FreeBSD/Linux Kernel Cross Reference
sys/powerpc/fpu/fpu_sqrt.c

     /*      $NetBSD: fpu_sqrt.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */
     
     /*-
      * SPDX-License-Identifier: BSD-3-Clause
      *
      * 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. 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/cdefs.h>
    __FBSDID("$FreeBSD$");
    
    #include <sys/types.h>
    #include <sys/systm.h>
    
    #include <machine/fpu.h>
    
    #include <powerpc/fpu/fpu_arith.h>
    #include <powerpc/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(struct fpemu *fe)
   {
           struct fpn *x = &fe->fe_f1;
           u_int bit, q, tt;
           u_int x0, x1, x2, x3;
           u_int y0, y1, y2, y3;
           u_int d0, d1, d2, d3;
           int e;
           FPU_DECL_CARRY;
   
           /*
            * 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).
            */
           DPRINTF(FPE_REG, ("fpu_sqer:\n"));
           DUMPFPN(FPE_REG, x);
           DPRINTF(FPE_REG, ("=>\n"));
           if (ISNAN(x)) {
                   fe->fe_cx |= FPSCR_VXSNAN;
                   DUMPFPN(FPE_REG, x);
                   return (x);
           }
           if (ISZERO(x)) {
                   fe->fe_cx |= FPSCR_ZX;
                   x->fp_class = FPC_INF;
                   DUMPFPN(FPE_REG, x);
                   return (x);
           }
           if (x->fp_sign) {
                   fe->fe_cx |= FPSCR_VXSQRT;
                   return (fpu_newnan(fe));
           }
           if (ISINF(x)) {
                   DUMPFPN(FPE_REG, 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 `(1U << 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);
           if ((int)d0 >= 0) {
                   x0 = d0, x1 = d1, x2 = d2; x3 = d3;
                   q = bit;
                   y2 |= 1;
           }
           ODD_DOUBLE;
           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; x3 = d3;
                           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;
           DUMPFPN(FPE_REG, x);
           return (x);
   }

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