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fpu_sqrt.c revision 1.3
      1 /*	$NetBSD: fpu_sqrt.c,v 1.3 2003/08/07 16:29:18 agc 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. 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 <sys/cdefs.h>
     48 __KERNEL_RCSID(0, "$NetBSD: fpu_sqrt.c,v 1.3 2003/08/07 16:29:18 agc Exp $");
     49 
     50 #include <sys/types.h>
     51 #if defined(DIAGNOSTIC)||defined(DEBUG)
     52 #include <sys/systm.h>
     53 #endif
     54 
     55 #include <machine/reg.h>
     56 #include <machine/fpu.h>
     57 
     58 #include <powerpc/fpu/fpu_arith.h>
     59 #include <powerpc/fpu/fpu_emu.h>
     60 
     61 /*
     62  * Our task is to calculate the square root of a floating point number x0.
     63  * This number x normally has the form:
     64  *
     65  *		    exp
     66  *	x = mant * 2		(where 1 <= mant < 2 and exp is an integer)
     67  *
     68  * This can be left as it stands, or the mantissa can be doubled and the
     69  * exponent decremented:
     70  *
     71  *			  exp-1
     72  *	x = (2 * mant) * 2	(where 2 <= 2 * mant < 4)
     73  *
     74  * If the exponent `exp' is even, the square root of the number is best
     75  * handled using the first form, and is by definition equal to:
     76  *
     77  *				exp/2
     78  *	sqrt(x) = sqrt(mant) * 2
     79  *
     80  * If exp is odd, on the other hand, it is convenient to use the second
     81  * form, giving:
     82  *
     83  *				    (exp-1)/2
     84  *	sqrt(x) = sqrt(2 * mant) * 2
     85  *
     86  * In the first case, we have
     87  *
     88  *	1 <= mant < 2
     89  *
     90  * and therefore
     91  *
     92  *	sqrt(1) <= sqrt(mant) < sqrt(2)
     93  *
     94  * while in the second case we have
     95  *
     96  *	2 <= 2*mant < 4
     97  *
     98  * and therefore
     99  *
    100  *	sqrt(2) <= sqrt(2*mant) < sqrt(4)
    101  *
    102  * so that in any case, we are sure that
    103  *
    104  *	sqrt(1) <= sqrt(n * mant) < sqrt(4),	n = 1 or 2
    105  *
    106  * or
    107  *
    108  *	1 <= sqrt(n * mant) < 2,		n = 1 or 2.
    109  *
    110  * This root is therefore a properly formed mantissa for a floating
    111  * point number.  The exponent of sqrt(x) is either exp/2 or (exp-1)/2
    112  * as above.  This leaves us with the problem of finding the square root
    113  * of a fixed-point number in the range [1..4).
    114  *
    115  * Though it may not be instantly obvious, the following square root
    116  * algorithm works for any integer x of an even number of bits, provided
    117  * that no overflows occur:
    118  *
    119  *	let q = 0
    120  *	for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
    121  *		x *= 2			-- multiply by radix, for next digit
    122  *		if x >= 2q + 2^k then	-- if adding 2^k does not
    123  *			x -= 2q + 2^k	-- exceed the correct root,
    124  *			q += 2^k	-- add 2^k and adjust x
    125  *		fi
    126  *	done
    127  *	sqrt = q / 2^(NBITS/2)		-- (and any remainder is in x)
    128  *
    129  * If NBITS is odd (so that k is initially even), we can just add another
    130  * zero bit at the top of x.  Doing so means that q is not going to acquire
    131  * a 1 bit in the first trip around the loop (since x0 < 2^NBITS).  If the
    132  * final value in x is not needed, or can be off by a factor of 2, this is
    133  * equivalant to moving the `x *= 2' step to the bottom of the loop:
    134  *
    135  *	for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
    136  *
    137  * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
    138  * (Since the algorithm is destructive on x, we will call x's initial
    139  * value, for which q is some power of two times its square root, x0.)
