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 43 /* 44 * Perform an FPU square root (return sqrt(x)). 45 */ 46 47 #include <sys/cdefs.h> 48 #include <sys/types.h> 49 #include <sys/systm.h> 50 51 #include <machine/fpu.h> 52 53 #include <powerpc/fpu/fpu_arith.h> 54 #include <powerpc/fpu/fpu_emu.h> 55 56 /* 57 * Our task is to calculate the square root of a floating point number x0. 58 * This number x normally has the form: 59 * 60 * exp 61 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 62 * 63 * This can be left as it stands, or the mantissa can be doubled and the 64 * exponent decremented: 65 * 66 * exp-1 67 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 68 * 69 * If the exponent `exp' is even, the square root of the number is best 70 * handled using the first form, and is by definition equal to: 71 * 72 * exp/2 73 * sqrt(x) = sqrt(mant) * 2 74 * 75 * If exp is odd, on the other hand, it is convenient to use the second 76 * form, giving: 77 * 78 * (exp-1)/2 79 * sqrt(x) = sqrt(2 * mant) * 2 80 * 81 * In the first case, we have 82 * 83 * 1 <= mant < 2 84 * 85 * and therefore 86 * 87 * sqrt(1) <= sqrt(mant) < sqrt(2) 88 * 89 * while in the second case we have 90 * 91 * 2 <= 2*mant < 4 92 * 93 * and therefore 94 * 95 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 96 * 97 * so that in any case, we are sure that 98 * 99 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 100 * 101 * or 102 * 103 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 104 * 105 * This root is therefore a properly formed mantissa for a floating 106 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 107 * as above. This leaves us with the problem of finding the square root 108 * of a fixed-point number in the range [1..4). 109 * 110 * Though it may not be instantly obvious, the following square root 111 * algorithm works for any integer x of an even number of bits, provided 112 * that no overflows occur: 113 * 114 * let q = 0 115 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 116 * x *= 2 -- multiply by radix, for next digit 117 * if x >= 2q + 2^k then -- if adding 2^k does not 118 * x -= 2q + 2^k -- exceed the correct root, 119 * q += 2^k -- add 2^k and adjust x 120 * fi 121 * done 122 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 123 * 124 * If NBITS is odd (so that k is initially even), we can just add another 125 * zero bit at the top of x. Doing so means that q is not going to acquire 126 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 127 * final value in x is not needed, or can be off by a factor of 2, this is 128 * equivalant to moving the `x *= 2' step to the bottom of the loop: 129 * 130 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 131 * 132 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 133 * (Since the algorithm is destructive on x, we will call x's initial 134 * value, for which q is some power of two times its square root, x0.) 135 * 136 * If we insert a loop invariant y = 2q, we can then rewrite this using 137 * C notation as: 138 * 139 * q = y = 0; x = x0; 140 * for (k = NBITS; --k >= 0;) { 141 * #if (NBITS is even) 142 * x *= 2; 143 * #endif 144 * t = y + (1 << k); 145 * if (x >= t) { 146 * x -= t; 147 * q += 1 << k; 148 * y += 1 << (k + 1); 149 * } 150 * #if (NBITS is odd) 151 * x *= 2; 152 * #endif 153 * } 154 * 155 * If x0 is fixed point, rather than an integer, we can simply alter the 156 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 157 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 158 * 159 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 160 * integers, which adds some complication. But note that q is built one 161 * bit at a time, from the top down, and is not used itself in the loop 162 * (we use 2q as held in y instead). This means we can build our answer 163 * in an integer, one word at a time, which saves a bit of work. Also, 164 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 165 * `new' bits in y and we can set them with an `or' operation rather than 166 * a full-blown multiword add. 167 * 168 * We are almost done, except for one snag. We must prove that none of our 169 * intermediate calculations can overflow. We know that x0 is in [1..4) 170 * and therefore the square root in q will be in [1..2), but what about x, 171 * y, and t? 172 * 173 * We know that y = 2q at the beginning of each loop. (The relation only 174 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 175 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 176 * Furthermore, we can prove with a bit of work that x never exceeds y by 177 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 178 * an exercise to the reader, mostly because I have become tired of working 179 * on this comment.) 180 * 181 * If our floating point mantissas (which are of the form 1.frac) occupy 182 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 183 * In fact, we want even one more bit (for a carry, to avoid compares), or 184 * three extra. There is a comment in fpu_emu.h reminding maintainers of 185 * this, so we have some justification in assuming it. 186 */ 187 struct fpn * 188 fpu_sqrt(struct fpemu *fe) 189 { 190 struct fpn *x = &fe->fe_f1; 191 u_int bit, q, tt; 192 u_int x0, x1, x2, x3; 193 u_int y0, y1, y2, y3; 194 u_int d0, d1, d2, d3; 195 int e; 196 FPU_DECL_CARRY; 197 198 /* 199 * Take care of special cases first. In order: 200 * 201 * sqrt(NaN) = NaN 202 * sqrt(+0) = +0 203 * sqrt(-0) = -0 204 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 205 * sqrt(+Inf) = +Inf 206 * 207 * Then all that remains are numbers with mantissas in [1..2). 208 */ 209 DPRINTF(FPE_REG, ("fpu_sqer:\n")); 210 DUMPFPN(FPE_REG, x); 211 DPRINTF(FPE_REG, ("=>\n")); 212 if (ISNAN(x)) { 213 fe->fe_cx |= FPSCR_VXSNAN; 214 DUMPFPN(FPE_REG, x); 215 return (x); 216 } 217 if (ISZERO(x)) { 218 fe->fe_cx |= FPSCR_ZX; 219 x->fp_class = FPC_INF; 220 DUMPFPN(FPE_REG, x); 221 return (x); 222 } 223 if (x->fp_sign) { 224 fe->fe_cx |= FPSCR_VXSQRT; 225 return (fpu_newnan(fe)); 226 } 227 if (ISINF(x)) { 228 DUMPFPN(FPE_REG, x); 229 return (x); 230 } 231 232 /* 233 * Calculate result exponent. As noted above, this may involve 234 * doubling the mantissa. We will also need to double x each 235 * time around the loop, so we define a macro for this here, and 236 * we break out the multiword mantissa. 237 */ 238 #ifdef FPU_SHL1_BY_ADD 239 #define DOUBLE_X { \ 240 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 241 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 242 } 243 #else 244 #define DOUBLE_X { \ 245 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 246 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 247 } 248 #endif 249 #if (FP_NMANT & 1) != 0 250 # define ODD_DOUBLE DOUBLE_X 251 # define EVEN_DOUBLE /* nothing */ 252 #else 253 # define ODD_DOUBLE /* nothing */ 254 # define EVEN_DOUBLE DOUBLE_X 255 #endif 256 x0 = x->fp_mant[0]; 257 x1 = x->fp_mant[1]; 258 x2 = x->fp_mant[2]; 259 x3 = x->fp_mant[3]; 260 e = x->fp_exp; 261 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 262 DOUBLE_X; 263 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 264 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 265 266 /* 267 * Now calculate the mantissa root. Since x is now in [1..4), 268 * we know that the first trip around the loop will definitely 269 * set the top bit in q, so we can do that manually and start 270 * the loop at the next bit down instead. We must be sure to 271 * double x correctly while doing the `known q=1.0'. 