How to use nan and inf in C?

There is no compiler independent way of doing this, as neither the C (nor the C++) standards say that the floating point math types must support NAN or INF.

Edit: I just checked the wording of the C++ standard, and it says that these functions (members of the templated class numeric_limits):

quiet_NaN() 
signalling_NaN()

wiill return NAN representations "if available". It doesn't expand on what "if available" means, but presumably something like "if the implementation's FP rep supports them". Similarly, there is a function:

infinity() 

which returns a positive INF rep "if available".

These are both defined in the <limits> header - I would guess that the C standard has something similar (probably also "if available") but I don't have a copy of the current C99 standard.

You can test if your implementation has it:

#include <math.h>
#ifdef NAN
/* NAN is supported */
#endif
#ifdef INFINITY
/* INFINITY is supported */
#endif

The existence of INFINITY is guaranteed by C99 (or the latest draft at least), and "expands to a constant expression of type float representing positive or unsigned infinity, if available; else to a positive constant of type float that overflows at translation time."

NAN may or may not be defined, and "is defined if and only if the implementation supports quiet NaNs for the float type. It expands to a constant expression of type float representing a quiet NaN."

Note that if you're comparing floating point values, and do:

a = NAN;

even then,

a == NAN;

is false. One way to check for NaN would be:

#include <math.h>
if (isnan(a)) { ... }

You can also do: a != a to test if a is NaN.

There is also isfinite(), isinf(), isnormal(), and signbit() macros in math.h in C99.

C99 also has nan functions:

#include <math.h>
double nan(const char *tagp);
float nanf(const char *tagp);
long double nanl(const char *tagp);

(Reference: n1256).

Docs INFINITY Docs NAN