CURRENT: CLANG 3.3 and -stad=c++11 and -stdlib=libc++: isnan()/isninf() oddity

Tijl Coosemans tijl at
Thu Jul 11 09:37:36 UTC 2013

On 2013-07-11 06:21, Bruce Evans wrote:
> On Wed, 10 Jul 2013, Garrett Wollman wrote:
>> <<On Wed, 10 Jul 2013 22:12:59 +0200, Tijl Coosemans <tijl at> said:
>>> I think isnan(double) and isinf(double) in math.h should only be
>>> visible if (_BSD_VISIBLE || _XSI_VISIBLE) && __ISO_C_VISIBLE < 1999.
>>> For C99 and higher there should only be the isnan/isinf macros.
>> I believe you are correct.  POSIX.1-2008 (which is aligned with C99)
>> consistently calls isnan() a "macro", and gives a pseudo-prototype of
>>     int isnan(real-floating x);
> Almost any macro may be implemented as a function, if no conforming
> program can tell the difference.  It is impossible for technical reasons
> to implement isnan() as a macro (except on weird implementations where
> all real-floating types are physically the same).  In the FreeBSD
> implementation, isnan() is a macro, but it is also a function, and
> the macro expands to the function in double precision:
> % #define    isnan(x)                    \
> %     ((sizeof (x) == sizeof (float)) ? __isnanf(x)    \
> %     : (sizeof (x) == sizeof (double)) ? isnan(x)    \
> %     : __isnanl(x))

The C99 standard says isnan is a macro. I would say that only means
defined(isnan) is true. Whether that macro then expands to function
calls or not is not important.

> I don't see how any conforming program can access the isnan() function
> directly.  It is just as protected as __isnan() would be.  (isnan)()
> gives the function (the function prototype uses this), but conforming
> programs can't do that since the function might not exist.

I don't think the standard allows a function to be declared with the same
name as a standard macro (it does allow the reverse: define a macro with
the same name as a standard function). I believe the following code is
C99 conforming but it currently does not compile with our math.h:

#include <math.h>
int (isnan)(int a, int b, int c) {
        return (a + b + c);

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