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1.1.1.2 ! root 1: /* $NetBSD: fpu_sqrt.c,v 1.9 2022/05/24 20:00:49 andvar Exp $ */ 1.1 root 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. Neither the name of the University nor the names of its contributors 25: * may be used to endorse or promote products derived from this software 26: * without specific prior written permission. 27: * 28: * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND 29: * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE 30: * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE 31: * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE 32: * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL 33: * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS 34: * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) 35: * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT 36: * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY 37: * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF 38: * SUCH DAMAGE. 39: * 40: * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93 41: */ 42: 43: /* 44: * Perform an FPU square root (return sqrt(x)). 45: */ 46: 47: #include "fpu_arith.h" 48: #include "fpu_emulate.h" 49: 50: /* 51: * Our task is to calculate the square root of a floating point number x0. 52: * This number x normally has the form: 53: * 54: * exp 55: * x = mant * 2 (where 1 <= mant < 2 and exp is an integer) 56: * 57: * This can be left as it stands, or the mantissa can be doubled and the 58: * exponent decremented: 59: * 60: * exp-1 61: * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4) 62: * 63: * If the exponent `exp' is even, the square root of the number is best 64: * handled using the first form, and is by definition equal to: 65: * 66: * exp/2 67: * sqrt(x) = sqrt(mant) * 2 68: * 69: * If exp is odd, on the other hand, it is convenient to use the second 70: * form, giving: 71: * 72: * (exp-1)/2 73: * sqrt(x) = sqrt(2 * mant) * 2 74: * 75: * In the first case, we have 76: * 77: * 1 <= mant < 2 78: * 79: * and therefore 80: * 81: * sqrt(1) <= sqrt(mant) < sqrt(2) 82: * 83: * while in the second case we have 84: * 85: * 2 <= 2*mant < 4 86: * 87: * and therefore 88: * 89: * sqrt(2) <= sqrt(2*mant) < sqrt(4) 90: * 91: * so that in any case, we are sure that 92: * 93: * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2 94: * 95: * or 96: * 97: * 1 <= sqrt(n * mant) < 2, n = 1 or 2. 98: * 99: * This root is therefore a properly formed mantissa for a floating 100: * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2 101: * as above. This leaves us with the problem of finding the square root 102: * of a fixed-point number in the range [1..4). 103: * 104: * Though it may not be instantly obvious, the following square root 105: * algorithm works for any integer x of an even number of bits, provided 106: * that no overflows occur: 107: * 108: * let q = 0 109: * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer... 110: * x *= 2 -- multiply by radix, for next digit 111: * if x >= 2q + 2^k then -- if adding 2^k does not 112: * x -= 2q + 2^k -- exceed the correct root, 113: * q += 2^k -- add 2^k and adjust x 114: * fi 115: * done 116: * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x) 117: * 118: * If NBITS is odd (so that k is initially even), we can just add another 119: * zero bit at the top of x. Doing so means that q is not going to acquire 120: * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the 121: * final value in x is not needed, or can be off by a factor of 2, this is 1.1.1.2 ! root 122: * equivalent to moving the `x *= 2' step to the bottom of the loop: 1.1 root 123: * 124: * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done 125: * 126: * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2). 