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/* Copyright (C) 2004 Free Software Foundation, Inc.

This file is free software; you can redistribute it and/or modify it

under the terms of the GNU General Public License as published by the

Free Software Foundation; either version 2, or (at your option) any

later version.

In addition to the permissions in the GNU General Public License, the

Free Software Foundation gives you unlimited permission to link the

compiled version of this file into combinations with other programs,

and to distribute those combinations without any restriction coming

from the use of this file.  (The General Public License restrictions

do apply in other respects; for example, they cover modification of

the file, and distribution when not linked into a combine

executable.)

This file is distributed in the hope that it will be useful, but

WITHOUT ANY WARRANTY; without even the implied warranty of

MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the GNU

General Public License for more details.

You should have received a copy of the GNU General Public License

along with this program; see the file COPYING.  If not, write to

the Free Software Foundation, 51 Franklin Street, Fifth Floor,

Boston, MA 02110-1301, USA.  */

/* Calculate division table for SH2..4 integer division

   Contributed by Joern Rernnecke

   joern.rennecke@superh.com  */

#include <stdio.h>

#include <math.h>

int

main ()

{

  int i, j;

  double q, r, err, max_err = 0, max_s_err = 0;

  puts("/* This table has been generated by divtab-sh4.c.  */");

  puts ("\t.balign 4");

  puts ("LOCAL(div_table_clz):");

  /* output some dummy number for 1/0.  */

  printf ("\t.byte\t%d\n", 0);

  for (i = 1; i <= 128; i++)

    {

      int n = 0;

      if (i == 128)

        puts ("\

/* Lookup table translating positive divisor to index into table of\n\

   normalized inverse.  N.B. the '0' entry is also the last entry of the\n\

 previous table, and causes an unaligned access for division by zero.  */\n\

LOCAL(div_table_ix):");

      for (j = i; j <= 128; j += j)

        n++;

      printf ("\t.byte\t%d\n", n - 7);

    }

  for (i = 1; i <= 128; i++)

    {

      j = i < 0 ? -i : i;

      while (j < 128)

        j += j;

      printf ("\t.byte\t%d\n", j * 2 - 96*4);

    }

  puts("\

/* 1/64 .. 1/127, normalized.  There is an implicit leading 1 in bit 32.  */\n\

        .balign 4\n\

LOCAL(zero_l):");

  for (i = 64; i < 128; i++)

    {

      if (i == 96)

        puts ("LOCAL(div_table):");

      q = 4.*(1<<30)*128/i;

      r = ceil (q);

      /* The value for 64 is actually differently scaled that it would

         appear from this calculation.  The implicit part is %01, not 10.

         Still, since the value in the table is 0 either way, this

         doesn't matter here.  Still, the 1/64 entry is effectively a 1/128

         entry.  */

      printf ("\t.long\t0x%X\n", (unsigned) r);

      err = r - q;

      if (err > max_err)

        max_err = err;

      err = err * i / 128;

      if (err > max_s_err)

        max_s_err = err;

    }

  printf ("\t/* maximum error: %f scaled: %f*/\n", max_err, max_s_err);

  exit (0);

}