44c01df173
shifts into the sign bit. Instead use (1U << 31) which gets the expected result. This fix is not ideal as it assumes a 32 bit int, but does fix the issue for most cases. A similar change was made in OpenBSD. Discussed with: -arch, rdivacky Reviewed by: cperciva
416 lines
12 KiB
C
416 lines
12 KiB
C
/* $NetBSD: fpu_sqrt.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */
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/*
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* Copyright (c) 1992, 1993
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* The Regents of the University of California. All rights reserved.
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*
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* This software was developed by the Computer Systems Engineering group
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* at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
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* contributed to Berkeley.
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*
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* All advertising materials mentioning features or use of this software
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* must display the following acknowledgement:
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* This product includes software developed by the University of
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* California, Lawrence Berkeley Laboratory.
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*
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* Redistribution and use in source and binary forms, with or without
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* modification, are permitted provided that the following conditions
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* are met:
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* 1. Redistributions of source code must retain the above copyright
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* notice, this list of conditions and the following disclaimer.
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* 2. Redistributions in binary form must reproduce the above copyright
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* notice, this list of conditions and the following disclaimer in the
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* documentation and/or other materials provided with the distribution.
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* 3. Neither the name of the University nor the names of its contributors
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* may be used to endorse or promote products derived from this software
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* without specific prior written permission.
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*
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* THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
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* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
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* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
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* ARE DISCLAIMED. IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
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* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
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* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
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* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
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* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
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* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
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* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
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* SUCH DAMAGE.
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*
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* @(#)fpu_sqrt.c 8.1 (Berkeley) 6/11/93
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*/
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/*
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* Perform an FPU square root (return sqrt(x)).
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*/
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#include <sys/cdefs.h>
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__FBSDID("$FreeBSD$");
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#include <sys/types.h>
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#include <sys/systm.h>
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#include <machine/fpu.h>
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#include <machine/reg.h>
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#include <powerpc/fpu/fpu_arith.h>
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#include <powerpc/fpu/fpu_emu.h>
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/*
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* Our task is to calculate the square root of a floating point number x0.
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* This number x normally has the form:
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*
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* exp
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* x = mant * 2 (where 1 <= mant < 2 and exp is an integer)
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*
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* This can be left as it stands, or the mantissa can be doubled and the
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* exponent decremented:
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*
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* exp-1
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* x = (2 * mant) * 2 (where 2 <= 2 * mant < 4)
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*
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* If the exponent `exp' is even, the square root of the number is best
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* handled using the first form, and is by definition equal to:
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*
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* exp/2
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* sqrt(x) = sqrt(mant) * 2
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*
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* If exp is odd, on the other hand, it is convenient to use the second
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* form, giving:
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*
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* (exp-1)/2
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* sqrt(x) = sqrt(2 * mant) * 2
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*
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* In the first case, we have
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*
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* 1 <= mant < 2
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*
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* and therefore
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*
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* sqrt(1) <= sqrt(mant) < sqrt(2)
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*
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* while in the second case we have
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*
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* 2 <= 2*mant < 4
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*
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* and therefore
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*
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* sqrt(2) <= sqrt(2*mant) < sqrt(4)
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*
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* so that in any case, we are sure that
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*
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* sqrt(1) <= sqrt(n * mant) < sqrt(4), n = 1 or 2
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*
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* or
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*
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* 1 <= sqrt(n * mant) < 2, n = 1 or 2.
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*
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* This root is therefore a properly formed mantissa for a floating
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* point number. The exponent of sqrt(x) is either exp/2 or (exp-1)/2
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* as above. This leaves us with the problem of finding the square root
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* of a fixed-point number in the range [1..4).
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*
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* Though it may not be instantly obvious, the following square root
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* algorithm works for any integer x of an even number of bits, provided
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* that no overflows occur:
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*
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* let q = 0
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* for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
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* x *= 2 -- multiply by radix, for next digit
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* if x >= 2q + 2^k then -- if adding 2^k does not
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* x -= 2q + 2^k -- exceed the correct root,
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* q += 2^k -- add 2^k and adjust x
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* fi
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* done
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* sqrt = q / 2^(NBITS/2) -- (and any remainder is in x)
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*
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* If NBITS is odd (so that k is initially even), we can just add another
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* zero bit at the top of x. Doing so means that q is not going to acquire
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* a 1 bit in the first trip around the loop (since x0 < 2^NBITS). If the
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* final value in x is not needed, or can be off by a factor of 2, this is
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* equivalant to moving the `x *= 2' step to the bottom of the loop:
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*
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* for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
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*
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* and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
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* (Since the algorithm is destructive on x, we will call x's initial
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* value, for which q is some power of two times its square root, x0.)