127: * (Since the algorithm is destructive on x, we will call x's initial 128: * value, for which q is some power of two times its square root, x0.) 129: * 130: * If we insert a loop invariant y = 2q, we can then rewrite this using 131: * C notation as: 132: * 133: * q = y = 0; x = x0; 134: * for (k = NBITS; --k >= 0;) { 135: * #if (NBITS is even) 136: * x *= 2; 137: * #endif 138: * t = y + (1 << k); 139: * if (x >= t) { 140: * x -= t; 141: * q += 1 << k; 142: * y += 1 << (k + 1); 143: * } 144: * #if (NBITS is odd) 145: * x *= 2; 146: * #endif 147: * } 148: * 149: * If x0 is fixed point, rather than an integer, we can simply alter the 150: * scale factor between q and sqrt(x0). As it happens, we can easily arrange 151: * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q. 152: * 153: * In our case, however, x0 (and therefore x, y, q, and t) are multiword 154: * integers, which adds some complication. But note that q is built one 155: * bit at a time, from the top down, and is not used itself in the loop 156: * (we use 2q as held in y instead). This means we can build our answer 157: * in an integer, one word at a time, which saves a bit of work. Also, 158: * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are 159: * `new' bits in y and we can set them with an `or' operation rather than 160: * a full-blown multiword add. 161: * 162: * We are almost done, except for one snag. We must prove that none of our 163: * intermediate calculations can overflow. We know that x0 is in [1..4) 164: * and therefore the square root in q will be in [1..2), but what about x, 165: * y, and t? 166: * 167: * We know that y = 2q at the beginning of each loop. (The relation only 168: * fails temporarily while y and q are being updated.) Since q < 2, y < 4. 169: * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and. 170: * Furthermore, we can prove with a bit of work that x never exceeds y by 171: * more than 2, so that even after doubling, 0 <= x < 8. (This is left as 172: * an exercise to the reader, mostly because I have become tired of working 173: * on this comment.) 174: * 175: * If our floating point mantissas (which are of the form 1.frac) occupy 176: * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra. 177: * In fact, we want even one more bit (for a carry, to avoid compares), or 178: * three extra. There is a comment in fpu_emu.h reminding maintainers of 179: * this, so we have some justification in assuming it. 180: */ 181: struct fpn * 182: fpu_sqrt(struct fpemu *fe) 183: { 184: struct fpn *x = &fe->fe_f2; 185: uint32_t bit, q, tt; 186: uint32_t x0, x1, x2; 187: uint32_t y0, y1, y2; 188: uint32_t d0, d1, d2; 189: int e; 190: FPU_DECL_CARRY 191: 192: /* 193: * Take care of special cases first. In order: 194: * 195: * sqrt(NaN) = NaN 196: * sqrt(+0) = +0 197: * sqrt(-0) = -0 198: * sqrt(x < 0) = NaN (including sqrt(-Inf)) 199: * sqrt(+Inf) = +Inf 200: * 201: * Then all that remains are numbers with mantissas in [1..2). 202: */ 203: if (ISNAN(x) || ISZERO(x)) 204: return (x); 205: if (x->fp_sign) 206: return (fpu_newnan(fe)); 207: if (ISINF(x)) 208: return (x); 209: 210: /* 211: * Calculate result exponent. As noted above, this may involve 212: * doubling the mantissa. We will also need to double x each 213: * time around the loop, so we define a macro for this here, and 214: * we break out the multiword mantissa. 