1 /* $OpenBSD: fpu_sqrt.c,v 1.7 2024/03/29 21:07:11 miod Exp $ */
2 /* $NetBSD: fpu_sqrt.c,v 1.2 1994/11/20 20:52:46 deraadt Exp $ */
3
4 /*
5 * Copyright (c) 1992, 1993
6 * The Regents of the University of California. All rights reserved.
7 *
8 * This software was developed by the Computer Systems Engineering group
9 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
10 * contributed to Berkeley.
11 *
12 * All advertising materials mentioning features or use of this software
13 * must display the following acknowledgement:
14 * This product includes software developed by the University of
15 * California, Lawrence Berkeley Laboratory.
16 *
17 * Redistribution and use in source and binary forms, with or without
18 * modification, are permitted provided that the following conditions
19 * are met:
20 * 1. Redistributions of source code must retain the above copyright
21 * notice, this list of conditions and the following disclaimer.
22 * 2. Redistributions in binary form must reproduce the above copyright
23 * notice, this list of conditions and the following disclaimer in the
24 * documentation and/or other materials provided with the distribution.
25 * 3. Neither the name of the University nor the names of its contributors
26 * may be used to endorse or promote products derived from this software
27 * without specific prior written permission.
28 *
29 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
30 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
31 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
32 * ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
33 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
34 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
35 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
36 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
37 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
38 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
39 * SUCH DAMAGE.
40 *
41 * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93
42 */
43
44 /*
45 * Perform an FPU square root (return sqrt(x)).
46 */
47
48 #include <sys/types.h>
49
50 #include <machine/reg.h>
51
52 #include <sparc64/fpu/fpu_arith.h>
53 #include <sparc64/fpu/fpu_emu.h>
54
55 /*
56 * Our task is to calculate the square root of a floating point number x0.
57 * This number x normally has the form:
58 *
59 * exp
60 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer)
61 *
62 * This can be left as it stands, or the mantissa can be doubled and the
63 * exponent decremented:
64 *
65 * exp-1
66 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4)
67 *
68 * If the exponent `exp' is even, the square root of the number is best
69 * handled using the first form, and is by definition equal to:
70 *
71 * exp/2
72 * sqrt(x) = sqrt(mant) * 2
73 *
74 * If exp is odd, on the other hand, it is convenient to use the second
75 * form, giving:
76 *
77 * (exp-1)/2
78 * sqrt(x) = sqrt(2 * mant) * 2
79 *
80 * In the first case, we have
81 *
82 * 1 <= mant < 2
83 *
84 * and therefore
85 *
86 * sqrt(1) <= sqrt(mant) < sqrt(2)
87 *
88 * while in the second case we have
89 *
90 * 2 <= 2*mant < 4
91 *
92 * and therefore
93 *
94 * sqrt(2) <= sqrt(2*mant) < sqrt(4)
95 *
96 * so that in any case, we are sure that
97 *
98 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2
99 *
100 * or
101 *
102 * 1 <= sqrt(n * mant) < 2, n = 1 or 2.
103 *
104 * This root is therefore a properly formed mantissa for a floating
105 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2
106 * as above. This leaves us with the problem of finding the square root
107 * of a fixed-point number in the range [1..4).
108 *
109 * Though it may not be instantly obvious, the following square root
110 * algorithm works for any integer x of an even number of bits, provided
111 * that no overflows occur:
112 *
113 * let q = 0
114 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
115 * x *= 2 -- multiply by radix, for next digit
116 * if x >= 2q + 2^k then -- if adding 2^k does not
117 * x -= 2q + 2^k -- exceed the correct root,
118 * q += 2^k -- add 2^k and adjust x
119 * fi
120 * done
121 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x)
122 *
123 * If NBITS is odd (so that k is initially even), we can just add another
124 * zero bit at the top of x. Doing so means that q is not going to acquire
125 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the
126 * final value in x is not needed, or can be off by a factor of 2, this is
127 * equivalent to moving the `x *= 2' step to the bottom of the loop:
128 *
129 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
130 *
131 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
132 * (Since the algorithm is destructive on x, we will call x's initial
133 * value, for which q is some power of two times its square root, x0.)
