What's the least possible denominator/divisor value?

Question

I'm writing a code to prevent the zero denominator/divisor to avoid NaN value as a result of the division.

I wonder what could be the least possible denominator value in double in C++ and how to find or what's the reason for that?

Answer 1

Well, the smallest positive^(a) normalised double in C++ can be obtained with std::numeric_limits<double>::min() (from the <limits> header).

However, while you may be able to use that to prevent NaN values^(b), it probably won't help with overflows. For example:

std::numeric_limits<double>::max() / 1.0 => ok
std::numeric_limits<double>::max() / 0.5 => overflow

Preventing that will depend on both the denominator and the numerator.

As for why that is the case, it's because C++ uses IEEE-754 double-precision format for its double types - it's a limitation of that format.

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^(a) I've chosen positive values here, the smallest value could be interpreted as the most negative, in which case it would be read as -std::numeric_limits<double>::max(). But, given your intent is to avoid NaN, I suspect my assumption is correct.

^(b) I'm not entirely sure how you intend to do this, which is why I also discuss overflows - you may want to make it clearer in your question.