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