initial import of NetBSD tree

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diff --git a/sys/arch/sparc/fpu/fpu_sqrt.c b/sys/arch/sparc/fpu/fpu_sqrt.c
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+/*	$NetBSD: fpu_sqrt.c,v 1.2 1994/11/20 20:52:46 deraadt Exp $ */
+
+/*
+ * Copyright (c) 1992, 1993
+ *	The Regents of the University of California.  All rights reserved.
+ *
+ * This software was developed by the Computer Systems Engineering group
+ * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
+ * contributed to Berkeley.
+ *
+ * All advertising materials mentioning features or use of this software
+ * must display the following acknowledgement:
+ *	This product includes software developed by the University of
+ *	California, Lawrence Berkeley Laboratory.
+ *
+ * Redistribution and use in source and binary forms, with or without
+ * modification, are permitted provided that the following conditions
+ * are met:
+ * 1. Redistributions of source code must retain the above copyright
+ *    notice, this list of conditions and the following disclaimer.
+ * 2. Redistributions in binary form must reproduce the above copyright
+ *    notice, this list of conditions and the following disclaimer in the
+ *    documentation and/or other materials provided with the distribution.
+ * 3. All advertising materials mentioning features or use of this software
+ *    must display the following acknowledgement:
+ *	This product includes software developed by the University of
+ *	California, Berkeley and its contributors.
+ * 4. Neither the name of the University nor the names of its contributors
+ *    may be used to endorse or promote products derived from this software
+ *    without specific prior written permission.
+ *
+ * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
+ * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
+ * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
+ * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
+ * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
+ * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
+ * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
+ * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
+ * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
+ * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
+ * SUCH DAMAGE.
+ *
+ *	@(#)fpu_sqrt.c	8.1 (Berkeley) 6/11/93
+ */
+
+/*
+ * Perform an FPU square root (return sqrt(x)).
+ */
+
+#include <sys/types.h>
+
+#include <machine/reg.h>
+
+#include <sparc/fpu/fpu_arith.h>
+#include <sparc/fpu/fpu_emu.h>
+
+/*
+ * Our task is to calculate the square root of a floating point number x0.
+ * This number x normally has the form:
+ *
+ *		    exp
+ *	x = mant * 2		(where 1 <= mant < 2 and exp is an integer)
+ *
+ * This can be left as it stands, or the mantissa can be doubled and the
+ * exponent decremented:
+ *
+ *			  exp-1
+ *	x = (2 * mant) * 2	(where 2 <= 2 * mant < 4)
+ *
+ * If the exponent `exp' is even, the square root of the number is best
+ * handled using the first form, and is by definition equal to:
+ *
+ *				exp/2
+ *	sqrt(x) = sqrt(mant) * 2
+ *
+ * If exp is odd, on the other hand, it is convenient to use the second
+ * form, giving:
+ *
+ *				    (exp-1)/2
+ *	sqrt(x) = sqrt(2 * mant) * 2
+ *
+ * In the first case, we have
+ *
+ *	1 <= mant < 2
+ *
+ * and therefore
+ *
+ *	sqrt(1) <= sqrt(mant) < sqrt(2)
+ *
+ * while in the second case we have
+ *
+ *	2 <= 2*mant < 4
+ *
+ * and therefore
+ *
+ *	sqrt(2) <= sqrt(2*mant) < sqrt(4)
+ *
+ * so that in any case, we are sure that
+ *
+ *	sqrt(1) <= sqrt(n * mant) < sqrt(4),	n = 1 or 2
+ *
+ * or
+ *
+ *	1 <= sqrt(n * mant) < 2,		n = 1 or 2.
+ *
+ * This root is therefore a properly formed mantissa for a floating
+ * point number.  The exponent of sqrt(x) is either exp/2 or (exp-1)/2
+ * as above.  This leaves us with the problem of finding the square root
+ * of a fixed-point number in the range [1..4).
+ *
+ * Though it may not be instantly obvious, the following square root
+ * algorithm works for any integer x of an even number of bits, provided
+ * that no overflows occur:
+ *
+ *	let q = 0
+ *	for k = NBITS-1 to 0 step -1 do -- for each digit in the answer...
