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fpu_sqrt.c revision 1.1
      1 /*	$NetBSD: fpu_sqrt.c,v 1.1 2001/06/13 06:01:47 simonb 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 #if defined(DIAGNOSTIC)||defined(DEBUG)
     53 #include <sys/systm.h>
     54 #endif
     55 
     56 #include <machine/reg.h>
     57 #include <machine/fpu.h>
     58 
     59 #include <powerpc/fpu/fpu_arith.h>
     60 #include <powerpc/fpu/fpu_emu.h>
     61 
     62 /*
     63  * Our task is to calculate the square root of a floating point number x0.
     64  * This number x normally has the form:
     65  *
     66  *		    exp
     67  *	x = mant * 2		(where 1 <= mant < 2 and exp is an integer)
     68  *
     69  * This can be left as it stands, or the mantissa can be doubled and the
     70  * exponent decremented:
     71  *
     72  *			  exp-1
     73  *	x = (2 * mant) * 2	(where 2 <= 2 * mant < 4)
     74  *
     75  * If the exponent `exp' is even, the square root of the number is best
     76  * handled using the first form, and is by definition equal to:
     77  *
     78  *				exp/2
     79  *	sqrt(x) = sqrt(mant) * 2
     80  *
     81  * If exp is odd, on the other hand, it is convenient to use the second
     82  * form, giving:
     83  *
     84  *				    (exp-1)/2
     85  *	sqrt(x) = sqrt(2 * mant) * 2
     86  *
     87  * In the first case, we have
     88  *
     89  *	1 <= mant < 2
     90  *
     91  * and therefore
     92  *
     93  *	sqrt(1) <= sqrt(mant) < sqrt(2)
     94  *
     95  * while in the second case we have
     96  *
     97  *	2 <= 2*mant < 4
     98  *
     99  * and therefore
    100  *
    101  *	sqrt(2) <= sqrt(2*mant) < sqrt(4)
    102  *
    103  * so that in any case, we are sure that
    104  *
    105  *	sqrt(1) <= sqrt(n * mant) < sqrt(4),	n = 1 or 2
    106  *
    107  * or
    108  *
    109  *	1 <= sqrt(n * mant) < 2,		n = 1 or 2.
    110  *
    111  * This root is therefore a properly formed mantissa for a floating
    112  * point number.  The exponent of sqrt(x) is either exp/2 or (exp-1)/2
    113  * as above.  This leaves us with the problem of finding the square root
    114  * of a fixed-point number in the range [1..4).
    115  *
    116  * Though it may not be instantly obvious, the following square root
    117  * algorithm works for any integer x of an even number of bits, provided
    118  * that no overflows occur:
    119  *
    120  *	let q = 0
    121  *	for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
    122  *		x *= 2			-- multiply by radix, for next digit
    123  *		if x >= 2q + 2^k then	-- if adding 2^k does not
    124  *			x -= 2q + 2^k	-- exceed the correct root,
    125  *			q += 2^k	-- add 2^k and adjust x
    126  *		fi
    127  *	done
    128  *	sqrt = q / 2^(NBITS/2)		-- (and any remainder is in x)
    129  *
    130  * If NBITS is odd (so that k is initially even), we can just add another
    131  * zero bit at the top of x.  Doing so means that q is not going to acquire
    132  * a 1 bit in the first trip around the loop (since x0 < 2^NBITS).  If the
    133  * final value in x is not needed, or can be off by a factor of 2, this is
    134  * equivalant to moving the `x *= 2' step to the bottom of the loop:
    135  *
    136  *	for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
    137  *
    138  * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
    139  * (Since the algorithm is destructive on x, we will call x's initial
    140  * value, for which q is some power of two times its square root, x0.)
