C and infinity arithmetic

Question

#include <math.h>

void main() {
    double a = 0;
    a += 5e308;
    a -= 3.5e308;

    if (a == INFINITY) {
        printf("1   %e\n", a);
    }
    else
        printf("0   %e\n", a);
}

Here, the first print line is what I expect but why is the second print statement executed? I think a should be positive infinity as infinity -3.5e308 should be positive infinity, right in line 4?

I have already tried to execute it but the output is:

0   -1.#IND00e+000

I would like to know where my logic goes wrong.

Answer 1

The incorrect assumption that you seem to have is that (double) INFINITY - 3.5e308 would result in positive infinity.

Since 3.5e308 exceeds the range of double, it is a floating-point infinity. Both GCC and Clang, with no added compiler flags, provide a warning that both of those large values are too large for the type double.

The result is then effectively:

double a = (double) 0 + (double) INFINITY - (double) INFINITY;

It doesn't matter that, numerically, 3.5e308 is smaller than 5e308. Both are a double-precision infinity.

The subtraction of infinity from infinity is defined to result in a NaN.