    140  *
    141  * If we insert a loop invariant y = 2q, we can then rewrite this using
    142  * C notation as:
    143  *
    144  *	q = y = 0; x = x0;
    145  *	for (k = NBITS; --k >= 0;) {
    146  * #if (NBITS is even)
    147  *		x *= 2;
    148  * #endif
    149  *		t = y + (1 << k);
    150  *		if (x >= t) {
    151  *			x -= t;
    152  *			q += 1 << k;
    153  *			y += 1 << (k + 1);
    154  *		}
    155  * #if (NBITS is odd)
    156  *		x *= 2;
    157  * #endif
    158  *	}
    159  *
    160  * If x0 is fixed point, rather than an integer, we can simply alter the
    161  * scale factor between q and sqrt(x0).  As it happens, we can easily arrange
    162  * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
    163  *
    164  * In our case, however, x0 (and therefore x, y, q, and t) are multiword
    165  * integers, which adds some complication.  But note that q is built one
    166  * bit at a time, from the top down, and is not used itself in the loop
    167  * (we use 2q as held in y instead).  This means we can build our answer
    168  * in an integer, one word at a time, which saves a bit of work.  Also,
    169  * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
    170  * `new' bits in y and we can set them with an `or' operation rather than
    171  * a full-blown multiword add.
    172  *
    173  * We are almost done, except for one snag.  We must prove that none of our
    174  * intermediate calculations can overflow.  We know that x0 is in [1..4)
    175  * and therefore the square root in q will be in [1..2), but what about x,
    176  * y, and t?
    177  *
    178  * We know that y = 2q at the beginning of each loop.  (The relation only
    179  * fails temporarily while y and q are being updated.)  Since q < 2, y < 4.
    180  * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
    181  * Furthermore, we can prove with a bit of work that x never exceeds y by
    182  * more than 2, so that even after doubling, 0 <= x < 8.  (This is left as
    183  * an exercise to the reader, mostly because I have become tired of working
    184  * on this comment.)
    185  *
    186  * If our floating point mantissas (which are of the form 1.frac) occupy
    187  * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
    188  * In fact, we want even one more bit (for a carry, to avoid compares), or
    189  * three extra.  There is a comment in fpu_emu.h reminding maintainers of
    190  * this, so we have some justification in assuming it.
    191  */
    192 struct fpn *
    193 fpu_sqrt(struct fpemu *fe)
    194 {
    195 	struct fpn *x = &fe->fe_f1;
    196 	u_int bit, q, tt;
    197 	u_int x0, x1, x2, x3;
    198 	u_int y0, y1, y2, y3;
    199 	u_int d0, d1, d2, d3;
    200 	int e;
    201 	FPU_DECL_CARRY;
    202 
    203 	/*
    204 	 * Take care of special cases first.  In order:
    205 	 *
    206 	 *	sqrt(NaN) = NaN
    207 	 *	sqrt(+0) = +0
    208 	 *	sqrt(-0) = -0
    209 	 *	sqrt(x < 0) = NaN	(including sqrt(-Inf))
    210 	 *	sqrt(+Inf) = +Inf
    211 	 *
    212 	 * Then all that remains are numbers with mantissas in [1..2).
    213 	 */
    214 	DPRINTF(FPE_REG, ("fpu_sqer:\n"));
    215 	DUMPFPN(FPE_REG, x);
    216 	DPRINTF(FPE_REG, ("=>\n"));
    217 	if (ISNAN(x)) {
    218 		fe->fe_cx |= FPSCR_VXSNAN;
    219 		DUMPFPN(FPE_REG, x);
    220 		return (x);
    221 	}
    222 	if (ISZERO(x)) {
    223 		fe->fe_cx |= FPSCR_ZX;
    224 		x->fp_class = FPC_INF;
    225 		DUMPFPN(FPE_REG, x);
    226 		return (x);
    227 	}
    228 	if (x->fp_sign) {
    229 		return (fpu_newnan(fe));
    230 	}
    231 	if (ISINF(x)) {
    232 		fe->fe_cx |= FPSCR_VXSQRT;
    233 		DUMPFPN(FPE_REG, 0);
    234 		return (0);
    235 	}
    236 
    237 	/*
    238 	 * Calculate result exponent.  As noted above, this may involve
    239 	 * doubling the mantissa.  We will also need to double x each
    240 	 * time around the loop, so we define a macro for this here, and
    241 	 * we break out the multiword mantissa.
