The Design and Implementation of the FreeBSD Operating System, Second Edition
Now available: The Design and Implementation of the FreeBSD Operating System (Second Edition)


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

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    1 /*      $NetBSD: fpu_add.c,v 1.4 2005/12/11 12:18:42 christos Exp $ */
    2 
    3 /*-
    4  * SPDX-License-Identifier: BSD-3-Clause
    5  *
    6  * Copyright (c) 1992, 1993
    7  *      The Regents of the University of California.  All rights reserved.
    8  *
    9  * This software was developed by the Computer Systems Engineering group
   10  * at Lawrence Berkeley Laboratory under DARPA contract BG 91-66 and
   11  * contributed to Berkeley.
   12  *
   13  * All advertising materials mentioning features or use of this software
   14  * must display the following acknowledgement:
   15  *      This product includes software developed by the University of
   16  *      California, Lawrence Berkeley Laboratory.
   17  *
   18  * Redistribution and use in source and binary forms, with or without
   19  * modification, are permitted provided that the following conditions
   20  * are met:
   21  * 1. Redistributions of source code must retain the above copyright
   22  *    notice, this list of conditions and the following disclaimer.
   23  * 2. Redistributions in binary form must reproduce the above copyright
   24  *    notice, this list of conditions and the following disclaimer in the
   25  *    documentation and/or other materials provided with the distribution.
   26  * 3. Neither the name of the University nor the names of its contributors
   27  *    may be used to endorse or promote products derived from this software
   28  *    without specific prior written permission.
   29  *
   30  * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
   31  * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
   32  * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
   33  * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
   34  * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
   35  * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
   36  * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
   37  * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
   38  * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
   39  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
   40  * SUCH DAMAGE.
   41  *
   42  *      @(#)fpu_add.c   8.1 (Berkeley) 6/11/93
   43  */
   44 
   45 /*
   46  * Perform an FPU add (return x + y).
   47  *
   48  * To subtract, negate y and call add.
   49  */
   50 
   51 #include <sys/cdefs.h>
   52 __FBSDID("$FreeBSD$");
   53 
   54 #include <sys/types.h>
   55 #include <sys/systm.h>
   56 
   57 #include <machine/fpu.h>
   58 #include <machine/ieeefp.h>
   59 
   60 #include <powerpc/fpu/fpu_arith.h>
   61 #include <powerpc/fpu/fpu_emu.h>
   62 
   63 struct fpn *
   64 fpu_add(struct fpemu *fe)
   65 {
   66         struct fpn *x = &fe->fe_f1, *y = &fe->fe_f2, *r;
   67         u_int r0, r1, r2, r3;
   68         int rd;
   69 
   70         /*
   71          * Put the `heavier' operand on the right (see fpu_emu.h).
   72          * Then we will have one of the following cases, taken in the
   73          * following order:
   74          *
   75          *  - y = NaN.  Implied: if only one is a signalling NaN, y is.
   76          *      The result is y.
   77          *  - y = Inf.  Implied: x != NaN (is 0, number, or Inf: the NaN
   78          *    case was taken care of earlier).
   79          *      If x = -y, the result is NaN.  Otherwise the result
   80          *      is y (an Inf of whichever sign).
   81          *  - y is 0.  Implied: x = 0.
   82          *      If x and y differ in sign (one positive, one negative),
   83          *      the result is +0 except when rounding to -Inf.  If same:
   84          *      +0 + +0 = +0; -0 + -0 = -0.
   85          *  - x is 0.  Implied: y != 0.
   86          *      Result is y.
   87          *  - other.  Implied: both x and y are numbers.
   88          *      Do addition a la Hennessey & Patterson.
   89          */
   90         DPRINTF(FPE_REG, ("fpu_add:\n"));
   91         DUMPFPN(FPE_REG, x);
   92         DUMPFPN(FPE_REG, y);
   93         DPRINTF(FPE_REG, ("=>\n"));
   94         ORDER(x, y);
   95         if (ISNAN(y)) {
   96                 fe->fe_cx |= FPSCR_VXSNAN;
   97                 DUMPFPN(FPE_REG, y);
   98                 return (y);
   99         }
  100         if (ISINF(y)) {
  101                 if (ISINF(x) && x->fp_sign != y->fp_sign) {
  102                         fe->fe_cx |= FPSCR_VXISI;
  103                         return (fpu_newnan(fe));
  104                 }
  105                 DUMPFPN(FPE_REG, y);
  106                 return (y);
  107         }
  108         rd = ((fe->fe_fpscr) & FPSCR_RN);
  109         if (ISZERO(y)) {
  110                 if (rd != FP_RM)        /* only -0 + -0 gives -0 */
  111                         y->fp_sign &= x->fp_sign;
  112                 else                    /* any -0 operand gives -0 */
  113                         y->fp_sign |= x->fp_sign;
  114                 DUMPFPN(FPE_REG, y);
  115                 return (y);
  116         }
  117         if (ISZERO(x)) {
  118                 DUMPFPN(FPE_REG, y);
  119                 return (y);
  120         }
  121         /*
  122          * We really have two numbers to add, although their signs may