+ *		x *= 2			-- multiply by radix, for next digit
+ *		if x >= 2q + 2^k then	-- if adding 2^k does not
+ *			x -= 2q + 2^k	-- exceed the correct root,
+ *			q += 2^k	-- add 2^k and adjust x
+ *		fi
+ *	done
+ *	sqrt = q / 2^(NBITS/2)		-- (and any remainder is in x)
+ *
+ * If NBITS is odd (so that k is initially even), we can just add another
+ * zero bit at the top of x.  Doing so means that q is not going to acquire
+ * a 1 bit in the first trip around the loop (since x0 < 2^NBITS).  If the
+ * final value in x is not needed, or can be off by a factor of 2, this is
+ * equivalant to moving the `x *= 2' step to the bottom of the loop:
+ *
+ *	for k = NBITS-1 to 0 step -1 do if ... fi; x *= 2; done
+ *
+ * and the result q will then be sqrt(x0) * 2^floor(NBITS / 2).
+ * (Since the algorithm is destructive on x, we will call x's initial
+ * value, for which q is some power of two times its square root, x0.)
+ *
+ * If we insert a loop invariant y = 2q, we can then rewrite this using
+ * C notation as:
+ *
+ *	q = y = 0; x = x0;
+ *	for (k = NBITS; --k >= 0;) {
+ * #if (NBITS is even)
+ *		x *= 2;
+ * #endif
+ *		t = y + (1 << k);
+ *		if (x >= t) {
+ *			x -= t;
+ *			q += 1 << k;
+ *			y += 1 << (k + 1);
+ *		}
+ * #if (NBITS is odd)
+ *		x *= 2;
+ * #endif
+ *	}
+ *
+ * If x0 is fixed point, rather than an integer, we can simply alter the
+ * scale factor between q and sqrt(x0).  As it happens, we can easily arrange
+ * for the scale factor to be 2**0 or 1, so that sqrt(x0) == q.
+ *
+ * In our case, however, x0 (and therefore x, y, q, and t) are multiword
+ * integers, which adds some complication.  But note that q is built one
+ * bit at a time, from the top down, and is not used itself in the loop
+ * (we use 2q as held in y instead).  This means we can build our answer
+ * in an integer, one word at a time, which saves a bit of work.  Also,
+ * since 1 << k is always a `new' bit in q, 1 << k and 1 << (k+1) are
+ * `new' bits in y and we can set them with an `or' operation rather than
+ * a full-blown multiword add.
+ *
+ * We are almost done, except for one snag.  We must prove that none of our
+ * intermediate calculations can overflow.  We know that x0 is in [1..4)
+ * and therefore the square root in q will be in [1..2), but what about x,
+ * y, and t?
+ *
+ * We know that y = 2q at the beginning of each loop.  (The relation only
+ * fails temporarily while y and q are being updated.)  Since q < 2, y < 4.
+ * The sum in t can, in our case, be as much as y+(1<<1) = y+2 < 6, and.
+ * Furthermore, we can prove with a bit of work that x never exceeds y by
+ * more than 2, so that even after doubling, 0 <= x < 8.  (This is left as
+ * an exercise to the reader, mostly because I have become tired of working
+ * on this comment.)
+ *
+ * If our floating point mantissas (which are of the form 1.frac) occupy
+ * B+1 bits, our largest intermediary needs at most B+3 bits, or two extra.
+ * In fact, we want even one more bit (for a carry, to avoid compares), or
+ * three extra.  There is a comment in fpu_emu.h reminding maintainers of
+ * this, so we have some justification in assuming it.
