1 /* $NetBSD: fpu_sqrt.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */
2
3 /*-
4 * SPDX-License-Identifier: BSD-3-Clause
5 *
6 * Copyright (c) 1992, 1993
7 * The Regents of the University of California. All rights reserved.
8 *
9 * This software was developed by the Computer Systems Engineering group
10 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
11 * contributed to Berkeley.
12 *
13 * All advertising materials mentioning features or use of this software
14 * must display the following acknowledgement:
15 * This product includes software developed by the University of
16 * California, Lawrence Berkeley Laboratory.
17 *
18 * Redistribution and use in source and binary forms, with or without
19 * modification, are permitted provided that the following conditions
20 * are met:
21 * 1. Redistributions of source code must retain the above copyright
22 * notice, this list of conditions and the following disclaimer.
23 * 2. Redistributions in binary form must reproduce the above copyright
24 * notice, this list of conditions and the following disclaimer in the
25 * documentation and/or other materials provided with the distribution.
26 * 3. Neither the name of the University nor the names of its contributors
27 * may be used to endorse or promote products derived from this software
28 * without specific prior written permission.
29 *
30 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
31 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
32 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
33 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
34 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
35 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
36 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
37 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
38 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
39 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
40 * SUCH DAMAGE.
41 *
42 * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93
43 */
44
45 /*
46 * Perform an FPU square root (return sqrt(x)).
47 */
48
49 #include <sys/cdefs.h>
50 #include <sys/types.h>
51 #include <sys/systm.h>
52
53 #include <machine/fpu.h>
54
55 #include <powerpc/fpu/fpu_arith.h>
56 #include <powerpc/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 *
fpu_sqrt(struct fpemu * fe)190 fpu_sqrt(struct fpemu *fe)
191 {
192 struct fpn *x = &fe->fe_f1;
193 u_int bit, q, tt;
194 u_int x0, x1, x2, x3;
195 u_int y0, y1, y2, y3;
196 u_int d0, d1, d2, d3;
197 int e;
198 FPU_DECL_CARRY;
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 DPRINTF(FPE_REG, ("fpu_sqer:\n"));
212 DUMPFPN(FPE_REG, x);
213 DPRINTF(FPE_REG, ("=>\n"));
214 if (ISNAN(x)) {
215 fe->fe_cx |= FPSCR_VXSNAN;
216 DUMPFPN(FPE_REG, x);
217 return (x);
218 }
219 if (ISZERO(x)) {
220 fe->fe_cx |= FPSCR_ZX;
221 x->fp_class = FPC_INF;
222 DUMPFPN(FPE_REG, x);
223 return (x);
224 }
225 if (x->fp_sign) {
226 fe->fe_cx |= FPSCR_VXSQRT;
227 return (fpu_newnan(fe));
228 }
229 if (ISINF(x)) {
230 DUMPFPN(FPE_REG, x);
231 return (x);
232 }
233
234 /*
235 * Calculate result exponent. As noted above, this may involve
236 * doubling the mantissa. We will also need to double x each
237 * time around the loop, so we define a macro for this here, and
238 * we break out the multiword mantissa.
239 */
240 #ifdef FPU_SHL1_BY_ADD
241 #define DOUBLE_X { \
242 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
243 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
244 }
245 #else
246 #define DOUBLE_X { \
247 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
248 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
249 }
250 #endif
251 #if (FP_NMANT & 1) != 0
252 # define ODD_DOUBLE DOUBLE_X
253 # define EVEN_DOUBLE /* nothing */
254 #else
255 # define ODD_DOUBLE /* nothing */
256 # define EVEN_DOUBLE DOUBLE_X
257 #endif
258 x0 = x->fp_mant[0];
259 x1 = x->fp_mant[1];
260 x2 = x->fp_mant[2];
261 x3 = x->fp_mant[3];
262 e = x->fp_exp;
263 if (e & 1) /* exponent is odd; use sqrt(2mant) */
264 DOUBLE_X;
265 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
266 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
267
268 /*
269 * Now calculate the mantissa root. Since x is now in [1..4),
270 * we know that the first trip around the loop will definitely
271 * set the top bit in q, so we can do that manually and start
272 * the loop at the next bit down instead. We must be sure to
273 * double x correctly while doing the `known q=1.0'.
274 *
275 * We do this one mantissa-word at a time, as noted above, to
276 * save work. To avoid `(1U << 31) << 1', we also do the top bit
277 * outside of each per-word loop.
278 *
279 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
280 * t3 = y3, t? |= bit' for the appropriate word. Since the bit
281 * is always a `new' one, this means that three of the `t?'s are
282 * just the corresponding `y?'; we use `#define's here for this.
