1 /* $NetBSD: fpu_sqrt.c,v 1.14 2022/09/06 23:14:28 rin 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 1.14 2022/09/06 23:14:28 rin Exp $");
49
50 #include <sys/types.h>
51 #if defined(DIAGNOSTIC)||defined(DEBUG)
52 #include <sys/systm.h>
53 #endif
54
55 #include <machine/fpu.h>
56 #include <machine/reg.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 * equivalent 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 *
fpu_sqrt(struct fpemu * fe)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_sqrt:\n"));
215 DUMPFPN(FPE_REG, x);
216 DPRINTF(FPE_REG, ("=>\n"));
217 if (ISNAN(x)) {
218 if (ISSNAN(x))
219 fe->fe_cx |= FPSCR_VXSNAN;
220 DUMPFPN(FPE_REG, x);
221 return (x);
222 }
223 if (ISZERO(x)) {
224 DUMPFPN(FPE_REG, x);
225 return (x);
226 }
227 if (x->fp_sign) {
228 fe->fe_cx |= FPSCR_VXSQRT;
229 return (fpu_newnan(fe));
230 }
231 if (ISINF(x)) {
232 DUMPFPN(FPE_REG, x);
233 return (x);
234 }
235
236 /*
237 * Calculate result exponent. As noted above, this may involve
238 * doubling the mantissa. We will also need to double x each
239 * time around the loop, so we define a macro for this here, and
240 * we break out the multiword mantissa.
241 */
242 #ifdef FPU_SHL1_BY_ADD
243 #define DOUBLE_X { \
244 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
245 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
246 }
247 #else
248 #define DOUBLE_X { \
249 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
250 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
251 }
252 #endif
253 #if (FP_NMANT & 1) != 0
254 # define ODD_DOUBLE DOUBLE_X
255 # define EVEN_DOUBLE /* nothing */
256 #else
257 # define ODD_DOUBLE /* nothing */
258 # define EVEN_DOUBLE DOUBLE_X
259 #endif
260 x0 = x->fp_mant[0];
261 x1 = x->fp_mant[1];
262 x2 = x->fp_mant[2];
263 x3 = x->fp_mant[3];
264 e = x->fp_exp;
265 if (e & 1) /* exponent is odd; use sqrt(2mant) */
266 DOUBLE_X;
267 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
268 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
269
270 /*
271 * Now calculate the mantissa root. Since x is now in [1..4),
272 * we know that the first trip around the loop will definitely
273 * set the top bit in q, so we can do that manually and start
274 * the loop at the next bit down instead. We must be sure to
275 * double x correctly while doing the `known q=1.0'.
276 *
277 * We do this one mantissa-word at a time, as noted above, to
278 * save work. To avoid `(1 << 31) << 1', we also do the top bit
279 * outside of each per-word loop.
280 *
281 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
282 * t3 = y3, t? |= bit' for the appropriate word. Since the bit
283 * is always a `new' one, this means that three of the `t?'s are
284 * just the corresponding `y?'; we use `#define's here for this.
285 * The variable `tt' holds the actual `t?' variable.
286 */
287
288 /* calculate q0 */
289 #define t0 tt
290 bit = FP_1;
291 EVEN_DOUBLE;
292 /* if (x >= (t0 = y0 | bit)) { */ /* always true */
293 q = bit;
294 x0 -= bit;
295 y0 = bit << 1;
296 /* } */
297 ODD_DOUBLE;
298 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
299 EVEN_DOUBLE;
300 t0 = y0 | bit; /* t = y + bit */
301 if (x0 >= t0) { /* if x >= t then */
302 x0 -= t0; /* x -= t */
303 q |= bit; /* q += bit */
304 y0 |= bit << 1; /* y += bit << 1 */
305 }
306 ODD_DOUBLE;
307 }
308 x->fp_mant[0] = q;
309 #undef t0
310
311 /* calculate q1. note (y0&1)==0. */
312 #define t0 y0
313 #define t1 tt
314 q = 0;
315 y1 = 0;
316 bit = 1 << 31;
317 EVEN_DOUBLE;
318 t1 = bit;
319 FPU_SUBS(d1, x1, t1);
320 FPU_SUBC(d0, x0, t0); /* d = x - t */
321 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
322 x0 = d0, x1 = d1; /* x -= t */
323 q = bit; /* q += bit */
324 y0 |= 1; /* y += bit << 1 */
325 }
326 ODD_DOUBLE;
327 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
328 EVEN_DOUBLE; /* as before */
329 t1 = y1 | bit;
330 FPU_SUBS(d1, x1, t1);
331 FPU_SUBC(d0, x0, t0);
332 if ((int)d0 >= 0) {
333 x0 = d0, x1 = d1;
334 q |= bit;
335 y1 |= bit << 1;
336 }
337 ODD_DOUBLE;
338 }
339 x->fp_mant[1] = q;
340 #undef t1
341
342 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
343 #define t1 y1
344 #define t2 tt
345 q = 0;
346 y2 = 0;
347 bit = 1 << 31;
348 EVEN_DOUBLE;
349 t2 = bit;
350 FPU_SUBS(d2, x2, t2);
351 FPU_SUBCS(d1, x1, t1);
352 FPU_SUBC(d0, x0, t0);
353 if ((int)d0 >= 0) {
354 x0 = d0, x1 = d1, x2 = d2;
355 q = bit;
356 y1 |= 1; /* now t1, y1 are set in concrete */
357 }
358 ODD_DOUBLE;
359 while ((bit >>= 1) != 0) {
360 EVEN_DOUBLE;
361 t2 = y2 | bit;
362 FPU_SUBS(d2, x2, t2);
363 FPU_SUBCS(d1, x1, t1);
364 FPU_SUBC(d0, x0, t0);
365 if ((int)d0 >= 0) {
366 x0 = d0, x1 = d1, x2 = d2;
367 q |= bit;
368 y2 |= bit << 1;
369 }
370 ODD_DOUBLE;
371 }
372 x->fp_mant[2] = q;
373 #undef t2
374
375 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */
376 #define t2 y2
377 #define t3 tt
378 q = 0;
379 y3 = 0;
380 bit = 1 << 31;
381 EVEN_DOUBLE;
382 t3 = bit;
383 FPU_SUBS(d3, x3, t3);
384 FPU_SUBCS(d2, x2, t2);
385 FPU_SUBCS(d1, x1, t1);
386 FPU_SUBC(d0, x0, t0);
387 if ((int)d0 >= 0) {
388 x0 = d0, x1 = d1, x2 = d2; x3 = d3;
389 q = bit;
390 y2 |= 1;
391 }
392 ODD_DOUBLE;
393 while ((bit >>= 1) != 0) {
394 EVEN_DOUBLE;
395 t3 = y3 | bit;
396 FPU_SUBS(d3, x3, t3);
397 FPU_SUBCS(d2, x2, t2);
398 FPU_SUBCS(d1, x1, t1);
399 FPU_SUBC(d0, x0, t0);
400 if ((int)d0 >= 0) {
401 x0 = d0, x1 = d1, x2 = d2; x3 = d3;
402 q |= bit;
403 y3 |= bit << 1;
404 }
405 ODD_DOUBLE;
406 }
407 x->fp_mant[3] = q;
408
409 /*
410 * The result, which includes guard and round bits, is exact iff
411 * x is now zero; any nonzero bits in x represent sticky bits.
412 */
413 x->fp_sticky = x0 | x1 | x2 | x3;
414 DUMPFPN(FPE_REG, x);
415 return (x);
416 }
417