On: NaN does not equal NaN

[voidr]

1/0 is not equal to 2/0, they are both errors but not the same.

[cdavid → voidr]

actually, they are both equal to infinity:

    #include <stdio.h>

    int main()
    {
	float a = 1. / 0;
	float b = 2. / 0;

 	printf("%f == %f ? %d\n", a, b, a == b);
        return 0;
    }

will print:

    inf == inf ? 1

Moreover:

    #include <stdio.h>

    int main()
    {
	float a = 0. / 0;
	float b = a;

 	printf("%f == %f ? %d\n", a, b, a == b);
        return 0;
    }

    nan == nan ? 0

So this is certainly surprising without knowing quite a bit already about floats, because it means = is not only not transitive but even not reflexive.

[blinks → cdavid]

Actually, they're _not_ both equal to infinity, in the mathematical sense (which I believe you're discussing) -- X / 0 is undefined.

What's equal to infinity is the limit of 1 / x as x approaches 0, and that's only from the positive side. From the negative side, it's -infinity.

[cdavid → blinks]

Nobody is discussing about usual mathematical definitions: this is about the IEEE754 standard. In that context, x / 0 is defined (to positive infinity). I was pointing the error of the OP about the explanation about Nan != Nan.

[ajross → voidr]

Actually no. Just in case you were confused by cdavid's sample code: the concise explanation is that division by zero in IEEE floating point produces an infinity (there are two: positive and negative), not a NaN. Neither is necessarily an "error" in the sense of a runtime exception (the generation of which is programmable).