+ */
+struct fpn *
+fpu_sqrt(fe)
+	struct fpemu *fe;
+{
+	register struct fpn *x = &fe->fe_f1;
+	register u_int bit, q, tt;
+	register u_int x0, x1, x2, x3;
+	register u_int y0, y1, y2, y3;
+	register u_int d0, d1, d2, d3;
+	register int e;
+
+	/*
+	 * Take care of special cases first.  In order:
+	 *
+	 *	sqrt(NaN) = NaN
+	 *	sqrt(+0) = +0
+	 *	sqrt(-0) = -0
+	 *	sqrt(x < 0) = NaN	(including sqrt(-Inf))
+	 *	sqrt(+Inf) = +Inf
+	 *
+	 * Then all that remains are numbers with mantissas in [1..2).
+	 */
+	if (ISNAN(x) || ISZERO(x))
+		return (x);
+	if (x->fp_sign)
+		return (fpu_newnan(fe));
+	if (ISINF(x))
+		return (x);
+
+	/*
+	 * Calculate result exponent.  As noted above, this may involve
+	 * doubling the mantissa.  We will also need to double x each
+	 * time around the loop, so we define a macro for this here, and
+	 * we break out the multiword mantissa.
+	 */
+#ifdef FPU_SHL1_BY_ADD
+#define	DOUBLE_X { \
+	FPU_ADDS(x3, x3, x3); FPU_ADDCS(x2, x2, x2); \
+	FPU_ADDCS(x1, x1, x1); FPU_ADDC(x0, x0, x0); \
+}
+#else
+#define	DOUBLE_X { \
+	x0 = (x0 << 1) | (x1 >> 31); x1 = (x1 << 1) | (x2 >> 31); \
+	x2 = (x2 << 1) | (x3 >> 31); x3 <<= 1; \
+}
+#endif
+#if (FP_NMANT & 1) != 0
+# define ODD_DOUBLE	DOUBLE_X
+# define EVEN_DOUBLE	/* nothing */
+#else
+# define ODD_DOUBLE	/* nothing */
+# define EVEN_DOUBLE	DOUBLE_X
+#endif
+	x0 = x->fp_mant[0];
+	x1 = x->fp_mant[1];
+	x2 = x->fp_mant[2];
+	x3 = x->fp_mant[3];
+	e = x->fp_exp;
+	if (e & 1)		/* exponent is odd; use sqrt(2mant) */
+		DOUBLE_X;
+	/* THE FOLLOWING ASSUMES THAT RIGHT SHIFT DOES SIGN EXTENSION */
+	x->fp_exp = e >> 1;	/* calculates (e&1 ? (e-1)/2 : e/2 */
+
+	/*
+	 * Now calculate the mantissa root.  Since x is now in [1..4),
+	 * we know that the first trip around the loop will definitely
+	 * set the top bit in q, so we can do that manually and start
+	 * the loop at the next bit down instead.  We must be sure to
+	 * double x correctly while doing the `known q=1.0'.
+	 *
+	 * We do this one mantissa-word at a time, as noted above, to
+	 * save work.  To avoid `(1 << 31) << 1', we also do the top bit
+	 * outside of each per-word loop.
+	 *
+	 * The calculation `t = y + bit' breaks down into `t0 = y0, ...,
+	 * t3 = y3, t? |= bit' for the appropriate word.  Since the bit
+	 * is always a `new' one, this means that three of the `t?'s are
+	 * just the corresponding `y?'; we use `#define's here for this.
+	 * The variable `tt' holds the actual `t?' variable.