134 *
135 * If we insert a loop invariant y = 2q, we can then rewrite this using
136 * C notation as:
137 *
138 * q = y = 0; x = x0;
139 * for (k = NBITS; --k >= 0;) {
140 * #if (NBITS is even)
141 * x *= 2;
142 * #endif
143 * t = y + (1 << k);
144 * if (x >= t) {
145 * x -= t;
146 * q += 1 << k;
147 * y += 1 << (k + 1);
148 * }
149 * #if (NBITS is odd)
150 * x *= 2;
151 * #endif
152 * }
153 *
154 * If x0 is fixed point, rather than an integer, we can simply alter the
155 * scale factor between q and sqrt(x0). As it happens, we can easily arrange
156 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
157 *
158 * In our case, however, x0 (and therefore x, y, q, and t) are multiword
159 * integers, which adds some complication. But note that q is built one
160 * bit at a time, from the top down, and is not used itself in the loop
161 * (we use 2q as held in y instead). This means we can build our answer
162 * in an integer, one word at a time, which saves a bit of work. Also,
163 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
164 * `new' bits in y and we can set them with an `or' operation rather than
165 * a full-blown multiword add.
166 *
167 * We are almost done, except for one snag. We must prove that none of our
168 * intermediate calculations can overflow. We know that x0 is in [1..4)
169 * and therefore the square root in q will be in [1..2), but what about x,
170 * y, and t?
171 *
172 * We know that y = 2q at the beginning of each loop. (The relation only
173 * fails temporarily while y and q are being updated.) Since q < 2, y < 4.
174 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
175 * Furthermore, we can prove with a bit of work that x never exceeds y by
176 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as
177 * an exercise to the reader, mostly because I have become tired of working
178 * on this comment.)
179 *
180 * If our floating point mantissas (which are of the form 1.frac) occupy
181 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
182 * In fact, we want even one more bit (for a carry, to avoid compares), or
183 * three extra. There is a comment in fpu_emu.h reminding maintainers of
184 * this, so we have some justification in assuming it.
185 */
186 struct fpn *
fpu_sqrt(struct fpemu * fe)187 fpu_sqrt(struct fpemu *fe)
188 {
189 struct fpn *x = &fe->fe_f1;
190 u_int bit, q, tt;
191 u_int x0, x1, x2, x3;
192 u_int y0, y1, y2, y3;
193 u_int d0, d1, d2, d3;
194 int e;
195
196 /*
197 * Take care of special cases first. In order:
198 *
199 * sqrt(NaN) = NaN
200 * sqrt(+0) = +0
201 * sqrt(-0) = -0
202 * sqrt(x < 0) = NaN (including sqrt(-Inf))
203 * sqrt(+Inf) = +Inf
204 *
205 * Then all that remains are numbers with mantissas in [1..2).
206 */
207 if (ISNAN(x) || ISZERO(x))
208 return (x);
209 if (x->fp_sign)
210 return (fpu_newnan(fe));
211 if (ISINF(x))
212 return (x);
213
214 /*
215 * Calculate result exponent. As noted above, this may involve
216 * doubling the mantissa. We will also need to double x each
217 * time around the loop, so we define a macro for this here, and
218 * we break out the multiword mantissa.
219 */
220 #ifdef FPU_SHL1_BY_ADD
221 #define DOUBLE_X { \
222 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
223 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
224 }
225 #else
226 #define DOUBLE_X { \
227 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
228 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
229 }
230 #endif
231 #if (FP_NMANT & 1) != 0
232 # define ODD_DOUBLE DOUBLE_X
233 # define EVEN_DOUBLE /* nothing */
234 #else
235 # define ODD_DOUBLE /* nothing */
236 # define EVEN_DOUBLE DOUBLE_X
237 #endif
238 x0 = x->fp_mant[0];
239 x1 = x->fp_mant[1];
240 x2 = x->fp_mant[2];
241 x3 = x->fp_mant[3];
242 e = x->fp_exp;
243 if (e & 1) /* exponent is odd; use sqrt(2mant) */
244 DOUBLE_X;
245 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
246 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
247
248 /*
249 * Now calculate the mantissa root. Since x is now in [1..4),
250 * we know that the first trip around the loop will definitely
251 * set the top bit in q, so we can do that manually and start
252 * the loop at the next bit down instead. We must be sure to
253 * double x correctly while doing the `known q=1.0'.
254 *
255 * We do this one mantissa-word at a time, as noted above, to
256 * save work. To avoid `(1U << 31) << 1', we also do the top bit
257 * outside of each per-word loop.
258 *
259 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
260 * t3 = y3, t? |= bit' for the appropriate word. Since the bit
261 * is always a `new' one, this means that three of the `t?'s are
262 * just the corresponding `y?'; we use `#define's here for this.