    242 	 */
    243 #ifdef FPU_SHL1_BY_ADD
    244 #define	DOUBLE_X { \
    245 	FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
    246 	FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
    247 }
    248 #else
    249 #define	DOUBLE_X { \
    250 	x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
    251 	x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
    252 }
    253 #endif
    254 #if (FP_NMANT & 1) != 0
    255 # define ODD_DOUBLE	DOUBLE_X
    256 # define EVEN_DOUBLE	/* nothing */
    257 #else
    258 # define ODD_DOUBLE	/* nothing */
    259 # define EVEN_DOUBLE	DOUBLE_X
    260 #endif
    261 	x0 = x->fp_mant[0];
    262 	x1 = x->fp_mant[1];
    263 	x2 = x->fp_mant[2];
    264 	x3 = x->fp_mant[3];
    265 	e = x->fp_exp;
    266 	if (e & 1)		/* exponent is odd; use sqrt(2mant) */
    267 		DOUBLE_X;
    268 	/* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
    269 	x->fp_exp = e >> 1;	/* calculates (e&1 ? (e-1)/2 : e/2 */
    270 
    271 	/*
    272 	 * Now calculate the mantissa root.  Since x is now in [1..4),
    273 	 * we know that the first trip around the loop will definitely
    274 	 * set the top bit in q, so we can do that manually and start
    275 	 * the loop at the next bit down instead.  We must be sure to
    276 	 * double x correctly while doing the `known q=1.0'.
    277 	 *
    278 	 * We do this one mantissa-word at a time, as noted above, to
    279 	 * save work.  To avoid `(1 << 31) << 1', we also do the top bit
    280 	 * outside of each per-word loop.
    281 	 *
    282 	 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
    283 	 * t3 = y3, t? |= bit' for the appropriate word.  Since the bit
    284 	 * is always a `new' one, this means that three of the `t?'s are
    285 	 * just the corresponding `y?'; we use `#define's here for this.
    286 	 * The variable `tt' holds the actual `t?' variable.
    287 	 */
    288 
    289 	/* calculate q0 */
    290 #define	t0 tt
    291 	bit = FP_1;
    292 	EVEN_DOUBLE;
    293 	/* if (x >= (t0 = y0 | bit)) { */	/* always true */
    294 		q = bit;
    295 		x0 -= bit;
    296 		y0 = bit << 1;
    297 	/* } */
    298 	ODD_DOUBLE;
    299 	while ((bit >>= 1) != 0) {	/* for remaining bits in q0 */
    300 		EVEN_DOUBLE;
    301 		t0 = y0 | bit;		/* t = y + bit */
    302 		if (x0 >= t0) {		/* if x >= t then */
    303 			x0 -= t0;	/*	x -= t */
    304 			q |= bit;	/*	q += bit */
    305 			y0 |= bit << 1;	/*	y += bit << 1 */
    306 		}
    307 		ODD_DOUBLE;
    308 	}
    309 	x->fp_mant[0] = q;
    310 #undef t0
    311 
    312 	/* calculate q1.  note (y0&1)==0. */
    313 #define t0 y0
    314 #define t1 tt
    315 	q = 0;
    316 	y1 = 0;
    317 	bit = 1 << 31;
    318 	EVEN_DOUBLE;