    141  *
    142  * If we insert a loop invariant y = 2q, we can then rewrite this using
    143  * C notation as:
    144  *
    145  *	q = y = 0; x = x0;
    146  *	for (k = NBITS; --k >= 0;) {
    147  * #if (NBITS is even)
    148  *		x *= 2;
    149  * #endif
    150  *		t = y + (1 << k);
    151  *		if (x >= t) {
    152  *			x -= t;
    153  *			q += 1 << k;
    154  *			y += 1 << (k + 1);
    155  *		}
    156  * #if (NBITS is odd)
    157  *		x *= 2;
    158  * #endif
    159  *	}
    160  *
    161  * If x0 is fixed point, rather than an integer, we can simply alter the
    162  * scale factor between q and sqrt(x0).  As it happens, we can easily arrange
    163  * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
    164  *
    165  * In our case, however, x0 (and therefore x, y, q, and t) are multiword
    166  * integers, which adds some complication.  But note that q is built one
    167  * bit at a time, from the top down, and is not used itself in the loop
    168  * (we use 2q as held in y instead).  This means we can build our answer
    169  * in an integer, one word at a time, which saves a bit of work.  Also,
    170  * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
    171  * `new' bits in y and we can set them with an `or' operation rather than
    172  * a full-blown multiword add.
    173  *
    174  * We are almost done, except for one snag.  We must prove that none of our
    175  * intermediate calculations can overflow.  We know that x0 is in [1..4)
    176  * and therefore the square root in q will be in [1..2), but what about x,
    177  * y, and t?
    178  *
    179  * We know that y = 2q at the beginning of each loop.  (The relation only
    180  * fails temporarily while y and q are being updated.)  Since q < 2, y < 4.
    181  * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
    182  * Furthermore, we can prove with a bit of work that x never exceeds y by
    183  * more than 2, so that even after doubling, 0 <= x < 8.  (This is left as
    184  * an exercise to the reader, mostly because I have become tired of working
    185  * on this comment.)
    186  *
    187  * If our floating point mantissas (which are of the form 1.frac) occupy
    188  * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
    189  * In fact, we want even one more bit (for a carry, to avoid compares), or
    190  * three extra.  There is a comment in fpu_emu.h reminding maintainers of
    191  * this, so we have some justification in assuming it.
    192  */
    193 struct fpn *
    194 fpu_sqrt(struct fpemu *fe)
    195 {
    196 	struct fpn *x = &fe->fe_f1;
    197 	u_int bit, q, tt;
    198 	u_int x0, x1, x2, x3;
    199 	u_int y0, y1, y2, y3;
    200 	u_int d0, d1, d2, d3;
    201 	int e;
    202 	FPU_DECL_CARRY;
    203 
    204 	/*
    205 	 * Take care of special cases first.  In order:
    206 	 *
    207 	 *	sqrt(NaN) = NaN
    208 	 *	sqrt(+0) = +0
    209 	 *	sqrt(-0) = -0
    210 	 *	sqrt(x < 0) = NaN	(including sqrt(-Inf))
    211 	 *	sqrt(+Inf) = +Inf
    212 	 *
    213 	 * Then all that remains are numbers with mantissas in [1..2).
    214 	 */
    215 	DPRINTF(FPE_REG, ("fpu_sqer:\n"));
    216 	DUMPFPN(FPE_REG, x);
    217 	DPRINTF(FPE_REG, ("=>\n"));
    218 	if (ISNAN(x)) {
    219 		fe->fe_cx |= FPSCR_VXSNAN;
    220 		DUMPFPN(FPE_REG, x);
    221 		return (x);
    222 	}
    223 	if (ISZERO(x)) {
    224 		fe->fe_cx |= FPSCR_ZX;
    225 		x->fp_class = FPC_INF;
    226 		DUMPFPN(FPE_REG, x);
    227 		return (x);
    228 	}
    229 	if (x->fp_sign) {
    230 		return (fpu_newnan(fe));
    231 	}
    232 	if (ISINF(x)) {
    233 		fe->fe_cx |= FPSCR_VXSQRT;
    234 		DUMPFPN(FPE_REG, 0);
    235 		return (0);
    236 	}
    237 
    238 	/*
    239 	 * Calculate result exponent.  As noted above, this may involve
    240 	 * doubling the mantissa.  We will also need to double x each
    241 	 * time around the loop, so we define a macro for this here, and
    242 	 * we break out the multiword mantissa.
