sign of infinity and isinf()?

[Stefan Farfeleder]

On Sun, Jun 13, 2004 at 09:06:30AM -0700, Steve Kargl wrote:

> I don't have a copy of the C99 standard only Harbison and Steele,
> 5th Ed.  On page 440, they state: "The isinf macro returns a nonzero
> value if and only if its argument is infinite (with any sign)."

This is what C99 says:

# Description
#   The isinf macro determines whether its argument value is an infinity
#   (positive or negative). First, an argument represented in a format
#   wider than its semantic type is converted to its semantic type. Then
#   determination is based on the type of the argument.
# Returns
#   The isinf macro returns a nonzero value if and only if its argument
#   has an infinite value.

I agree that the text from H&S is a bit ambiguous.

> The program below does note return the sign.
> kargl[243] gcc -o a -O a.c
> kargl[244] ./a
> -inf
> 1

> #include <stdio.h>
> #include <math.h>
> 
> int main(void) {
> 	double x, y;
> 	x = -1.;
> 	y = 0.;
> 	x /= y;
> 	printf("%lf\n", x);
> 	printf("%d\n", isinf(x));
> }

Since 1 is a nonzero value, this is correct.

Cheers,
Stefan