1 /* 2 * Copyright (c) 1992, 1993 3 * The Regents of the University of California. All rights reserved. 4 * 5 * This software was developed by the Computer Systems Engineering group 6 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and 7 * contributed to Berkeley. 8 * 9 * All advertising materials mentioning features or use of this software 10 * must display the following acknowledgement: 11 * This product includes software developed by the University of 12 * California, Lawrence Berkeley Laboratory. 13 * 14 * Redistribution and use in source and binary forms, with or without 15 * modification, are permitted provided that the following conditions 16 * are met: 17 * 1. Redistributions of source code must retain the above copyright 18 * notice, this list of conditions and the following disclaimer. 19 * 2. Redistributions in binary form must reproduce the above copyright 20 * notice, this list of conditions and the following disclaimer in the 21 * documentation and/or other materials provided with the distribution. 22 * 3. All advertising materials mentioning features or use of this software 23 * must display the following acknowledgement: 24 * This product includes software developed by the University of 25 * California, Berkeley and its contributors. 26 * 4. 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 * from: Header: fpu_sqrt.c,v 1.3 92/11/26 01:39:51 torek Exp 45 * $Id: fpu_sqrt.c,v 1.1 1993/10/02 10:23:00 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/reg.h> 55 56 #include <sparc/fpu/fpu_arith.h> 57 #include <sparc/fpu/fpu_emu.h> 58 59 /* 60 * Our task is to calculate the square root of a floating point number x0. 61 * This number x normally has the form: 62 * 63 * exp 64 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 65 * 66 * This can be left as it stands, or the mantissa can be doubled and the 67 * exponent decremented: 68 * 69 * exp-1 70 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 71 * 72 * If the exponent `exp' is even, the square root of the number is best 73 * handled using the first form, and is by definition equal to: 74 * 75 * exp/2 76 * sqrt(x) = sqrt(mant) * 2 77 * 78 * If exp is odd, on the other hand, it is convenient to use the second 79 * form, giving: 80 * 81 * (exp-1)/2 82 * sqrt(x) = sqrt(2 * mant) * 2 83 * 84 * In the first case, we have 85 * 86 * 1 <= mant < 2 87 * 88 * and therefore 89 * 90 * sqrt(1) <= sqrt(mant) < sqrt(2) 91 * 92 * while in the second case we have 93 * 94 * 2 <= 2*mant < 4 95 * 96 * and therefore 97 * 98 * sqrt(2) <= sqrt(2*mant) < sqrt(4) 99 * 100 * so that in any case, we are sure that 101 * 102 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 103 * 104 * or 105 * 106 * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 107 * 108 * This root is therefore a properly formed mantissa for a floating 109 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 110 * as above. This leaves us with the problem of finding the square root 111 * of a fixed-point number in the range [1..4). 112 * 113 * Though it may not be instantly obvious, the following square root 114 * algorithm works for any integer x of an even number of bits, provided 115 * that no overflows occur: 116 * 117 * let q = 0 118 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 119 * x *= 2 -- multiply by radix, for next digit 120 * if x >= 2q + 2^k then -- if adding 2^k does not 121 * x -= 2q + 2^k -- exceed the correct root, 122 * q += 2^k -- add 2^k and adjust x 123 * fi 124 * done 125 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 126 * 127 * If NBITS is odd (so that k is initially even), we can just add another 128 * zero bit at the top of x. Doing so means that q is not going to acquire 129 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 130 * final value in x is not needed, or can be off by a factor of 2, this is 131 * equivalant to moving the `x *= 2' step to the bottom of the loop: 132 * 133 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 134 * 135 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 136 * (Since the algorithm is destructive on x, we will call x's initial 137 * value, for which q is some power of two times its square root, x0.) 