    243 	 */
    244 #ifdef FPU_SHL1_BY_ADD
    245 #define	DOUBLE_X { \
    246 	FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
    247 	FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
    248 }
    249 #else
    250 #define	DOUBLE_X { \
    251 	x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
    252 	x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
    253 }
    254 #endif
    255 #if (FP_NMANT & 1) != 0
    256 # define ODD_DOUBLE	DOUBLE_X
    257 # define EVEN_DOUBLE	/* nothing */
    258 #else
    259 # define ODD_DOUBLE	/* nothing */
    260 # define EVEN_DOUBLE	DOUBLE_X
    261 #endif
    262 	x0 = x->fp_mant[0];
    263 	x1 = x->fp_mant[1];
    264 	x2 = x->fp_mant[2];
    265 	x3 = x->fp_mant[3];
    266 	e = x->fp_exp;
    267 	if (e & 1)		/* exponent is odd; use sqrt(2mant) */
    268 		DOUBLE_X;
    269 	/* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
    270 	x->fp_exp = e >> 1;	/* calculates (e&1 ? (e-1)/2 : e/2 */
    271 
    272 	/*
    273 	 * Now calculate the mantissa root.  Since x is now in [1..4),
    274 	 * we know that the first trip around the loop will definitely
    275 	 * set the top bit in q, so we can do that manually and start
    276 	 * the loop at the next bit down instead.  We must be sure to
    277 	 * double x correctly while doing the `known q=1.0'.
    278 	 *
    279 	 * We do this one mantissa-word at a time, as noted above, to
    280 	 * save work.  To avoid `(1 << 31) << 1', we also do the top bit
    281 	 * outside of each per-word loop.
    282 	 *
    283 	 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
    284 	 * t3 = y3, t? |= bit' for the appropriate word.  Since the bit
    285 	 * is always a `new' one, this means that three of the `t?'s are
    286 	 * just the corresponding `y?'; we use `#define's here for this.
    287 	 * The variable `tt' holds the actual `t?' variable.
    288 	 */
    289 
    290 	/* calculate q0 */
    291 #define	t0 tt
    292 	bit = FP_1;
    293 	EVEN_DOUBLE;
    294 	/* if (x >= (t0 = y0 | bit)) { */	/* always true */
    295 		q = bit;
    296 		x0 -= bit;
    297 		y0 = bit << 1;
    298 	/* } */
    299 	ODD_DOUBLE;
    300 	while ((bit >>= 1) != 0) {	/* for remaining bits in q0 */
    301 		EVEN_DOUBLE;
    302 		t0 = y0 | bit;		/* t = y + bit */
    303 		if (x0 >= t0) {		/* if x >= t then */
    304 			x0 -= t0;	/*	x -= t */
    305 			q |= bit;	/*	q += bit */
    306 			y0 |= bit << 1;	/*	y += bit << 1 */
    307 		}
    308 		ODD_DOUBLE;
    309 	}
    310 	x->fp_mant[0] = q;
    311 #undef t0
    312 
    313 	/* calculate q1.  note (y0&1)==0. */
    314 #define t0 y0
    315 #define t1 tt
    316 	q = 0;
    317 	y1 = 0;
    318 	bit = 1 << 31;
    319 	EVEN_DOUBLE;
