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jeremybenn |
/* Copyright (C) 2004, 2009 Free Software Foundation, Inc.
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This file is free software; you can redistribute it and/or modify it
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under the terms of the GNU General Public License as published by the
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Free Software Foundation; either version 3, or (at your option) any
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later version.
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This file is distributed in the hope that it will be useful, but
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WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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General Public License for more details.
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Under Section 7 of GPL version 3, you are granted additional
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permissions described in the GCC Runtime Library Exception, version
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3.1, as published by the Free Software Foundation.
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You should have received a copy of the GNU General Public License and
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a copy of the GCC Runtime Library Exception along with this program;
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see the files COPYING3 and COPYING.RUNTIME respectively. If not, see
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<http://www.gnu.org/licenses/>. */
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/* Calculate division table for SH2..4 integer division
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Contributed by Joern Rernnecke
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joern.rennecke@superh.com */
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#include <stdio.h>
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#include <math.h>
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int
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main ()
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{
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int i, j;
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double q, r, err, max_err = 0, max_s_err = 0;
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puts("/* This table has been generated by divtab-sh4.c. */");
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puts ("\t.balign 4");
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puts ("LOCAL(div_table_clz):");
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/* output some dummy number for 1/0. */
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printf ("\t.byte\t%d\n", 0);
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for (i = 1; i <= 128; i++)
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{
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int n = 0;
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if (i == 128)
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puts ("\
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/* Lookup table translating positive divisor to index into table of\n\
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normalized inverse. N.B. the '0' entry is also the last entry of the\n\
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previous table, and causes an unaligned access for division by zero. */\n\
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LOCAL(div_table_ix):");
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for (j = i; j <= 128; j += j)
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n++;
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printf ("\t.byte\t%d\n", n - 7);
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}
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for (i = 1; i <= 128; i++)
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{
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j = i < 0 ? -i : i;
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while (j < 128)
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j += j;
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printf ("\t.byte\t%d\n", j * 2 - 96*4);
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}
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puts("\
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/* 1/64 .. 1/127, normalized. There is an implicit leading 1 in bit 32. */\n\
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.balign 4\n\
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LOCAL(zero_l):");
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for (i = 64; i < 128; i++)
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{
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if (i == 96)
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puts ("LOCAL(div_table):");
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q = 4.*(1<<30)*128/i;
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r = ceil (q);
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/* The value for 64 is actually differently scaled that it would
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appear from this calculation. The implicit part is %01, not 10.
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Still, since the value in the table is 0 either way, this
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doesn't matter here. Still, the 1/64 entry is effectively a 1/128
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entry. */
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printf ("\t.long\t0x%X\n", (unsigned) r);
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err = r - q;
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if (err > max_err)
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max_err = err;
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err = err * i / 128;
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if (err > max_s_err)
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max_s_err = err;
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}
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printf ("\t/* maximum error: %f scaled: %f*/\n", max_err, max_s_err);
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exit (0);
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}
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