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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?

erip
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ctanakul
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  • Are you asking how to detect if a division will result in overflow? https://stackoverflow.com/questions/15655070/how-to-detect-double-precision-floating-point-overflow-and-underflow – Jive Dadson Jul 12 '18 at 01:06
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    @It'scominghome, Epsilon isn't quite right. The step to the next value after 1 isn't the same thing as the step to the next value after 0. For me, the difference between the two is quite pronounced (e-16 vs. e-308). – chris Jul 12 '18 at 01:06
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    What exactly are you doing to avoid NaN? Simply replacing a zero divisor with a small (positive) value is not necessarily a correct approach (it depends on the context). A more correct approach might be to deal with NaN rather than avoiding it. – JaMiT Jul 12 '18 at 04:47
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    @JaMiT Thanks for pointing this out. For my application, just avoiding NaN is clearly not enough. It moves result out of the proper path. – ctanakul Jul 12 '18 at 16:10

2 Answers2

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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.

(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.

paxdiablo
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The smallest possible float is 1.17549e-38. To expand upon It's coming home's comment, see the answer from here:

#include <limits>

//...

std::numeric_limits<float>::max();  // 3.40282e+38
std::numeric_limits<float>::min();  // 1.17549e-38
std::numeric_limits<float>::infinity();

The float in the above code can be replaced by any data type you want, ie:

std::numeric_limits<int>::min();
Nathan majicvr.com
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    The exact maximum and minimum for 'float' isn't set by the standard. Implementations can use any conforming floating point representation. – François Andrieux Jul 12 '18 at 03:26