I am compiling the following code with the -ffast-math option:
#include <limits>
#include <开发者_如何学Gocmath>
#include <iostream>
int main() {
std::cout << std::isnan(std::numeric_limits<double>::quiet_NaN() ) << std::endl;
}
I am getting 0 as output. How can my code tell whether a floating point number is NaN when it is compiled with -ffast-math?
Note: On linux, std::isnan works even with -ffast-math.
Since -ffast-math instructs GCC not to handle NaNs, it is expected that isnan() has an undefined behaviour. Returning 0 is therefore valid.
You can use the following fast replacement for isnan():
#if defined __FAST_MATH__
# undef isnan
#endif
#if !defined isnan
# define isnan isnan
# include <stdint.h>
static inline int isnan(float f)
{
union { float f; uint32_t x; } u = { f };
return (u.x << 1) > 0xff000000u;
}
#endif
On linux, the gcc flag -ffast-math breaks isnan(), isinf() and isfinite() - there may be other related functions that are also broken that I have not tested.
The trick of wrapping the function/macro in parentheses also did not work (ie. (isnan)(x))
Removing -ffast-math works ;-)
加载中,请稍侯......
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