263 * The variable `tt' holds the actual `t?' variable.
264 */
265
266 /* calculate q0 */
267 #define t0 tt
268 bit = FP_1;
269 EVEN_DOUBLE;
270 /* if (x >= (t0 = y0 | bit)) { */ /* always true */
271 q = bit;
272 x0 -= bit;
273 y0 = bit << 1;
274 /* } */
275 ODD_DOUBLE;
276 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
277 EVEN_DOUBLE;
278 t0 = y0 | bit; /* t = y + bit */
279 if (x0 >= t0) { /* if x >= t then */
280 x0 -= t0; /* x -= t */
281 q |= bit; /* q += bit */
282 y0 |= bit << 1; /* y += bit << 1 */
283 }
284 ODD_DOUBLE;
285 }
286 x->fp_mant[0] = q;
287 #undef t0
288
289 /* calculate q1. note (y0&1)==0. */
290 #define t0 y0
291 #define t1 tt
292 q = 0;
293 y1 = 0;
294 bit = 1U << 31;
295 EVEN_DOUBLE;
296 t1 = bit;
297 FPU_SUBS(d1, x1, t1);
298 FPU_SUBC(d0, x0, t0); /* d = x - t */
299 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
300 x0 = d0, x1 = d1; /* x -= t */
301 q = bit; /* q += bit */
302 y0 |= 1; /* y += bit << 1 */
303 }
304 ODD_DOUBLE;
305 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
306 EVEN_DOUBLE; /* as before */
307 t1 = y1 | bit;
308 FPU_SUBS(d1, x1, t1);
309 FPU_SUBC(d0, x0, t0);
310 if ((int)d0 >= 0) {
311 x0 = d0, x1 = d1;
312 q |= bit;
313 y1 |= bit << 1;
314 }
315 ODD_DOUBLE;
316 }
317 x->fp_mant[1] = q;
318 #undef t1
319
320 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
321 #define t1 y1
322 #define t2 tt
323 q = 0;
324 y2 = 0;
325 bit = 1U << 31;
326 EVEN_DOUBLE;
327 t2 = bit;
328 FPU_SUBS(d2, x2, t2);
329 FPU_SUBCS(d1, x1, t1);
330 FPU_SUBC(d0, x0, t0);
331 if ((int)d0 >= 0) {
332 x0 = d0, x1 = d1, x2 = d2;
333 q |= bit;
334 y1 |= 1; /* now t1, y1 are set in concrete */
335 }
336 ODD_DOUBLE;
337 while ((bit >>= 1) != 0) {
338 EVEN_DOUBLE;
339 t2 = y2 | bit;
340 FPU_SUBS(d2, x2, t2);
341 FPU_SUBCS(d1, x1, t1);
342 FPU_SUBC(d0, x0, t0);
343 if ((int)d0 >= 0) {
344 x0 = d0, x1 = d1, x2 = d2;
345 q |= bit;
346 y2 |= bit << 1;
347 }
348 ODD_DOUBLE;
349 }
350 x->fp_mant[2] = q;
351 #undef t2
352
353 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */
354 #define t2 y2
355 #define t3 tt
356 q = 0;
357 y3 = 0;
358 bit = 1U << 31;
359 EVEN_DOUBLE;
360 t3 = bit;
361 FPU_SUBS(d3, x3, t3);
362 FPU_SUBCS(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, x3 = d3;
367 q |= bit;
368 y2 |= 1;
369 }
370 ODD_DOUBLE;
371 while ((bit >>= 1) != 0) {
372 EVEN_DOUBLE;
373 t3 = y3 | bit;
374 FPU_SUBS(d3, x3, t3);
375 FPU_SUBCS(d2, x2, t2);
376 FPU_SUBCS(d1, x1, t1);
377 FPU_SUBC(d0, x0, t0);
378 if ((int)d0 >= 0) {
379 x0 = d0, x1 = d1, x2 = d2, x3 = d3;
380 q |= bit;
381 y3 |= bit << 1;
382 }
383 ODD_DOUBLE;
384 }
385 x->fp_mant[3] = q;
386
387 /*
388 * The result, which includes guard and round bits, is exact iff
389 * x is now zero; any nonzero bits in x represent sticky bits.
390 */
391 x->fp_sticky = x0 | x1 | x2 | x3;
392 return (x);
393 }
394