    319 	t1 = bit;
    320 	FPU_SUBS(d1, x1, t1);
    321 	FPU_SUBC(d0, x0, t0);		/* d = x - t */
    322 	if ((int)d0 >= 0) {		/* if d >= 0 (i.e., x >= t) then */
    323 		x0 = d0, x1 = d1;	/*	x -= t */
    324 		q = bit;		/*	q += bit */
    325 		y0 |= 1;		/*	y += bit << 1 */
    326 	}
    327 	ODD_DOUBLE;
    328 	while ((bit >>= 1) != 0) {	/* for remaining bits in q1 */
    329 		EVEN_DOUBLE;		/* as before */
    330 		t1 = y1 | bit;
    331 		FPU_SUBS(d1, x1, t1);
    332 		FPU_SUBC(d0, x0, t0);
    333 		if ((int)d0 >= 0) {
    334 			x0 = d0, x1 = d1;
    335 			q |= bit;
    336 			y1 |= bit << 1;
    337 		}
    338 		ODD_DOUBLE;
    339 	}
    340 	x->fp_mant[1] = q;
    341 #undef t1
    342 
    343 	/* calculate q2.  note (y1&1)==0; y0 (aka t0) is fixed. */
    344 #define t1 y1
    345 #define t2 tt
    346 	q = 0;
    347 	y2 = 0;
    348 	bit = 1 << 31;
    349 	EVEN_DOUBLE;
    350 	t2 = bit;
    351 	FPU_SUBS(d2, x2, t2);
    352 	FPU_SUBCS(d1, x1, t1);
    353 	FPU_SUBC(d0, x0, t0);
    354 	if ((int)d0 >= 0) {
    355 		x0 = d0, x1 = d1, x2 = d2;
    356 		q |= bit;
    357 		y1 |= 1;		/* now t1, y1 are set in concrete */
    358 	}
    359 	ODD_DOUBLE;
    360 	while ((bit >>= 1) != 0) {
    361 		EVEN_DOUBLE;
    362 		t2 = y2 | bit;
    363 		FPU_SUBS(d2, x2, t2);
    364 		FPU_SUBCS(d1, x1, t1);
    365 		FPU_SUBC(d0, x0, t0);
    366 		if ((int)d0 >= 0) {
    367 			x0 = d0, x1 = d1, x2 = d2;
    368 			q |= bit;
    369 			y2 |= bit << 1;
    370 		}
    371 		ODD_DOUBLE;
    372 	}
    373 	x->fp_mant[2] = q;
    374 #undef t2
    375 
    376 	/* calculate q3.  y0, t0, y1, t1 all fixed; y2, t2, almost done. */
    377 #define t2 y2
    378 #define t3 tt
    379 	q = 0;
    380 	y3 = 0;
    381 	bit = 1 << 31;
    382 	EVEN_DOUBLE;
    383 	t3 = bit;
    384 	FPU_SUBS(d3, x3, t3);
    385 	FPU_SUBCS(d2, x2, t2);
    386 	FPU_SUBCS(d1, x1, t1);
    387 	FPU_SUBC(d0, x0, t0);
    388 	ODD_DOUBLE;
    389 	if ((int)d0 >= 0) {
    390 		x0 = d0, x1 = d1, x2 = d2;
    391 		q |= bit;
    392 		y2 |= 1;
    393 	}
    394 	while ((bit >>= 1) != 0) {
    395 		EVEN_DOUBLE;
    396 		t3 = y3 | bit;
    397 		FPU_SUBS(d3, x3, t3);
    398 		FPU_SUBCS(d2, x2, t2);
    399 		FPU_SUBCS(d1, x1, t1);
    400 		FPU_SUBC(d0, x0, t0);
    401 		if ((int)d0 >= 0) {
    402 			x0 = d0, x1 = d1, x2 = d2;
    403 			q |= bit;
    404 			y3 |= bit << 1;
    405 		}
    406 		ODD_DOUBLE;
    407 	}
    408 	x->fp_mant[3] = q;
    409 
    410 	/*
    411 	 * The result, which includes guard and round bits, is exact iff
    412 	 * x is now zero; any nonzero bits in x represent sticky bits.
    413 	 */
    414 	x->fp_sticky = x0 | x1 | x2 | x3;
    415 	DUMPFPN(FPE_REG, x);
    416 	return (x);
    417 }
    418