138 * 139 * If we insert a loop invariant y = 2q, we can then rewrite this using 140 * C notation as: 141 * 142 * q = y = 0; x = x0; 143 * for (k = NBITS; --k >= 0;) { 144 * #if (NBITS is even) 145 * x *= 2; 146 * #endif 147 * t = y + (1 << k); 148 * if (x >= t) { 149 * x -= t; 150 * q += 1 << k; 151 * y += 1 << (k + 1); 152 * } 153 * #if (NBITS is odd) 154 * x *= 2; 155 * #endif 156 * } 157 * 158 * If x0 is fixed point, rather than an integer, we can simply alter the 159 * scale factor between q and sqrt(x0). As it happens, we can easily arrange 160 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 161 * 162 * In our case, however, x0 (and therefore x, y, q, and t) are multiword 163 * integers, which adds some complication. But note that q is built one 164 * bit at a time, from the top down, and is not used itself in the loop 165 * (we use 2q as held in y instead). This means we can build our answer 166 * in an integer, one word at a time, which saves a bit of work. Also, 167 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 168 * `new' bits in y and we can set them with an `or' operation rather than 169 * a full-blown multiword add. 170 * 171 * We are almost done, except for one snag. We must prove that none of our 172 * intermediate calculations can overflow. We know that x0 is in [1..4) 173 * and therefore the square root in q will be in [1..2), but what about x, 174 * y, and t? 175 * 176 * We know that y = 2q at the beginning of each loop. (The relation only 177 * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 178 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 179 * Furthermore, we can prove with a bit of work that x never exceeds y by 180 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 181 * an exercise to the reader, mostly because I have become tired of working 182 * on this comment.) 183 * 184 * If our floating point mantissas (which are of the form 1.frac) occupy 185 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 186 * In fact, we want even one more bit (for a carry, to avoid compares), or 187 * three extra. There is a comment in fpu_emu.h reminding maintainers of 188 * this, so we have some justification in assuming it. 189 */ 190 struct fpn * 191 fpu_sqrt(fe) 192 struct fpemu *fe; 193 { 194 register struct fpn *x = &fe->fe_f1; 195 register u_int bit, q, tt; 196 register u_int x0, x1, x2, x3; 197 register u_int y0, y1, y2, y3; 198 register u_int d0, d1, d2, d3; 199 register int e; 200 201 /* 202 * Take care of special cases first. In order: 203 * 204 * sqrt(NaN) = NaN 205 * sqrt(+0) = +0 206 * sqrt(-0) = -0 207 * sqrt(x < 0) = NaN (including sqrt(-Inf)) 208 * sqrt(+Inf) = +Inf 209 * 210 * Then all that remains are numbers with mantissas in [1..2). 211 */ 212 if (ISNAN(x) || ISZERO(x)) 213 return (x); 214 if (x->fp_sign) 215 return (fpu_newnan(fe)); 216 if (ISINF(x)) 217 return (x); 218 219 /* 220 * Calculate result exponent. As noted above, this may involve 221 * doubling the mantissa. We will also need to double x each 222 * time around the loop, so we define a macro for this here, and 223 * we break out the multiword mantissa. 