2

I am compiling the following code with the -ffast-math option:

#include <limits>
#include <cmath>
#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.

einpoklum
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André Puel
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2 Answers2

10

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
sam hocevar
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1

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 ;-)

Rupert
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