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*
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* If we insert a loop invariant y = 2q, we can then rewrite this using
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* C notation as:
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*
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* q = y = 0; x = x0;
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* for (k = NBITS; --k >= 0;) {
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* #if (NBITS is even)
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* x *= 2;
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* #endif
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* t = y + (1 << k);
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* if (x >= t) {
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* x -= t;
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* q += 1 << k;
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* y += 1 << (k + 1);
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* }
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* #if (NBITS is odd)
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* x *= 2;
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* #endif
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* }
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*
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* If x0 is fixed point, rather than an integer, we can simply alter the
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* scale factor between q and sqrt(x0). As it happens, we can easily arrange
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* for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
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*
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* In our case, however, x0 (and therefore x, y, q, and t) are multiword
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* integers, which adds some complication. But note that q is built one
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* bit at a time, from the top down, and is not used itself in the loop
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* (we use 2q as held in y instead). This means we can build our answer
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* in an integer, one word at a time, which saves a bit of work. Also,
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* since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
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* `new' bits in y and we can set them with an `or' operation rather than
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* a full-blown multiword add.
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*
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* We are almost done, except for one snag. We must prove that none of our
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* intermediate calculations can overflow. We know that x0 is in [1..4)
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* and therefore the square root in q will be in [1..2), but what about x,
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* y, and t?
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*
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* We know that y = 2q at the beginning of each loop. (The relation only
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* fails temporarily while y and q are being updated.) Since q < 2, y < 4.
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* The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
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* Furthermore, we can prove with a bit of work that x never exceeds y by
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* more than 2, so that even after doubling, 0 <= x < 8. (This is left as
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* an exercise to the reader, mostly because I have become tired of working
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* on this comment.)
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*
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* If our floating point mantissas (which are of the form 1.frac) occupy
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* B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
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* In fact, we want even one more bit (for a carry, to avoid compares), or
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* three extra. There is a comment in fpu_emu.h reminding maintainers of
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* this, so we have some justification in assuming it.
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*/
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struct fpn *
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fpu_sqrt(struct fpemu *fe)
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{
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struct fpn *x = &fe->fe_f1;
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u_int bit, q, tt;
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u_int x0, x1, x2, x3;
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u_int y0, y1, y2, y3;
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u_int d0, d1, d2, d3;
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int e;
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FPU_DECL_CARRY;
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/*
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* Take care of special cases first. In order:
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*
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* sqrt(NaN) = NaN
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* sqrt(+0) = +0
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* sqrt(-0) = -0
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* sqrt(x < 0) = NaN (including sqrt(-Inf))
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* sqrt(+Inf) = +Inf
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*
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* Then all that remains are numbers with mantissas in [1..2).
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*/
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DPRINTF(FPE_REG, ("fpu_sqer:\n"));
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DUMPFPN(FPE_REG, x);
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DPRINTF(FPE_REG, ("=>\n"));
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if (ISNAN(x)) {
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fe->fe_cx |= FPSCR_VXSNAN;
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DUMPFPN(FPE_REG, x);
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return (x);
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}
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if (ISZERO(x)) {
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fe->fe_cx |= FPSCR_ZX;
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x->fp_class = FPC_INF;
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DUMPFPN(FPE_REG, x);
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return (x);
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}
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if (x->fp_sign) {
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return (fpu_newnan(fe));
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}
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if (ISINF(x)) {
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fe->fe_cx |= FPSCR_VXSQRT;
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DUMPFPN(FPE_REG, 0);
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return (0);
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}
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/*
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* Calculate result exponent. As noted above, this may involve
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* doubling the mantissa. We will also need to double x each
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* time around the loop, so we define a macro for this here, and
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* we break out the multiword mantissa.
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*/
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#ifdef FPU_SHL1_BY_ADD
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#define DOUBLE_X { \
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FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
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FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
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}
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#else
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#define DOUBLE_X { \
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x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
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x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
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}
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#endif
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#if (FP_NMANT & 1) != 0
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# define ODD_DOUBLE DOUBLE_X
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# define EVEN_DOUBLE /* nothing */
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#else
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# define ODD_DOUBLE /* nothing */
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# define EVEN_DOUBLE DOUBLE_X
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#endif
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x0 = x->fp_mant[0];
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x1 = x->fp_mant[1];
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x2 = x->fp_mant[2];
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x3 = x->fp_mant[3];
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e = x->fp_exp;
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if (e & 1) /* exponent is odd; use sqrt(2mant) */
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DOUBLE_X;
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/* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
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x->fp_exp = e >> 1; /* calculates (e&1 ? (e-1)/2 : e/2 */
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/*
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* Now calculate the mantissa root. Since x is now in [1..4),
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* we know that the first trip around the loop will definitely
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* set the top bit in q, so we can do that manually and start
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* the loop at the next bit down instead. We must be sure to
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* double x correctly while doing the `known q=1.0'.
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*
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* We do this one mantissa-word at a time, as noted above, to
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* save work. To avoid `(1U << 31) << 1', we also do the top bit
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* outside of each per-word loop.