  123          * differ.  Make the exponents match, by shifting the smaller
  124          * number right (e.g., 1.011 => 0.1011) and increasing its
  125          * exponent (2^3 => 2^4).  Note that we do not alter the exponents
  126          * of x and y here.
  127          */
  128         r = &fe->fe_f3;
  129         r->fp_class = FPC_NUM;
  130         if (x->fp_exp == y->fp_exp) {
  131                 r->fp_exp = x->fp_exp;
  132                 r->fp_sticky = 0;
  133         } else {
  134                 if (x->fp_exp < y->fp_exp) {
  135                         /*
  136                          * Try to avoid subtract case iii (see below).
  137                          * This also guarantees that x->fp_sticky = 0.
  138                          */
  139                         SWAP(x, y);
  140                 }
  141                 /* now x->fp_exp > y->fp_exp */
  142                 r->fp_exp = x->fp_exp;
  143                 r->fp_sticky = fpu_shr(y, x->fp_exp - y->fp_exp);
  144         }
  145         r->fp_sign = x->fp_sign;
  146         if (x->fp_sign == y->fp_sign) {
  147                 FPU_DECL_CARRY
  148 
  149                 /*
  150                  * The signs match, so we simply add the numbers.  The result
  151                  * may be `supernormal' (as big as 1.111...1 + 1.111...1, or
  152                  * 11.111...0).  If so, a single bit shift-right will fix it
  153                  * (but remember to adjust the exponent).
  154                  */
  155                 /* r->fp_mant = x->fp_mant + y->fp_mant */
  156                 FPU_ADDS(r->fp_mant[3], x->fp_mant[3], y->fp_mant[3]);
  157                 FPU_ADDCS(r->fp_mant[2], x->fp_mant[2], y->fp_mant[2]);
  158                 FPU_ADDCS(r->fp_mant[1], x->fp_mant[1], y->fp_mant[1]);
  159                 FPU_ADDC(r0, x->fp_mant[0], y->fp_mant[0]);
  160                 if ((r->fp_mant[0] = r0) >= FP_2) {
  161                         (void) fpu_shr(r, 1);
  162                         r->fp_exp++;
  163                 }
  164         } else {
  165                 FPU_DECL_CARRY
  166 
  167                 /*
  168                  * The signs differ, so things are rather more difficult.
  169                  * H&P would have us negate the negative operand and add;
  170                  * this is the same as subtracting the negative operand.
  171                  * This is quite a headache.  Instead, we will subtract
  172                  * y from x, regardless of whether y itself is the negative
  173                  * operand.  When this is done one of three conditions will
  174                  * hold, depending on the magnitudes of x and y:
  175                  *   case i)   |x| > |y|.  The result is just x - y,
  176                  *      with x's sign, but it may need to be normalized.
  177                  *   case ii)  |x| = |y|.  The result is 0 (maybe -0)
  178                  *      so must be fixed up.
  179                  *   case iii) |x| < |y|.  We goofed; the result should
  180                  *      be (y - x), with the same sign as y.
  181                  * We could compare |x| and |y| here and avoid case iii,
  182                  * but that would take just as much work as the subtract.
  183                  * We can tell case iii has occurred by an overflow.
  184                  *
  185                  * N.B.: since x->fp_exp >= y->fp_exp, x->fp_sticky = 0.
  186                  */
  187                 /* r->fp_mant = x->fp_mant - y->fp_mant */
  188                 FPU_SET_CARRY(y->fp_sticky);
  189                 FPU_SUBCS(r3, x->fp_mant[3], y->fp_mant[3]);
  190                 FPU_SUBCS(r2, x->fp_mant[2], y->fp_mant[2]);
  191                 FPU_SUBCS(r1, x->fp_mant[1], y->fp_mant[1]);
  192                 FPU_SUBC(r0, x->fp_mant[0], y->fp_mant[0]);
  193                 if (r0 < FP_2) {
  194                         /* cases i and ii */
  195                         if ((r0 | r1 | r2 | r3) == 0) {
  196                                 /* case ii */
  197                                 r->fp_class = FPC_ZERO;
  198                                 r->fp_sign = rd == FP_RM;
  199                                 return (r);
  200                         }
  201                 } else {
  202                         /*
  203                          * Oops, case iii.  This can only occur when the
  204                          * exponents were equal, in which case neither
  205                          * x nor y have sticky bits set.  Flip the sign
  206                          * (to y's sign) and negate the result to get y - x.
  207                          */
  208 #ifdef DIAGNOSTIC
  209                         if (x->fp_exp != y->fp_exp || r->fp_sticky)
  210                                 panic("fpu_add");
  211 #endif
  212                         r->fp_sign = y->fp_sign;
  213                         FPU_SUBS(r3, 0, r3);
  214                         FPU_SUBCS(r2, 0, r2);
  215                         FPU_SUBCS(r1, 0, r1);
  216                         FPU_SUBC(r0, 0, r0);
  217                 }
  218                 r->fp_mant[3] = r3;
  219                 r->fp_mant[2] = r2;
  220                 r->fp_mant[1] = r1;
  221                 r->fp_mant[0] = r0;
  222                 if (r0 < FP_1)
  223                         fpu_norm(r);
  224         }
  225         DUMPFPN(FPE_REG, r);
  226         return (r);
  227 }

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