blob: 00c845052fbbcec154f70803efb677aa1c7f1b92 [file] [log] [blame]
/* Remainder.
Copyright (C) 2012-2020 Free Software Foundation, Inc.
This program 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 3 of the License, or
(at your option) any later version.
This program 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. If not, see <https://www.gnu.org/licenses/>. */
#if ! (defined USE_LONG_DOUBLE || defined USE_FLOAT)
# include <config.h>
#endif
/* Specification. */
#include <math.h>
#ifdef USE_LONG_DOUBLE
# define REMAINDER remainderl
# define DOUBLE long double
# define L_(literal) literal##L
# define FABS fabsl
# define FMOD fmodl
# define ISNAN isnanl
#elif ! defined USE_FLOAT
# define REMAINDER remainder
# define DOUBLE double
# define L_(literal) literal
# define FABS fabs
# define FMOD fmod
# define ISNAN isnand
#else /* defined USE_FLOAT */
# define REMAINDER remainderf
# define DOUBLE float
# define L_(literal) literal##f
# define FABS fabsf
# define FMOD fmodf
# define ISNAN isnanf
#endif
#undef NAN
#if defined _MSC_VER
static DOUBLE zero;
# define NAN (zero / zero)
#else
# define NAN (L_(0.0) / L_(0.0))
#endif
DOUBLE
REMAINDER (DOUBLE x, DOUBLE y)
{
if (isfinite (x) && isfinite (y) && y != L_(0.0))
{
if (x == L_(0.0))
/* Return x, regardless of the sign of y. */
return x;
{
int negate = ((!signbit (x)) ^ (!signbit (y)));
DOUBLE r;
/* Take the absolute value of x and y. */
x = FABS (x);
y = FABS (y);
/* Trivial case that requires no computation. */
if (x <= L_(0.5) * y)
return (negate ? - x : x);
/* With a fixed y, the function x -> remainder(x,y) has a period 2*y.
Therefore we can reduce the argument x modulo 2*y. And it's no
problem if 2*y overflows, since fmod(x,Inf) = x. */
x = FMOD (x, L_(2.0) * y);
/* Consider the 3 cases:
0 <= x <= 0.5 * y
0.5 * y < x < 1.5 * y
1.5 * y <= x <= 2.0 * y */
if (x <= L_(0.5) * y)
r = x;
else
{
r = x - y;
if (r > L_(0.5) * y)
r = x - L_(2.0) * y;
}
return (negate ? - r : r);
}
}
else
{
if (ISNAN (x) || ISNAN (y))
return x + y; /* NaN */
else if (isinf (y))
return x;
else
/* x infinite or y zero */
return NAN;
}
}