+	 */
+
+	/* calculate q0 */
+#define	t0 tt
+	bit = FP_1;
+	EVEN_DOUBLE;
+	/* if (x >= (t0 = y0 | bit)) { */	/* always true */
+		q = bit;
+		x0 -= bit;
+		y0 = bit << 1;
+	/* } */
+	ODD_DOUBLE;
+	while ((bit >>= 1) != 0) {	/* for remaining bits in q0 */
+		EVEN_DOUBLE;
+		t0 = y0 | bit;		/* t = y + bit */
+		if (x0 >= t0) {		/* if x >= t then */
+			x0 -= t0;	/*	x -= t */
+			q |= bit;	/*	q += bit */
+			y0 |= bit << 1;	/*	y += bit << 1 */
+		}
+		ODD_DOUBLE;
+	}
+	x->fp_mant[0] = q;
+#undef t0
+
+	/* calculate q1.  note (y0&1)==0. */
+#define t0 y0
+#define t1 tt
+	q = 0;
+	y1 = 0;
+	bit = 1 << 31;
+	EVEN_DOUBLE;
+	t1 = bit;
+	FPU_SUBS(d1, x1, t1);
+	FPU_SUBC(d0, x0, t0);		/* d = x - t */
+	if ((int)d0 >= 0) {		/* if d >= 0 (i.e., x >= t) then */
+		x0 = d0, x1 = d1;	/*	x -= t */
+		q = bit;		/*	q += bit */
+		y0 |= 1;		/*	y += bit << 1 */
+	}
+	ODD_DOUBLE;
+	while ((bit >>= 1) != 0) {	/* for remaining bits in q1 */
+		EVEN_DOUBLE;		/* as before */
+		t1 = y1 | bit;
+		FPU_SUBS(d1, x1, t1);
+		FPU_SUBC(d0, x0, t0);
+		if ((int)d0 >= 0) {
+			x0 = d0, x1 = d1;
+			q |= bit;
+			y1 |= bit << 1;
+		}
+		ODD_DOUBLE;
+	}
+	x->fp_mant[1] = q;
+#undef t1
+
+	/* calculate q2.  note (y1&1)==0; y0 (aka t0) is fixed. */
+#define t1 y1
+#define t2 tt
+	q = 0;
+	y2 = 0;
+	bit = 1 << 31;
+	EVEN_DOUBLE;
+	t2 = bit;
+	FPU_SUBS(d2, x2, t2);
+	FPU_SUBCS(d1, x1, t1);
+	FPU_SUBC(d0, x0, t0);
+	if ((int)d0 >= 0) {
+		x0 = d0, x1 = d1, x2 = d2;
+		q |= bit;
+		y1 |= 1;		/* now t1, y1 are set in concrete */
+	}
+	ODD_DOUBLE;
+	while ((bit >>= 1) != 0) {
+		EVEN_DOUBLE;
+		t2 = y2 | bit;
+		FPU_SUBS(d2, x2, t2);
+		FPU_SUBCS(d1, x1, t1);
+		FPU_SUBC(d0, x0, t0);
+		if ((int)d0 >= 0) {
+			x0 = d0, x1 = d1, x2 = d2;
+			q |= bit;
+			y2 |= bit << 1;
+		}
+		ODD_DOUBLE;
+	}
+	x->fp_mant[2] = q;
+#undef t2
+
+	/* calculate q3.  y0, t0, y1, t1 all fixed; y2, t2, almost done. */
+#define t2 y2
+#define t3 tt
+	q = 0;
+	y3 = 0;
+	bit = 1 << 31;
+	EVEN_DOUBLE;
+	t3 = bit;
+	FPU_SUBS(d3, x3, t3);
+	FPU_SUBCS(d2, x2, t2);
+	FPU_SUBCS(d1, x1, t1);
+	FPU_SUBC(d0, x0, t0);
+	ODD_DOUBLE;
+	if ((int)d0 >= 0) {
+		x0 = d0, x1 = d1, x2 = d2;
+		q |= bit;
+		y2 |= 1;
+	}
+	while ((bit >>= 1) != 0) {
+		EVEN_DOUBLE;
+		t3 = y3 | bit;
+		FPU_SUBS(d3, x3, t3);
+		FPU_SUBCS(d2, x2, t2);
+		FPU_SUBCS(d1, x1, t1);
+		FPU_SUBC(d0, x0, t0);
+		if ((int)d0 >= 0) {
+			x0 = d0, x1 = d1, x2 = d2;
+			q |= bit;
+			y3 |= bit << 1;
+		}
+		ODD_DOUBLE;
+	}
+	x->fp_mant[3] = q;
+
+	/*
+	 * The result, which includes guard and round bits, is exact iff
+	 * x is now zero; any nonzero bits in x represent sticky bits.
+	 */
+	x->fp_sticky = x0 | x1 | x2 | x3;
+	return (x);
+}