224 */ 225 #ifdef FPU_SHL1_BY_ADD 226 #define DOUBLE_X { \ 227 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 228 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 229 } 230 #else 231 #define DOUBLE_X { \ 232 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 233 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 234 } 235 #endif 236 #if (FP_NMANT & 1) != 0 237 # define ODD_DOUBLE DOUBLE_X 238 # define EVEN_DOUBLE /* nothing */ 239 #else 240 # define ODD_DOUBLE /* nothing */ 241 # define EVEN_DOUBLE DOUBLE_X 242 #endif 243 x0 = x->fp_mant[0]; 244 x1 = x->fp_mant[1]; 245 x2 = x->fp_mant[2]; 246 x3 = x->fp_mant[3]; 247 e = x->fp_exp; 248 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 249 DOUBLE_X; 250 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 251 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 252 253 /* 254 * Now calculate the mantissa root. Since x is now in [1..4), 255 * we know that the first trip around the loop will definitely 256 * set the top bit in q, so we can do that manually and start 257 * the loop at the next bit down instead. We must be sure to 258 * double x correctly while doing the `known q=1.0'. 259 * 260 * We do this one mantissa-word at a time, as noted above, to 261 * save work. To avoid `(1 << 31) << 1', we also do the top bit 262 * outside of each per-word loop. 263 * 264 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 265 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 266 * is always a `new' one, this means that three of the `t?'s are 267 * just the corresponding `y?'; we use `#define's here for this. 268 * The variable `tt' holds the actual `t?' variable. 269 */ 270 271 /* calculate q0 */ 272 #define t0 tt 273 bit = FP_1; 274 EVEN_DOUBLE; 275 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 276 q = bit; 277 x0 -= bit; 278 y0 = bit << 1; 279 /* } */ 280 ODD_DOUBLE; 281 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 282 EVEN_DOUBLE; 283 t0 = y0 | bit; /* t = y + bit */ 284 if (x0 >= t0) { /* if x >= t then */ 285 x0 -= t0; /* x -= t */ 286 q |= bit; /* q += bit */ 287 y0 |= bit << 1; /* y += bit << 1 */ 288 } 289 ODD_DOUBLE; 290 } 291 x->fp_mant[0] = q; 292 #undef t0 293 294 /* calculate q1. note (y0&1)==0. */ 295 #define t0 y0 296 #define t1 tt 297 q = 0; 298 y1 = 0; 299 bit = 1 << 31; 300 EVEN_DOUBLE; 301 t1 = bit; 302 FPU_SUBS(d1, x1, t1); 303 FPU_SUBC(d0, x0, t0); /* d = x - t */ 304 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 305 x0 = d0, x1 = d1; /* x -= t */ 306 q = bit; /* q += bit */ 307 y0 |= 1; /* y += bit << 1 */ 308 } 309 ODD_DOUBLE; 310 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 311 EVEN_DOUBLE; /* as before */ 312 t1 = y1 | bit; 313 FPU_SUBS(d1, x1, t1); 314 FPU_SUBC(d0, x0, t0); 315 if ((int)d0 >= 0) { 316 x0 = d0, x1 = d1; 317 q |= bit; 318 y1 |= bit << 1; 319 } 320 ODD_DOUBLE; 321 } 322 x->fp_mant[1] = q; 323 #undef t1 324 325 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 326 #define t1 y1 327 #define t2 tt 328 q = 0; 329 y2 = 0; 330 bit = 1 << 31; 331 EVEN_DOUBLE; 332 t2 = bit; 333 FPU_SUBS(d2, x2, t2); 334 FPU_SUBCS(d1, x1, t1); 335 FPU_SUBC(d0, x0, t0); 336 if ((int)d0 >= 0) { 337 x0 = d0, x1 = d1, x2 = d2; 338 q |= bit; 339 y1 |= 1; /* now t1, y1 are set in concrete */ 340 } 341 ODD_DOUBLE; 342 while ((bit >>= 1) != 0) { 343 EVEN_DOUBLE; 344 t2 = y2 | bit; 345 FPU_SUBS(d2, x2, t2); 346 FPU_SUBCS(d1, x1, t1); 347 FPU_SUBC(d0, x0, t0); 348 if ((int)d0 >= 0) { 349 x0 = d0, x1 = d1, x2 = d2; 350 q |= bit; 351 y2 |= bit << 1; 352 } 353 ODD_DOUBLE; 354 } 355 x->fp_mant[2] = q; 356 #undef t2 357 358 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 359 #define t2 y2 360 #define t3 tt 361 q = 0; 362 y3 = 0; 363 bit = 1 << 31; 364 EVEN_DOUBLE; 365 t3 = bit; 366 FPU_SUBS(d3, x3, t3); 367 FPU_SUBCS(d2, x2, t2); 368 FPU_SUBCS(d1, x1, t1); 369 FPU_SUBC(d0, x0, t0); 370 ODD_DOUBLE; 371 if ((int)d0 >= 0) { 372 x0 = d0, x1 = d1, x2 = d2; 373 q |= bit; 374 y2 |= 1; 375 } 376 while ((bit >>= 1) != 0) { 377 EVEN_DOUBLE; 378 t3 = y3 | 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; 385 q |= bit; 386 y3 |= bit << 1; 387 } 388 ODD_DOUBLE; 389 } 390 x->fp_mant[3] = q; 391 392 /* 393 * The result, which includes guard and round bits, is exact iff 394 * x is now zero; any nonzero bits in x represent sticky bits. 395 */ 396 x->fp_sticky = x0 | x1 | x2 | x3; 397 return (x); 398 } 399