272 * 273 * We do this one mantissa-word at a time, as noted above, to 274 * save work. To avoid `(1U << 31) << 1', we also do the top bit 275 * outside of each per-word loop. 276 * 277 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 278 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 279 * is always a `new' one, this means that three of the `t?'s are 280 * just the corresponding `y?'; we use `#define's here for this. 281 * The variable `tt' holds the actual `t?' variable. 282 */ 283 284 /* calculate q0 */ 285 #define t0 tt 286 bit = FP_1; 287 EVEN_DOUBLE; 288 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 289 q = bit; 290 x0 -= bit; 291 y0 = bit << 1; 292 /* } */ 293 ODD_DOUBLE; 294 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 295 EVEN_DOUBLE; 296 t0 = y0 | bit; /* t = y + bit */ 297 if (x0 >= t0) { /* if x >= t then */ 298 x0 -= t0; /* x -= t */ 299 q |= bit; /* q += bit */ 300 y0 |= bit << 1; /* y += bit << 1 */ 301 } 302 ODD_DOUBLE; 303 } 304 x->fp_mant[0] = q; 305 #undef t0 306 307 /* calculate q1. note (y0&1)==0. */ 308 #define t0 y0 309 #define t1 tt 310 q = 0; 311 y1 = 0; 312 bit = 1 << 31; 313 EVEN_DOUBLE; 314 t1 = bit; 315 FPU_SUBS(d1, x1, t1); 316 FPU_SUBC(d0, x0, t0); /* d = x - t */ 317 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 318 x0 = d0, x1 = d1; /* x -= t */ 319 q = bit; /* q += bit */ 320 y0 |= 1; /* y += bit << 1 */ 321 } 322 ODD_DOUBLE; 323 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 324 EVEN_DOUBLE; /* as before */ 325 t1 = y1 | bit; 326 FPU_SUBS(d1, x1, t1); 327 FPU_SUBC(d0, x0, t0); 328 if ((int)d0 >= 0) { 329 x0 = d0, x1 = d1; 330 q |= bit; 331 y1 |= bit << 1; 332 } 333 ODD_DOUBLE; 334 } 335 x->fp_mant[1] = q; 336 #undef t1 337 338 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 339 #define t1 y1 340 #define t2 tt 341 q = 0; 342 y2 = 0; 343 bit = 1 << 31; 344 EVEN_DOUBLE; 345 t2 = 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 y1 |= 1; /* now t1, y1 are set in concrete */ 353 } 354 ODD_DOUBLE; 355 while ((bit >>= 1) != 0) { 356 EVEN_DOUBLE; 357 t2 = y2 | bit; 358 FPU_SUBS(d2, x2, t2); 359 FPU_SUBCS(d1, x1, t1); 360 FPU_SUBC(d0, x0, t0); 361 if ((int)d0 >= 0) { 362 x0 = d0, x1 = d1, x2 = d2; 363 q |= bit; 364 y2 |= bit << 1; 365 } 366 ODD_DOUBLE; 367 } 368 x->fp_mant[2] = q; 369 #undef t2 370 371 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 372 #define t2 y2 373 #define t3 tt 374 q = 0; 375 y3 = 0; 376 bit = 1 << 31; 377 EVEN_DOUBLE; 378 t3 = bit; 379 FPU_SUBS(d3, x3, t3); 380 FPU_SUBCS(d2, x2, t2); 381 FPU_SUBCS(d1, x1, t1); 382 FPU_SUBC(d0, x0, t0); 383 if ((int)d0 >= 0) { 384 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 385 q = bit; 386 y2 |= 1; 387 } 388 ODD_DOUBLE; 389 while ((bit >>= 1) != 0) { 390 EVEN_DOUBLE; 391 t3 = y3 | bit; 392 FPU_SUBS(d3, x3, t3); 393 FPU_SUBCS(d2, x2, t2); 394 FPU_SUBCS(d1, x1, t1); 395 FPU_SUBC(d0, x0, t0); 396 if ((int)d0 >= 0) { 397 x0 = d0, x1 = d1, x2 = d2; x3 = d3; 398 q |= bit; 399 y3 |= bit << 1; 400 } 401 ODD_DOUBLE; 402 } 403 x->fp_mant[3] = q; 404 405 /* 406 * The result, which includes guard and round bits, is exact iff 407 * x is now zero; any nonzero bits in x represent sticky bits. 408 */ 409 x->fp_sticky = x0 | x1 | x2 | x3; 410 DUMPFPN(FPE_REG, x); 411 return (x); 412 } 413