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*
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* The calculation `t = y + bit' breaks down into `t0 = y0, ...,
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* t3 = y3, t? |= bit' for the appropriate word. Since the bit
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* is always a `new' one, this means that three of the `t?'s are
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* just the corresponding `y?'; we use `#define's here for this.
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* The variable `tt' holds the actual `t?' variable.
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*/
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/* calculate q0 */
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#define t0 tt
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bit = FP_1;
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EVEN_DOUBLE;
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/* if (x >= (t0 = y0 | bit)) { */ /* always true */
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q = bit;
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x0 -= bit;
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y0 = bit << 1;
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/* } */
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ODD_DOUBLE;
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while ((bit >>= 1) != 0) { /* for remaining bits in q0 */
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EVEN_DOUBLE;
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t0 = y0 | bit; /* t = y + bit */
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if (x0 >= t0) { /* if x >= t then */
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x0 -= t0; /* x -= t */
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q |= bit; /* q += bit */
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y0 |= bit << 1; /* y += bit << 1 */
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}
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ODD_DOUBLE;
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}
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x->fp_mant[0] = q;
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#undef t0
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/* calculate q1. note (y0&1)==0. */
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#define t0 y0
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#define t1 tt
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q = 0;
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y1 = 0;
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bit = 1 << 31;
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EVEN_DOUBLE;
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t1 = bit;
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FPU_SUBS(d1, x1, t1);
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FPU_SUBC(d0, x0, t0); /* d = x - t */
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if ((int)d0 >= 0) { /* if d >= 0 (i.e., x >= t) then */
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x0 = d0, x1 = d1; /* x -= t */
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q = bit; /* q += bit */
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y0 |= 1; /* y += bit << 1 */
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}
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ODD_DOUBLE;
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while ((bit >>= 1) != 0) { /* for remaining bits in q1 */
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EVEN_DOUBLE; /* as before */
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t1 = y1 | bit;
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FPU_SUBS(d1, x1, t1);
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FPU_SUBC(d0, x0, t0);
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if ((int)d0 >= 0) {
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x0 = d0, x1 = d1;
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q |= bit;
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y1 |= bit << 1;
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}
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ODD_DOUBLE;
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}
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x->fp_mant[1] = q;
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#undef t1
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/* calculate q2. note (y1&1)==0; y0 (aka t0) is fixed. */
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#define t1 y1
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#define t2 tt
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q = 0;
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y2 = 0;
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bit = 1 << 31;
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EVEN_DOUBLE;
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t2 = bit;
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FPU_SUBS(d2, x2, t2);
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FPU_SUBCS(d1, x1, t1);
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FPU_SUBC(d0, x0, t0);
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if ((int)d0 >= 0) {
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x0 = d0, x1 = d1, x2 = d2;
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q |= bit;
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y1 |= 1; /* now t1, y1 are set in concrete */
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}
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ODD_DOUBLE;
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while ((bit >>= 1) != 0) {
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EVEN_DOUBLE;
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t2 = y2 | bit;
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FPU_SUBS(d2, x2, t2);
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FPU_SUBCS(d1, x1, t1);
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FPU_SUBC(d0, x0, t0);
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if ((int)d0 >= 0) {
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x0 = d0, x1 = d1, x2 = d2;
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q |= bit;
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y2 |= bit << 1;
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}
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ODD_DOUBLE;
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}
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x->fp_mant[2] = q;
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#undef t2
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/* calculate q3. y0, t0, y1, t1 all fixed; y2, t2, almost done. */
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#define t2 y2
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#define t3 tt
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q = 0;
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y3 = 0;
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bit = 1 << 31;
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EVEN_DOUBLE;
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t3 = bit;
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FPU_SUBS(d3, x3, t3);
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FPU_SUBCS(d2, x2, t2);
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FPU_SUBCS(d1, x1, t1);
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FPU_SUBC(d0, x0, t0);
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ODD_DOUBLE;
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if ((int)d0 >= 0) {
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x0 = d0, x1 = d1, x2 = d2;
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q |= bit;
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y2 |= 1;
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}
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while ((bit >>= 1) != 0) {
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EVEN_DOUBLE;
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t3 = y3 | bit;
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FPU_SUBS(d3, x3, t3);
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FPU_SUBCS(d2, x2, t2);
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FPU_SUBCS(d1, x1, t1);
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FPU_SUBC(d0, x0, t0);
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if ((int)d0 >= 0) {
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x0 = d0, x1 = d1, x2 = d2;
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q |= bit;
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y3 |= bit << 1;
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}
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ODD_DOUBLE;
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}
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x->fp_mant[3] = q;
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/*
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* The result, which includes guard and round bits, is exact iff
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* x is now zero; any nonzero bits in x represent sticky bits.
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*/
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x->fp_sticky = x0 | x1 | x2 | x3;
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DUMPFPN(FPE_REG, x);
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return (x);
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}
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