1 /* $NetBSD: fpu_sqrt.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */
4 * Copyright (c) 1992, 1993
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7 * This software was developed by the Computer Systems Engineering group
8 * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
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40 * @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93
44 * Perform an FPU square root (return sqrt(x)).
47 #include <sys/cdefs.h>
48 __FBSDID("$FreeBSD$");
50 #include <sys/types.h>
51 #include <sys/systm.h>
53 #include <machine/fpu.h>
54 #include <machine/reg.h>
56 #include <powerpc/fpu/fpu_arith.h>
57 #include <powerpc/fpu/fpu_emu.h>
60 * Our task is to calculate the square root of a floating point number x0.
61 * This number x normally has the form:
64 * x = mant * 2 (where 1 <= mant < 2 and exp is an integer)
66 * This can be left as it stands, or the mantissa can be doubled and the
67 * exponent decremented:
70 * x = (2 * mant) * 2 (where 2 <= 2 * mant < 4)
72 * If the exponent `exp' is even, the square root of the number is best
73 * handled using the first form, and is by definition equal to:
76 * sqrt(x) = sqrt(mant) * 2
78 * If exp is odd, on the other hand, it is convenient to use the second
82 * sqrt(x) = sqrt(2 * mant) * 2
84 * In the first case, we have
90 * sqrt(1) <= sqrt(mant) < sqrt(2)
92 * while in the second case we have
98 * sqrt(2) <= sqrt(2*mant) < sqrt(4)
100 * so that in any case, we are sure that
102 * sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2
106 * 1 <= sqrt(n * mant) < 2, n = 1 or 2.
108 * This root is therefore a properly formed mantissa for a floating
109 * point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2
110 * as above. This leaves us with the problem of finding the square root
111 * of a fixed-point number in the range [1..4).
113 * Though it may not be instantly obvious, the following square root
114 * algorithm works for any integer x of an even number of bits, provided
115 * that no overflows occur:
118 * for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
119 * x *= 2 -- multiply by radix, for next digit
120 * if x >= 2q + 2^k then -- if adding 2^k does not
121 * x -= 2q + 2^k -- exceed the correct root,
122 * q += 2^k -- add 2^k and adjust x
125 * sqrt = q / 2^(NBITS/2) -- (and any remainder is in x)
127 * If NBITS is odd (so that k is initially even), we can just add another
128 * zero bit at the top of x. Doing so means that q is not going to acquire
129 * a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the
130 * final value in x is not needed, or can be off by a factor of 2, this is
131 * equivalant to moving the `x *= 2' step to the bottom of the loop:
133 * for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
135 * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
136 * (Since the algorithm is destructive on x, we will call x's initial
137 * value, for which q is some power of two times its square root, x0.)
139 * If we insert a loop invariant y = 2q, we can then rewrite this using
143 * for (k = NBITS; --k >= 0;) {
144 * #if (NBITS is even)
158 * If x0 is fixed point, rather than an integer, we can simply alter the
159 * scale factor between q and sqrt(x0). As it happens, we can easily arrange
160 * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
162 * In our case, however, x0 (and therefore x, y, q, and t) are multiword
163 * integers, which adds some complication. But note that q is built one
164 * bit at a time, from the top down, and is not used itself in the loop
165 * (we use 2q as held in y instead). This means we can build our answer
166 * in an integer, one word at a time, which saves a bit of work. Also,
167 * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
168 * `new' bits in y and we can set them with an `or' operation rather than
169 * a full-blown multiword add.
171 * We are almost done, except for one snag. We must prove that none of our
172 * intermediate calculations can overflow. We know that x0 is in [1..4)
173 * and therefore the square root in q will be in [1..2), but what about x,
176 * We know that y = 2q at the beginning of each loop. (The relation only
177 * fails temporarily while y and q are being updated.) Since q < 2, y < 4.
178 * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
179 * Furthermore, we can prove with a bit of work that x never exceeds y by
180 * more than 2, so that even after doubling, 0 <= x < 8. (This is left as
181 * an exercise to the reader, mostly because I have become tired of working
184 * If our floating point mantissas (which are of the form 1.frac) occupy
185 * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
186 * In fact, we want even one more bit (for a carry, to avoid compares), or
187 * three extra. There is a comment in fpu_emu.h reminding maintainers of
188 * this, so we have some justification in assuming it.