215: */ 216: #ifdef FPU_SHL1_BY_ADD 217: #define DOUBLE_X { \ 218: FPU_ADDS(x2, x2, x2); \ 219: FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \ 220: } 221: #else 222: #define DOUBLE_X { \ 223: x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \ 224: x2 <<= 1; \ 225: } 226: #endif 227: #if (FP_NMANT & 1) != 0 228: # define ODD_DOUBLE DOUBLE_X 229: # define EVEN_DOUBLE /* nothing */ 230: #else 231: # define ODD_DOUBLE /* nothing */ 232: # define EVEN_DOUBLE DOUBLE_X 233: #endif 234: x0 = x->fp_mant[0]; 235: x1 = x->fp_mant[1]; 236: x2 = x->fp_mant[2]; 237: e = x->fp_exp; 238: if (e & 1) /* exponent is odd; use sqrt(2mant) */ 239: DOUBLE_X; 240: /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */ 241: x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */ 242: 243: /* 244: * Now calculate the mantissa root. Since x is now in [1..4), 245: * we know that the first trip around the loop will definitely 246: * set the top bit in q, so we can do that manually and start 247: * the loop at the next bit down instead. We must be sure to 248: * double x correctly while doing the `known q=1.0'. 249: * 250: * We do this one mantissa-word at a time, as noted above, to 251: * save work. To avoid `(1 << 31) << 1', we also do the top bit 252: * outside of each per-word loop. 253: * 254: * The calculation `t = y + bit' breaks down into `t0 = y0, ..., 255: * t2 = y2, t? |= bit' for the appropriate word. Since the bit 256: * is always a `new' one, this means that three of the `t?'s are 257: * just the corresponding `y?'; we use `#define's here for this. 258: * The variable `tt' holds the actual `t?' variable. 259: */ 260: 261: /* calculate q0 */ 262: #define t0 tt 263: bit = FP_1; 264: EVEN_DOUBLE; 265: /* if (x >= (t0 = y0 | bit)) { */ /* always true */ 266: q = bit; 267: x0 -= bit; 268: y0 = bit << 1; 269: /* } */ 270: ODD_DOUBLE; 271: while ((bit >>= 1) != 0) { /* for remaining bits in q0 */ 272: EVEN_DOUBLE; 273: t0 = y0 | bit; /* t = y + bit */ 274: if (x0 >= t0) { /* if x >= t then */ 275: x0 -= t0; /* x -= t */ 276: q |= bit; /* q += bit */ 277: y0 |= bit << 1; /* y += bit << 1 */ 278: } 279: ODD_DOUBLE; 280: } 281: x->fp_mant[0] = q; 282: #undef t0 283: 284: /* calculate q1. note (y0&1)==0. */ 285: #define t0 y0 286: #define t1 tt 287: q = 0; 288: y1 = 0; 289: bit = 1 << 31; 290: EVEN_DOUBLE; 291: t1 = bit; 292: FPU_SUBS(d1, x1, t1); 293: FPU_SUBC(d0, x0, t0); /* d = x - t */ 294: if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */ 295: x0 = d0, x1 = d1; /* x -= t */ 296: q = bit; /* q += bit */ 297: y0 |= 1; /* y += bit << 1 */ 298: } 299: ODD_DOUBLE; 300: while ((bit >>= 1) != 0) { /* for remaining bits in q1 */ 301: EVEN_DOUBLE; /* as before */ 302: t1 = y1 | bit; 303: FPU_SUBS(d1, x1, t1); 304: FPU_SUBC(d0, x0, t0); 305: if ((int)d0 >= 0) { 306: x0 = d0, x1 = d1; 307: q |= bit; 308: y1 |= bit << 1; 309: } 310: ODD_DOUBLE; 311: } 312: x->fp_mant[1] = q; 313: #undef t1 314: 315: /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */ 316: #define t1 y1 317: #define t2 tt 318: q = 0; 319: y2 = 0; 320: bit = 1 << 31; 321: EVEN_DOUBLE; 322: t2 = bit; 323: FPU_SUBS(d2, x2, t2); 324: FPU_SUBCS(d1, x1, t1); 325: FPU_SUBC(d0, x0, t0); 326: if ((int)d0 >= 0) { 327: x0 = d0, x1 = d1, x2 = d2; 328: q |= bit; 329: y1 |= 1; /* now t1, y1 are set in concrete */ 330: } 331: ODD_DOUBLE; 332: while ((bit >>= 1) != 0) { 333: EVEN_DOUBLE; 334: t2 = y2 | bit; 335: FPU_SUBS(d2, x2, t2); 336: FPU_SUBCS(d1, x1, t1); 337: FPU_SUBC(d0, x0, t0); 338: if ((int)d0 >= 0) { 339: x0 = d0, x1 = d1, x2 = d2; 340: q |= bit; 341: y2 |= bit << 1; 342: } 343: ODD_DOUBLE; 344: } 345: x->fp_mant[2] = q; 346: #undef t2 347: 348: /* 349: * The result, which includes guard and round bits, is exact iff 350: * x is now zero; any nonzero bits in x represent sticky bits. 351: */ 352: x->fp_sticky = x0 | x1 | x2; 353: return (x); 354: }
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