Why does GCC implement isnan() more efficiently for C++ than C <math.h>?

Question

Here\'s my code:

int f(double x)
{
  return isnan(x);
}

If I #include I get this assembly:

xorl          


        

Answer 1

Looking at <cmath> for libstdc++ shipped with gcc 4.9 you get this:

  constexpr bool
  isnan(double __x)
  { return __builtin_isnan(__x); }

A constexpr function could be aggressively inlined and, of course, the function just delegates the work over to __builtin_isnan.

The <math.h> header doesn't use __builtin_isnan, rather it uses an __isnan implementation which is kind of long to paste here but it's lines 430 of math.h on my machine™. Since the C99 standard requires using a macro for isnan et al (section 7.12 of the C99 standard) the 'function' is defined as follows:

#define isnan(x) (sizeof (x) == sizeof (float) ? __isnanf (x)   \
  : sizeof (x) == sizeof (double) ? __isnan (x) \
  : __isnanl (x))

However, I see no reason why it can't use __builtin_isnan instead of __isnan so I suspect it's an oversight. As Marc Glisse points out in the comments, there is a relevant bug report for a similar issue using isinf instead of isnan.