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107 lines
2.8 KiB
C
107 lines
2.8 KiB
C
/* Remainder.
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Copyright (C) 2012-2021 Free Software Foundation, Inc.
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This file is free software: you can redistribute it and/or modify
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it under the terms of the GNU Lesser General Public License as
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published by the Free Software Foundation; either version 3 of the
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License, or (at your option) any later version.
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This file is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public License
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along with this program. If not, see <https://www.gnu.org/licenses/>. */
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#if ! (defined USE_LONG_DOUBLE || defined USE_FLOAT)
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# include <config.h>
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#endif
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/* Specification. */
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#include <math.h>
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#ifdef USE_LONG_DOUBLE
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# define REMAINDER remainderl
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# define DOUBLE long double
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# define L_(literal) literal##L
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# define FABS fabsl
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# define FMOD fmodl
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# define ISNAN isnanl
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#elif ! defined USE_FLOAT
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# define REMAINDER remainder
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# define DOUBLE double
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# define L_(literal) literal
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# define FABS fabs
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# define FMOD fmod
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# define ISNAN isnand
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#else /* defined USE_FLOAT */
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# define REMAINDER remainderf
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# define DOUBLE float
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# define L_(literal) literal##f
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# define FABS fabsf
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# define FMOD fmodf
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# define ISNAN isnanf
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#endif
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#undef NAN
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#if defined _MSC_VER
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static DOUBLE zero;
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# define NAN (zero / zero)
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#else
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# define NAN (L_(0.0) / L_(0.0))
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#endif
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DOUBLE
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REMAINDER (DOUBLE x, DOUBLE y)
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{
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if (isfinite (x) && isfinite (y) && y != L_(0.0))
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{
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if (x == L_(0.0))
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/* Return x, regardless of the sign of y. */
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return x;
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{
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int negate = ((!signbit (x)) ^ (!signbit (y)));
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DOUBLE r;
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/* Take the absolute value of x and y. */
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x = FABS (x);
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y = FABS (y);
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/* Trivial case that requires no computation. */
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if (x <= L_(0.5) * y)
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return (negate ? - x : x);
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/* With a fixed y, the function x -> remainder(x,y) has a period 2*y.
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Therefore we can reduce the argument x modulo 2*y. And it's no
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problem if 2*y overflows, since fmod(x,Inf) = x. */
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x = FMOD (x, L_(2.0) * y);
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/* Consider the 3 cases:
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0 <= x <= 0.5 * y
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0.5 * y < x < 1.5 * y
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1.5 * y <= x <= 2.0 * y */
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if (x <= L_(0.5) * y)
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r = x;
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else
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{
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r = x - y;
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if (r > L_(0.5) * y)
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r = x - L_(2.0) * y;
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}
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return (negate ? - r : r);
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}
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}
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else
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{
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if (ISNAN (x) || ISNAN (y))
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return x + y; /* NaN */
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else if (isinf (y))
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return x;
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else
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/* x infinite or y zero */
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return NAN;
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}
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}
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