= bit; /* q += bit */ 302 y0 |= bit << 1; /* y += bit << 1 */ 303 } 304 ODD_DOUBLE; 305 } 306 x->fp_mant[0] = q; 307 #undef t0 308 309 /* calculate q1. note (y0&1)==0. */ 310 #define t0 y0 311 #define t1 tt 312 q = 0; 313 y1 = 0; 314 bit = 1 << 31; 315 EVEN_DOUBLE; 316 t1 = bit; 317 FPU_SUBS(d1, x1, t1); 318 FPU_SUBC(d0, x0, t0); /* d = x - t */ 319 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 320 x0 = d0, x1 = d1; /* x -= t */ 321 q = bit; /* q += bit */ 322 y0 |= 1; /* y += bit << 1 */ 323 } 324 ODD_DOUBLE; 325 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 326 EVEN_DOUBLE; /* as before */ 327 t1 = y1 | bit; 328 FPU_SUBS(d1, x1, t1); 329 FPU_SUBC(d0, x0, t0); 330 if ((int)d0 >= 0) { 331 x0 = d0, x1 = d1; 332 q |= bit; 333 y1 |= bit << 1; 334 } 335 ODD_DOUBLE; 336 } 337 x->fp_mant[1] = q; 338 #undef t1 339 340 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 341 #define t1 y1 342 #define t2 tt 343 q = 0; 344 y2 = 0; 345 bit = 1 << 31; 346 EVEN_DOUBLE; 347 t2 = 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 y1 |= 1; /* now t1, y1 are set in concrete */ 355 } 356 ODD_DOUBLE; 357 while ((bit >>= 1) != 0) { 358 EVEN_DOUBLE; 359 t2 = y2 | bit; 360 FPU_SUBS(d2, x2, t2); 361 FPU_SUBCS(d1, x1, t1); 362 FPU_SUBC(d0, x0, t0); 363 if ((int)d0 >= 0) { 364 x0 = d0, x1 = d1, x2 = d2; 365 q |= bit; 366 y2 |= bit << 1; 367 } 368 ODD_DOUBLE; 369 } 370 x->fp_mant[2] = q; 371 #undef t2 372 373 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 374 #define t2 y2 375 #define t3 tt 376 q = 0; 377 y3 = 0; 378 bit = 1 << 31; 379 EVEN_DOUBLE; 380 t3 = bit; 381 FPU_SUBS(d3, x3, t3); 382 FPU_SUBCS(d2, x2, t2); 383 FPU_SUBCS(d1, x1, t1); 384 FPU_SUBC(d0, x0, t0); 385 if ((int)d0 >= 0) { 386 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 387 q = bit; 388 y2 |= 1; 389 } 390 ODD_DOUBLE; 391 while ((bit >>= 1) != 0) { 392 EVEN_DOUBLE; 393 t3 = y3 | bit; 394 FPU_SUBS(d3, x3, t3); 395 FPU_SUBCS(d2, x2, t2); 396 FPU_SUBCS(d1, x1, t1); 397 FPU_SUBC(d0, x0, t0); 398 if ((int)d0 >= 0) { 399 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 400 q |= bit; 401 y3 |= bit << 1; 402 } 403 ODD_DOUBLE; 404 } 405 x->fp_mant[3] = q; 406 407 /* 408 * The result, which includes guard and round bits, is exact iff 409 * x is now zero; any nonzero bits in x represent sticky bits. 410 */ 411 x->fp_sticky = x0 | x1 | x2 | x3; 412 DUMPFPN(FPE_REG, x); 413 return (x); 414 } 415