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coreutils/gnulib-tests/infinity.h
Daniel Baumann c08a8f7410
Adding upstream version 9.7.
Signed-off-by: Daniel Baumann <daniel.baumann@progress-linux.org>
2025-06-21 07:57:52 +02:00

66 lines
1.9 KiB
C

/* Macros for infinity.
Copyright (C) 2011-2025 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/>. */
/* Infinityf () returns a 'float' +Infinity. */
/* The Microsoft MSVC 9 compiler chokes on the expression 1.0f / 0.0f.
The IBM XL C compiler on z/OS complains.
PGI 16.10 complains. */
#if defined _MSC_VER || (defined __MVS__ && defined __IBMC__) || defined __PGI
static float
Infinityf ()
{
static float zero = 0.0f;
return 1.0f / zero;
}
#else
# define Infinityf() (1.0f / 0.0f)
#endif
/* Infinityd () returns a 'double' +Infinity. */
/* The Microsoft MSVC 9 compiler chokes on the expression 1.0 / 0.0.
The IBM XL C compiler on z/OS complains.
PGI 16.10 complains. */
#if defined _MSC_VER || (defined __MVS__ && defined __IBMC__) || defined __PGI
static double
Infinityd ()
{
static double zero = 0.0;
return 1.0 / zero;
}
#else
# define Infinityd() (1.0 / 0.0)
#endif
/* Infinityl () returns a 'long double' +Infinity. */
/* The Microsoft MSVC 9 compiler chokes on the expression 1.0L / 0.0L.
The IBM XL C compiler on z/OS complains.
PGI 16.10 complains. */
#if defined _MSC_VER || (defined __MVS__ && defined __IBMC__) || defined __PGI
static long double
Infinityl ()
{
static long double zero = 0.0L;
return 1.0L / zero;
}
#else
# define Infinityl() (1.0L / 0.0L)
#endif