    320 	t1 = bit;
    321 	FPU_SUBS(d1, x1, t1);
    322 	FPU_SUBC(d0, x0, t0);		/* d = x - t */
    323 	if ((int)d0 >= 0) {		/* if d >= 0 (i.e., x >= t) then */
    324 		x0 = d0, x1 = d1;	/*	x -= t */
    325 		q = bit;		/*	q += bit */
    326 		y0 |= 1;		/*	y += bit << 1 */
    327 	}
    328 	ODD_DOUBLE;
    329 	while ((bit >>= 1) != 0) {	/* for remaining bits in q1 */
    330 		EVEN_DOUBLE;		/* as before */
    331 		t1 = y1 | bit;
    332 		FPU_SUBS(d1, x1, t1);
    333 		FPU_SUBC(d0, x0, t0);
    334 		if ((int)d0 >= 0) {
    335 			x0 = d0, x1 = d1;
    336 			q |= bit;
    337 			y1 |= bit << 1;
    338 		}
    339 		ODD_DOUBLE;
    340 	}
    341 	x->fp_mant[1] = q;
    342 #undef t1
    343 
    344 	/* calculate q2.  note (y1&1)==0; y0 (aka t0) is fixed. */
    345 #define t1 y1
    346 #define t2 tt
    347 	q = 0;
    348 	y2 = 0;
    349 	bit = 1 << 31;
    350 	EVEN_DOUBLE;
    351 	t2 = bit;
    352 	FPU_SUBS(d2, x2, t2);
    353 	FPU_SUBCS(d1, x1, t1);
    354 	FPU_SUBC(d0, x0, t0);
    355 	if ((int)d0 >= 0) {
    356 		x0 = d0, x1 = d1, x2 = d2;
    357 		q |= bit;
    358 		y1 |= 1;		/* now t1, y1 are set in concrete */
    359 	}
    360 	ODD_DOUBLE;
    361 	while ((bit >>= 1) != 0) {
    362 		EVEN_DOUBLE;
    363 		t2 = y2 | bit;
    364 		FPU_SUBS(d2, x2, t2);
    365 		FPU_SUBCS(d1, x1, t1);
    366 		FPU_SUBC(d0, x0, t0);
    367 		if ((int)d0 >= 0) {
    368 			x0 = d0, x1 = d1, x2 = d2;
    369 			q |= bit;
    370 			y2 |= bit << 1;
    371 		}
    372 		ODD_DOUBLE;
    373 	}
    374 	x->fp_mant[2] = q;
    375 #undef t2
    376 
    377 	/* calculate q3.  y0, t0, y1, t1 all fixed; y2, t2, almost done. */
    378 #define t2 y2
    379 #define t3 tt
    380 	q = 0;
    381 	y3 = 0;
    382 	bit = 1 << 31;
    383 	EVEN_DOUBLE;
    384 	t3 = bit;
    385 	FPU_SUBS(d3, x3, t3);
    386 	FPU_SUBCS(d2, x2, t2);
    387 	FPU_SUBCS(d1, x1, t1);
    388 	FPU_SUBC(d0, x0, t0);
    389 	ODD_DOUBLE;
    390 	if ((int)d0 >= 0) {
    391 		x0 = d0, x1 = d1, x2 = d2;
    392 		q |= bit;
    393 		y2 |= 1;
    394 	}
    395 	while ((bit >>= 1) != 0) {
    396 		EVEN_DOUBLE;
    397 		t3 = y3 | bit;
    398 		FPU_SUBS(d3, x3, t3);
    399 		FPU_SUBCS(d2, x2, t2);
    400 		FPU_SUBCS(d1, x1, t1);
    401 		FPU_SUBC(d0, x0, t0);
    402 		if ((int)d0 >= 0) {
    403 			x0 = d0, x1 = d1, x2 = d2;
    404 			q |= bit;
    405 			y3 |= bit << 1;
    406 		}
    407 		ODD_DOUBLE;
    408 	}
    409 	x->fp_mant[3] = q;
    410 
    411 	/*
    412 	 * The result, which includes guard and round bits, is exact iff
    413 	 * x is now zero; any nonzero bits in x represent sticky bits.
    414 	 */
    415 	x->fp_sticky = x0 | x1 | x2 | x3;
    416 	DUMPFPN(FPE_REG, x);
    417 	return (x);
    418 }
    419