Problems casting NAN floats to int

Question

Ignoring why I would want to do this, the 754 IEEE fp standard doesn\'t define the behavior for the following:

float h = NAN;
printf(\"%x %d\\n\", (int)h, (         


        

Answer 1

First, a NAN is everything not considered a float number according to the IEEE standard. So it can be several things. In the compiler I work with there is NAN and -NAN, so it's not about only one value.

Second, every compiler has its isnan set of functions to test for this case, so the programmer doesn't have to deal with the bits himself. To summarize, I don't think peeking at the value makes any difference. You might peek the value to see its IEEE construction, like sign, mantissa and exponent, but, again, each compiler gives its own functions (or better say, library) to deal with it.

I do have more to say about your testing, however.

float h = NAN;
printf("%x %d\n", (int)h, (int)h);

The casting you did trucates the float for converting it to an int. If you want to get the integer represented by the float, do the following

printf("%x %d\n", *(int *)&h, *(int *)&h);

That is, you take the address of the float, then refer to it as a pointer to int, and eventually take the int value. This way the bit representation is preserved.