1 /* $NetBSD: fpu_sqrt.c,v 1.2 1994/11/20 20:52:46 deraadt 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 */ 46 47 /* 48 * Perform an FPU square root (return sqrt(x)). 49 */ 50 51 #include <sys/types.h> 52 53 #include <machine/reg.h> 54 55 #include <sparc/fpu/fpu_arith.h> 56 #include <sparc/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(fe) 191 struct fpemu *fe; 192 { 193 register struct fpn *x = &fe->fe_f1; 194 register u_int bit, q, tt; 195 register u_int x0, x1, x2, x3; 196 register u_int y0, y1, y2, y3; 197 register u_int d0, d1, d2, d3; 198 register int e; 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 if (ISNAN(x) || ISZERO(x)) 212 return (x); 213 if (x->fp_sign) 214 return (fpu_newnan(fe)); 215 if (ISINF(x)) 216 return (x); 217 218 /* 219 * Calculate result exponent. As noted above, this may involve 220 * doubling the mantissa. We will also need to double x each 221 * time around the loop, so we define a macro for this here, and 222 * we break out the multiword mantissa. 223 */ 224 #ifdef FPU_SHL1_BY_ADD 225 #define DOUBLE_X { \ 226 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \ 227 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 228 } 229 #else 230 #define DOUBLE_X { \ 231 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 232 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \ 233 } 234 #endif 235 #if (FP_NMANT & 1) != 0 236 # define ODD_DOUBLE DOUBLE_X 237 # define EVEN_DOUBLE /* nothing */ 238 #else 239 # define ODD_DOUBLE /* nothing */ 240 # define EVEN_DOUBLE DOUBLE_X 241 #endif 242 x0 = x->fp_mant[0]; 243 x1 = x->fp_mant[1]; 244 x2 = x->fp_mant[2]; 245 x3 = x->fp_mant[3]; 246 e = x->fp_exp; 247 if (e & 1) /* exponent is odd; use sqrt(2mant) */ 248 DOUBLE_X; 249 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 250 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 251 252 /* 253 * Now calculate the mantissa root. Since x is now in [1..4), 254 * we know that the first trip around the loop will definitely 255 * set the top bit in q, so we can do that manually and start 256 * the loop at the next bit down instead. We must be sure to 257 * double x correctly while doing the `known q=1.0'. 258 * 259 * We do this one mantissa-word at a time, as noted above, to 260 * save work. To avoid `(1 << 31) << 1', we also do the top bit 261 * outside of each per-word loop. 262 * 263 * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 264 * t3 = y3, t? |= bit' for the appropriate word. Since the bit 265 * is always a `new' one, this means that three of the `t?'s are 266 * just the corresponding `y?'; we use `#define's here for this. 267 * The variable `tt' holds the actual `t?' variable. 268 */ 269 270 /* calculate q0 */ 271 #define t0 tt 272 bit = FP_1; 273 EVEN_DOUBLE; 274 /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 275 q = bit; 276 x0 -= bit; 277 y0 = bit << 1; 278 /* } */ 279 ODD_DOUBLE; 280 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 281 EVEN_DOUBLE; 282 t0 = y0 | bit; /* t = y + bit */ 283 if (x0 >= t0) { /* if x >= t then */ 284 x0 -= t0; /* x -= t */ 285 q |= bit; /* q += bit */ 286 y0 |= bit << 1; /* y += bit << 1 */ 287 } 288 ODD_DOUBLE; 289 } 290 x->fp_mant[0] = q; 291 #undef t0 292 293 /* calculate q1. note (y0&1)==0. */ 294 #define t0 y0 295 #define t1 tt 296 q = 0; 297 y1 = 0; 298 bit = 1 << 31; 299 EVEN_DOUBLE; 300 t1 = bit; 301 FPU_SUBS(d1, x1, t1); 302 FPU_SUBC(d0, x0, t0); /* d = x - t */ 303 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 304 x0 = d0, x1 = d1; /* x -= t */ 305 q = bit; /* q += bit */ 306 y0 |= 1; /* y += bit << 1 */ 307 } 308 ODD_DOUBLE; 309 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 310 EVEN_DOUBLE; /* as before */ 311 t1 = y1 | bit; 312 FPU_SUBS(d1, x1, t1); 313 FPU_SUBC(d0, x0, t0); 314 if ((int)d0 >= 0) { 315 x0 = d0, x1 = d1; 316 q |= bit; 317 y1 |= bit << 1; 318 } 319 ODD_DOUBLE; 320 } 321 x->fp_mant[1] = q; 322 #undef t1 323 324 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 325 #define t1 y1 326 #define t2 tt 327 q = 0; 328 y2 = 0; 329 bit = 1 << 31; 330 EVEN_DOUBLE; 331 t2 = bit; 332 FPU_SUBS(d2, x2, t2); 333 FPU_SUBCS(d1, x1, t1); 334 FPU_SUBC(d0, x0, t0); 335 if ((int)d0 >= 0) { 336 x0 = d0, x1 = d1, x2 = d2; 337 q |= bit; 338 y1 |= 1; /* now t1, y1 are set in concrete */ 339 } 340 ODD_DOUBLE; 341 while ((bit >>= 1) != 0) { 342 EVEN_DOUBLE; 343 t2 = y2 | bit; 344 FPU_SUBS(d2, x2, t2); 345 FPU_SUBCS(d1, x1, t1); 346 FPU_SUBC(d0, x0, t0); 347 if ((int)d0 >= 0) { 348 x0 = d0, x1 = d1, x2 = d2; 349 q |= bit; 350 y2 |= bit << 1; 351 } 352 ODD_DOUBLE; 353 } 354 x->fp_mant[2] = q; 355 #undef t2 356 357 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */ 358 #define t2 y2 359 #define t3 tt 360 q = 0; 361 y3 = 0; 362 bit = 1 << 31; 363 EVEN_DOUBLE; 364 t3 = bit; 365 FPU_SUBS(d3, x3, t3); 366 FPU_SUBCS(d2, x2, t2); 367 FPU_SUBCS(d1, x1, t1); 368 FPU_SUBC(d0, x0, t0); 369 ODD_DOUBLE; 370 if ((int)d0 >= 0) { 371 x0 = d0, x1 = d1, x2 = d2; 372 q |= bit; 373 y2 |= 1; 374 } 375 while ((bit >>= 1) != 0) { 376 EVEN_DOUBLE; 377 t3 = y3 | bit; 378 FPU_SUBS(d3, x3, t3); 379 FPU_SUBCS(d2, x2, t2); 380 FPU_SUBCS(d1, x1, t1); 381 FPU_SUBC(d0, x0, t0); 382 if ((int)d0 >= 0) { 383 x0 = d0, x1 = d1, x2 = d2; 384 q |= bit; 385 y3 |= bit << 1; 386 } 387 ODD_DOUBLE; 388 } 389 x->fp_mant[3] = q; 390 391 /* 392 * The result, which includes guard and round bits, is exact iff 393 * x is now zero; any nonzero bits in x represent sticky bits. 394 */ 395 x->fp_sticky = x0 | x1 | x2 | x3; 396 return (x); 397 } 398