283 * The variable `tt' holds the actual `t?' variable.
284 */
285
286 /* calculate q0 */
287 #define t0 tt
288 bit = FP_1;
289 EVEN_DOUBLE;
290 /* if (x >= (t0 = y0 | bit)) { */ /* always true */
291 q = bit;
292 x0 -= bit;
293 y0 = bit << 1;
294 /* } */
295 ODD_DOUBLE;
296 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
297 EVEN_DOUBLE;
298 t0 = y0 | bit; /* t = y + bit */
299 if (x0 >= t0) { /* if x >= t then */
300 x0 -= t0; /* x -= t */
301 q |= bit; /* q += bit */
302 y0 |= bit << 1; /* y += bit << 1 */
303 }
304 ODD_DOUBLE;
305 }
306 x->fp_mant[0] = q;
307 #undef t0
308
309 /* calculate q1. note (y0&1)==0. */
310 #define t0 y0
311 #define t1 tt
312 q = 0;
313 y1 = 0;
314 bit = 1 << 31;
315 EVEN_DOUBLE;
316 t1 = bit;
317 FPU_SUBS(d1, x1, t1);
318 FPU_SUBC(d0, x0, t0); /* d = x - t */
319 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
320 x0 = d0, x1 = d1; /* x -= t */
321 q = bit; /* q += bit */
322 y0 |= 1; /* y += bit << 1 */
323 }
324 ODD_DOUBLE;
325 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
326 EVEN_DOUBLE; /* as before */
327 t1 = y1 | bit;
328 FPU_SUBS(d1, x1, t1);
329 FPU_SUBC(d0, x0, t0);
330 if ((int)d0 >= 0) {
331 x0 = d0, x1 = d1;
332 q |= bit;
333 y1 |= bit << 1;
334 }
335 ODD_DOUBLE;
336 }
337 x->fp_mant[1] = q;
338 #undef t1
339
340 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
341 #define t1 y1
342 #define t2 tt
343 q = 0;
344 y2 = 0;
345 bit = 1 << 31;
346 EVEN_DOUBLE;
347 t2 = bit;
348 FPU_SUBS(d2, x2, t2);
349 FPU_SUBCS(d1, x1, t1);
350 FPU_SUBC(d0, x0, t0);
351 if ((int)d0 >= 0) {
352 x0 = d0, x1 = d1, x2 = d2;
353 q = bit;
354 y1 |= 1; /* now t1, y1 are set in concrete */
355 }
356 ODD_DOUBLE;
357 while ((bit >>= 1) != 0) {
358 EVEN_DOUBLE;
359 t2 = y2 | bit;
360 FPU_SUBS(d2, x2, t2);
361 FPU_SUBCS(d1, x1, t1);
362 FPU_SUBC(d0, x0, t0);
363 if ((int)d0 >= 0) {
364 x0 = d0, x1 = d1, x2 = d2;
365 q |= bit;
366 y2 |= bit << 1;
367 }
368 ODD_DOUBLE;
369 }
370 x->fp_mant[2] = q;
371 #undef t2
372
373 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */
374 #define t2 y2
375 #define t3 tt
376 q = 0;
377 y3 = 0;
378 bit = 1 << 31;
379 EVEN_DOUBLE;
380 t3 = bit;
381 FPU_SUBS(d3, x3, t3);
382 FPU_SUBCS(d2, x2, t2);
383 FPU_SUBCS(d1, x1, t1);
384 FPU_SUBC(d0, x0, t0);
385 if ((int)d0 >= 0) {
386 x0 = d0, x1 = d1, x2 = d2; x3 = d3;
387 q = bit;
388 y2 |= 1;
389 }
390 ODD_DOUBLE;
391 while ((bit >>= 1) != 0) {
392 EVEN_DOUBLE;
393 t3 = y3 | bit;
394 FPU_SUBS(d3, x3, t3);
395 FPU_SUBCS(d2, x2, t2);
396 FPU_SUBCS(d1, x1, t1);
397 FPU_SUBC(d0, x0, t0);
398 if ((int)d0 >= 0) {
399 x0 = d0, x1 = d1, x2 = d2; x3 = d3;
400 q |= bit;
401 y3 |= bit << 1;
402 }
403 ODD_DOUBLE;
404 }
405 x->fp_mant[3] = q;
406
407 /*
408 * The result, which includes guard and round bits, is exact iff
409 * x is now zero; any nonzero bits in x represent sticky bits.
410 */
411 x->fp_sticky = x0 | x1 | x2 | x3;
412 DUMPFPN(FPE_REG, x);
413 return (x);
414 }
415