191 fpu_sqrt(struct fpemu *fe)
193 struct fpn *x = &fe->fe_f1;
195 u_int x0, x1, x2, x3;
196 u_int y0, y1, y2, y3;
197 u_int d0, d1, d2, d3;
202 * Take care of special cases first. In order:
207 * sqrt(x < 0) = NaN (including sqrt(-Inf))
210 * Then all that remains are numbers with mantissas in [1..2).
212 DPRINTF(FPE_REG, ("fpu_sqer:\n"));
214 DPRINTF(FPE_REG, ("=>\n"));
216 fe->fe_cx |= FPSCR_VXSNAN;
221 fe->fe_cx |= FPSCR_ZX;
222 x->fp_class = FPC_INF;
227 return (fpu_newnan(fe));
230 fe->fe_cx |= FPSCR_VXSQRT;
236 * Calculate result exponent. As noted above, this may involve
237 * doubling the mantissa. We will also need to double x each
238 * time around the loop, so we define a macro for this here, and
239 * we break out the multiword mantissa.
241 #ifdef FPU_SHL1_BY_ADD
243 FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
244 FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
248 x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
249 x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
252 #if (FP_NMANT & 1) != 0
253 # define ODD_DOUBLE DOUBLE_X
254 # define EVEN_DOUBLE /* nothing */
256 # define ODD_DOUBLE /* nothing */
257 # define EVEN_DOUBLE DOUBLE_X
264 if (e & 1) /* exponent is odd; use sqrt(2mant) */
266 /* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
267 x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
270 * Now calculate the mantissa root. Since x is now in [1..4),
271 * we know that the first trip around the loop will definitely
272 * set the top bit in q, so we can do that manually and start
273 * the loop at the next bit down instead. We must be sure to
274 * double x correctly while doing the `known q=1.0'.
276 * We do this one mantissa-word at a time, as noted above, to
277 * save work. To avoid `(1U << 31) << 1', we also do the top bit
278 * outside of each per-word loop.
280 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
281 * t3 = y3, t? |= bit' for the appropriate word. Since the bit
282 * is always a `new' one, this means that three of the `t?'s are
283 * just the corresponding `y?'; we use `#define's here for this.
284 * The variable `tt' holds the actual `t?' variable.
291 /* if (x >= (t0 = y0 | bit)) { */ /* always true */
297 while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
299 t0 = y0 | bit; /* t = y + bit */
300 if (x0 >= t0) { /* if x >= t then */
301 x0 -= t0; /* x -= t */
302 q |= bit; /* q += bit */
303 y0 |= bit << 1; /* y += bit << 1 */
310 /* calculate q1. note (y0&1)==0. */
318 FPU_SUBS(d1, x1, t1);
319 FPU_SUBC(d0, x0, t0); /* d = x - t */
320 if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
321 x0 = d0, x1 = d1; /* x -= t */
322 q = bit; /* q += bit */
323 y0 |= 1; /* y += bit << 1 */
326 while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
327 EVEN_DOUBLE; /* as before */
329 FPU_SUBS(d1, x1, t1);
330 FPU_SUBC(d0, x0, t0);
341 /* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
349 FPU_SUBS(d2, x2, t2);
350 FPU_SUBCS(d1, x1, t1);
351 FPU_SUBC(d0, x0, t0);
353 x0 = d0, x1 = d1, x2 = d2;
355 y1 |= 1; /* now t1, y1 are set in concrete */
358 while ((bit >>= 1) != 0) {
361 FPU_SUBS(d2, x2, t2);
362 FPU_SUBCS(d1, x1, t1);
363 FPU_SUBC(d0, x0, t0);
365 x0 = d0, x1 = d1, x2 = d2;
374 /* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */
382 FPU_SUBS(d3, x3, t3);
383 FPU_SUBCS(d2, x2, t2);
384 FPU_SUBCS(d1, x1, t1);
385 FPU_SUBC(d0, x0, t0);
388 x0 = d0, x1 = d1, x2 = d2;
392 while ((bit >>= 1) != 0) {
395 FPU_SUBS(d3, x3, t3);
396 FPU_SUBCS(d2, x2, t2);
397 FPU_SUBCS(d1, x1, t1);
398 FPU_SUBC(d0, x0, t0);
400 x0 = d0, x1 = d1, x2 = d2;
409 * The result, which includes guard and round bits, is exact iff
410 * x is now zero; any nonzero bits in x represent sticky bits.
412 x->fp_sticky = x0 | x1 | x2 | x3;